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maximum Power Transfer Theorem

Although applicable to all branches of electrical engineering, this theorem is particularly useful
for analysing communication networks. The overall efficiency of a network supplying maximum
power to any branch is 50 per cent. For this reason, the application of this theorem to power transmission
and distribution networks is limited because, in their case, the goal is high efficiency and not
maximum power transfer.
However, in the case of electronic and communication networks, very often, the goal is either to
receive or transmit maximum power (through at reduced efficiency) specially when power involved
is only a few milliwatts or microwatts. Frequently, the problem of maximum power transfer is of
crucial significance in the operation of transmission lines and antennas.
As applied to d.c. networks, this theorem may be stated as follows :
A resistive load will abstract maximum power from a network when the load resistance is equal
to the resistance of the network as viewed from the output terminals, with all energy sources removed
leavin g behind their internal resistances. , , A
W'
 a load resistanceof RLis connectedacross I I R
the terminals A and B of a network which consists of a generator
of e.m.f. E and internal resistance Rg and a series resistance R
which, in fact, represents the lumped resistance of the connecting
wires. Let Rj =Rg + R =internalresistanceof the networkas
viewed from A and B.
According to this theorem, RL will abstract maximum power
from the network when RL=Rj'
Let us consider an a.c. source of internal impedance (R) + j X) supplying power to a load
impedance (RL + jXJ. It can be proved that maximum power transfer will take place when the
modulesof the loadimpedanceis equalto the modulusof the sourceimpedancei.e. I ZLI = I Z) I
Where there is a completely free choice about the load, the maximum power transfer is obtained
when load impedance is the complex conjugate of the source impedance. For example, if source
impedance is (R) +jX), then maximum transfer power occurs, when load impedance is (R) - jX). It
can be shown that under this condition, the load power is = E-/4R).The variation of 11with RL is shown in . The maximum value of 11is unity when
RL = 00and has a value of 0.5 when RL =Rj' It means that under maximum power transfer conditions,
the power transfer efficiency is only 50%. As mentioned above, maximum power tr-ansfer condition
is important in communication applications but in most power systems applications, a 50% efficiency
is undesirable because of the wasted energy. Often, a compromise has to be made between
the load power and the power transfer efficiency. For example, if we make RL = 2 Rj' then

How To Nortonize a Given Circuit?

This procedure is based on the first statement of the theorem given above.
1. Remove the resistance (if any) across the two given tenninals and put a short-circuit across
them.
2. Compute the short-circuit current Isc-
3. Remove all voltage sources but retain their internal resistances, if any. Similarly, remove
all current sources and replace ..'1~n> Jpen-circuits i.e. by infinite resistance.
4. Next, find the resistance RI (also called RN)of the network as looked into from the given
terminals. It is exactly the same as R'h
5. The current source (Isc)joined in parallel across Rj between the two terminals gives Norton's
equivalent circuit. .

(b) Norton Equivalent Circuit
A short placed across tenninals A and B will short out R2 as well. Hence, Isc =EIR1. The
Norton equivalent resistance is exactly the same as Thevenin resistance except that it is connected in
parallel with the current source as shown in
 Using Norton's theorem, find the constant-current equivalent of the circuit

Solution. When tenninals A and B are short-circuited as shown, total resistance
of the circuit, as seen by the battery, consists of a 10 Q resistance in series with a parallel
combination of 10 Q and 15 Q resistances.This current is divided into two parts at point C of .
Current through A B is Isc =6.25 x 10/25=2.5 A
Since the battery has no internal resistance, the input resistance of the network when viewed
fromA and B consists of a 15.0 resistance in series with the parallel combination of 10.0 and 10.0.
Hence, RI =15 + (10/2) =20 .a
Hence, the equivalent constant-current source is as shown in .
. Apply Norton's theorem to calculate currentflowing through 5 - Q resistor of

Solution. (i) Remove 5 - .a resistor and put a short across terminals A and B as shown in
. As seen, 10 -.a resistor also becomes short-circuited.
(ii) Let us now find Isc' The battery sees a parallel combination of 4.0 and 8.0 in series with
a 4 .a resistance. Total resistance seen by the battery =4 + 4 II8 =20/3.0. Hence, 1= 20 + 20/3 =
3 A. This current divides at point C of Fig. 2.205 (b). Current going along path CAB gives Isc- Its
value =3 x 4/12 = 1 A
The negative sign shows that the actual direction of flow of 12is opposite to that shown in Fig.

As Applicable to Voltage Sources
This Theorem is a combination of Thevenin's and Norton's theorems. It is used for finding the
common voltage across any network which contains a number of parallel voltage sources as shown
in . Then common voltage VABwhich appears across the output terminals A and B is
affected by the voltage sources EI' E2 and E3' The value of the voltage is given by
E)IR) + EzlRz +E31R3 _ I) +/z +/3 _ ~
VAB = --
11R) + 11Rz+ 11R3 G) + Gz + G3 1:G
This voltage represents the Thevenin's voltage Vth. The resistance Rthcan be found, as usual, by
replacing each voltage source by a short circuit. If there is a load resistance RL across the terminals
A and B, then load current IL is given by
IL = VtJ(Rth + RJ
 a branch does not contain any voltage source, the same procedure
is used except that the value of the voltage for that branch is equated to zero as illustrated in . Hence, ISh =IN =12 = - 5/4 A i.e. current flows from point B to A.
After the terminals A and B are open-circuited and the three batteries are replaced by shortcircuits
(since their internal resistances are zero), the internal resistance of the circuit, as viewed
from these terminals' is
Rj = RN=2+4114=4Q
The Norton's equivalent circuit consists of a constant current source of 5/4 A in parallel with a
resistance of 4 Q as shown in . When 6 Q resistance is connected across the equivalent
circuit, current through it can be found by the current-divider rule (Art).
Current through 6 Q resistor = % x 1~ =0.5 from B to A
General instructions For Finding Norton Equivalent Circuit As Applicable to Current Sources
This theorem is applicable to a mixture of parallel voltage and current sources that are reduced
to a single flam equivalent source which is either a constant current or a constant voltage source.
This theorem can be stated as follows :
Any number of constant current sources which are directly connected in parallel can be converted
into a single current source whose current is the algebraic sum of the individual source currents and
whose total internal resistances equals the combined individual source resistances in parallel

Procedure for fmding Norton equivalent circuit of a given network has already been given in
Art. That procedure applies to circuits which contain resistors and independent voltage or current
sources. Similar procedures for circuits which contain both dependent and independent sources or
only dependent sources are given below:
(a) Circuits Containing Both Dependent and Independent Sources
(i) Find the open-circuit voltage Vo<with all the sources activated or 'alive'.
(ii) Find short-circuit current ishby short-circuiting the terminals a and b but with all sources
activated.
(iii) RN=Vo!ish
(b) Circuits Containing Dependent Sources Only
(i) ish= o.
(ii) Connect I A source to the terminals a and b calculate v00.
(iii) RN = v0011.

Delta/Star* Transformation

In solving networks (having considerable number of branches) by the application of Kirchhoff s
Laws, one sometimes experiences great difficulty due to a large number of simultaneous equations
that have to be solved. However, such complicated network can be simplified by successively
replacing delta meshes by equivalent star system and vice versa.
Suppose we are given three resistance RI2' R23and R31connected in delta fashion between
terminals 1,2 and 3 as in Fig. 2.185 (a). So far as the respective terminals are concerned, these three
given resistances can be replaced by the three resistances RJ' R2and R3connected in star as shown in
Fig. 2.185 (b).
These two arrangements will be electrically equivalent if the resistance as measured between
any pair of terminals is the same in both the arrangements. Let us find this condition.
First, take delta connection: Between terminals 1 and 2, there are two parallel paths; one having
a resistance of RI2 and the other having a resistance of (R12+R31).
. R x (R + R )
.. ResIstance between terminals 1 and 2 is = n1l22+ (~33+ ;131
Now, take star connection: The resistance between the same terminals 1 and 2 is (R1+ R2).
As terminal resistances have to be the same
.. R1+ R2- _ R12~ ~~~!!"_R31)
RI2 + R23 + R31
Similarly, for terminals 2 and 3 and terminals 3 and I, we get
~3 x (R31+ R12)
R2 + R3 = -.-.
RI2 + R23 + R31
R +R - -R-31 x (R12 + R-2-3)
3 1 - RI2 + R23 + R31
Now, subtracting (ii) from (i) and adding the result to (;;i), we get
R R R R R R
RI=--1L JI_ ; R2= 1Ll.L - and R3= 31 23
RI2+ R23+ R31 RI2+ ~3 + R31 RI2+ R23+ R31How to Remember?
It is seen from above that each numerator is the product of the two sides of the delta which meet
at the point in star. Hence, it should be remembered that: resistance of each arm of the star is given
by the product of the resistances of the two delta sides that meet at its end divided by the sum of the
three delta resistances.
Compensation Theorem

~ortoo's Th1:9FeJr..
This theorem is an alternative to the Thevenin's theorem. In fact, it is the dual of Thevenin's
theorem. Whereas Thevenin's theorem reduces a two-terminal active network of linear resistances
and generators to an equivalent constant-voltage source and series resistance, Norton's theorem
replaces the network by an equivalent constant-current source and a parallel resistance.
This theorem is particularly usful for the following two purposes :
(a) For analysing those networks where the values of the branch elements are varied and for
studying the effect of tolerance on such values.
(b) For calculating the sensitivity of bridge network.
As applied to d.c. circuits, it may be stated in the following to ways :
(i) In its simplest form. this theorem asserts that any resistance R in a branch of a network in
which a current I isflowing can be replaced,for the purposes of calculations, by a voltage
equal to - IR.
OR
(ii) If the resistance of any branch of network is changedfrom R to (R + M) where the current This theorem may be stated as follows :
(i) Any two-terminal active network containinll voltalle sources and resistance when viewed
from its oUtpUIle.,IfUtUlI,:iI,)equIvalent to a consrant-.c-urrentsource and a Darallei resistance. The --~...
constant current is equaf to the current which would flow m a short-circuit placed across the
terminals and parallel resistance is the resistance of the network when viewed from these open-
, circuited terminals after all voltage and current sources have been removed and replaced by their
internal resistances.flowing originally is I, the change of current at any other place in the network may be
calculated by assuming that an e.m.f. - I. M has been injectedinto the modifiedbranch
while all other sources have their e.m.f.s. suppressed and are represented by their internal
resistances only As seen from Fig. 2.202 (a), a short is placed across the terminals A and B of the network with
all its energy sources present. The short-circuit current Isc gives the value of constant-current
source.
For finding Rj' all sources have been removed as shown in Fig. 2.202 (b). The resistance of the
network when looked into from terminals A and B gives Rj'
The Norton's equivalent circuit is shown in Fig. 2.202 (c). It consists of an ideal constantcurrent
source of infinite internal resistance (Art. 2.16) having a resistance of Rj connected in parallel
with it. Solved Examples 2.96, 2.97 and 2.98 etc. illustrate this procedure.
(ii) Another useful generalized form of this theorem is as follows:
The voltage between any two points in a network is equal to ISc. Rj where Iscis the short-circuit
current between the two points and Rj is the resistance of the network as vil;!wedfrom these points
with all voltage sources being replaced by their internal resistances (if any) and current sources
replaced by open-circuits.
Suppose, it is required to find the voltage across resistance R3and hence current through if [Fig.
2.202 (d)]. If short-circuit is placed between A and B, then current in it due to battery of e.m.f. EI is
E/RI and due to the o~er battery is E.jR2'

General Instructions for Finding Thevenin Equivalent Circuit

So far, we have considered circuits which consisted of resistors and independent current or
voltage sources only. However, we often come across circuits which contain both independent and
dependent sources or circuits which contain only dependent sources. Procedure for finding the
value of Vlhand Rlhin such cases is detalied below :
(a) When Circuit Contains Both Dependent and Independent Sources
(i) The open-circuit voltage Vocis determined as usual with the sources activated or 'alive'.
(ii) A short-circuit is applied across the terminals a and b and the value of short-circuit
current ilhis found as usual.
(iii) Thevenin resistance Rlh = vo/ish' It is the same procedureas adoptedfor Norton's
theorem. Solved examples 2.81 to 2.85 illustrate this procedure.
(b) When Circuit Contains Dependent Sources Only
(i) In this case, voe =0
(ii) We connect 1A source to the terminals a and b and calculate the value of vab.
(iii) Rlh=VaJ I 11
The above procedure is illustrated by solved Examples.
Example 2.81. Find Thevenin equivalent circuit for the network shown in Fig. 2.162 (a) which
contains a current controlled voltage source (CCVS).
Solution. For finding Voeavailable across open-circuit terminals a and b, we will apply KVL to
the closed loop.
.. 12 - 4 i + 2 i - 4 i = 0 :. i =2 A
Hence, Voe= drop across 4 11resistor =4 x 2 =8 V. It is so because there is no current through
the 2 11 resistor.
For finging Rlh, we will put a short-circuit across terminals a and b and calculate Ish' as shown in
Fig. 2.162 (b). Using the two mesh currents, we have
12- 4 i( + 2 i - 4(i\ - i2) =0 and - 8 i2 - 4 (i2- i\) =O. Substituting i=(i\ - i2)and Simplifying
the above equations, we have
12 - 4 i} + 2 (i} - i2)- 4 (i} - i2)=0 or 3 i} - i2=6 ...(i)
Similarly, from the second equation, we get i\ =3 i2. Hence, i2=3/4 and Rlh= Vo/lsh= 8/(3/4)
= 32/311. The Thevenin equivalent circuit is as shown in Fig. 2.162 (c).
Example 2.82. Find the Thevenin equivalent circuit which respect to tenninals a and b of the
network shown in Fig. 2.163 (a).
Solution. It will be seen that with terminals a and b open, current through the 8 11resistor is
vaJ4 and potential of point A is the same of that of point a (because there is no current through 4 11
resistor). Applying KVL to the closed loop of Fig. 2.163 (a), we get It is also the value of the open-circuit voltage voc'
For finding short-circuit current ish'we short-circuit the terminals a and b as shown in Fig. 2.163
(b). Since with a and b short-circuited, vab=0, the dependent current source also becomes zero.
Hence, it is replaced by an open-circuit as shown. Going around the closed loop, we get
12 - ish (8 + 4) = 0 or ish=6/12 =0.5 A
Hence, the Thevenin equivalent is as shown in Fig. 2.163 (c).
Example 2.83. Find the Thevenin equivalent circuit for the network shown in Fig. 2.164 (a)
which contains only a dependent source.
Solution. Since circuit contains no indepenedent source, i=0 when terminals a and b are open.
Hence, Voc=O. Moreover, ishis zero since Voc= O.
Consequently, RShcannot be found from the relation R'h=v~ish' Hence, as per Art. 2.20, we
will connect a I A current source to terminals a and b as shown in Fig. 2.164 (b). Then by finding the
value of v00' we will be able to calculate R'h =vmJl.
It should be noted that potential of point A is the same as that of point a t.e. voltages across 120.
resistor is vab. Applying KCL to point A, we get
2i-voo voo
-12+1 = 0 or 4i-3vab=-12
Since i = vmJl2, we have 4 (vmJI2) - 3 voo=- 12 or voo=4.5 V :. R'h =vmJI =4.5/1 =4.50..
The Thevenin equivalent circuit is shown in Fig. 2.164 (c).
Example 2.84. Determine the Thevenins equivalent circuit as viewed from the open-circuit
terminals a and b of the network shown in Fig. 2.165 (a). All resistances are in ohms.
Solution. It would be seec from Fig. 2.165(a) that potential of node A equals the open-circuit
terminal voltage voc' Also, i = (vs - voc)/(80 + 20) = (6 - voc)/Ioo.
Applying KCL to node, A we get other branch, then the same e.m.f. E acting in the second bamch would produce the same current I
in thefirst branch.
In other words, it simply means that E and I are mutually transferrable. The ratio Ell is known
as the transfer resistance (or impedance in a.c. systems). Another way of stating the above is that the
receiving point and the sending point in a network are interchangebale. It also means that interchange
of an ideal voltage sources and an ideal ammeter in any network will not change the ammeter
reading. Same is the case with the interchange of an ideal current source and an ideal voltmeter.
The new circuit having 2 -V battery connected in the branch BD is shown in Fig. 2.169. According
to the Principle of Superposition,the new current in the 1-volt battery circuit is due to the superposition
of two currents; one due to 1 - volts battery and the other due to the 2 - volt battery when each acts
independently.
The current in the external circuit due to 1 - volt battery when 2 - battery is not there, as found
above, is 0.0723 A.
Now, according to Reciprocity theorem; if 1 - volt battery were tansferred to the branch BD
(where it produced a current of 0.0049 A), then it would produce a current of 0.0049 A in the branch
CEA (where it was before). Hence, a battery of 2 -V would produce a current of (- 2 x 0.0049) =-
0.0098 A (by proportion). The negative sign is used because the 2 - volt battery has been so connected
as to oppose the current in branch BD,

How to Thevenize a Given Circuit?

It is clear from above that any network of resistors and voltage sources (and current sources as
well) when viewed from any points A and B in the network, can be replaced by a single voltage
source and a single resistance* in series with the voltage source.
After this replacement of the network by a single voltage source with a series resistance has
been accomplished, i~is easy to find current in any load resistance joined across terminals A and B.
This theorem is valid even for those linear networks which have a nonlinear load.
Hence, Thevenin' s theorem, as applied to d.c. circuits, may be stated as under :
The current flowing through a load resistance RL connected across any two tenninals A and B
of a linear, acti~e bilateral network is given by VocII(Rj + RJ where Vocis the open-circuit voltage
(i.e. voltage across the two tenninals when RL is removed) and Rj is the internal resistance of tM
network as viewed back into the open-circuited network from tenninals A and B with all voltage
sources replaced by their internal resistance (if any) and current sources by infinite resistance.
2.19. How to Thevenize a Given Circuit?
I A
B
Fig. 2.128 "
1. Temporarily remove the resistance (called load resistance RJ whose current is required.
2. Find the open-circuit voltage Vocwhich appears across the two terminals from where
resistance has been removed. It is also called Thevenin voltage Vth.
3. Compute the resistance of the whose network as looked into from these two terminals after
all voltage sources have been removed leaving behind their internal resistances (if any) and
current sources have been replaced by open-circuit i.e. infinite resistance. It is also called
Thevenin resistance Rthor Tj'
4. Replace the entire network by a single Thevenin source, whose voltage is Vthor Vocand
whose internal resistance is Rthor Ri.
5. Connect RL back to its terminals from where it was previously removed.
6. Finally, calculate the current flowing through RLby using the equation
polarity as shown in Fig. 2.129 (a).
.. VAB = V,h =+ 10 + 5/3 =35/3V.
Incid~ntly, we could also fmd VABwhile going along the parallel route BFEA.
Drop across 8 Q resistor =8 x 5/12 =10/3 V. VAB equal the algebraic sum of voltages met on the
way from B to A. Hence, VAB=(- 10/3) + 15 =35/3 V.
As shown in Fig. 2.129 (b), the single voltage source has a voltage of 35/3 V.
For findign R'h' we will replace the two voltage sources by short-circuits. In that case, R'h =RAB
=4118=8/3Q. Now, we will find Rthi.e. equivalent resistnace of the network as looked back into the opencircuited
terminals A and B. For this purpose, we will replace
both the voltage and current sources. Since voltage source
has no internalresistance, it would be replacedby a short circuit
i.e. zeroresistance. However, current source would be removed
and replaced by an 'open' i.e. infinite resistance (Art. 1.18).
In that case, the circuit becomes as shown in Fig. 2.133 (c).
As seen from Fig. 2.133 (d), Fth= 6 II3 + 2 = 4 Q. Hence,
Thevenin's equivalent circuit consists of a voltage source of
12V and a series resistance of 4 Q as shown in Fig. 2.134 (a).
When 4 Q resistor is connected a The negative sign shows that point A is negative with respect to point B or which is the same
thing, point B is positive with respect to point A.
For finding RAS=Rth'we replace the batteries by short-circuits as shown in Fig. 2.128 (c).
:. RAS = Rth =8 + 2 II5=9.43 Q
Hence, the equivalent Thevenin' s source with respect to tenninals A and B is an shown in Fig.
2.136. When 10Q resistance is reconnected across A and B, current through it is 1= 6.24/9.43 + 10)
= 0.32A (I) When galvanometer is removed from Fig. 2.139 (a), we get the circuit of Fig.
2.139 (b).
(ii) Let us next find the open-circuit voltage VDC (also called Thevenin voltage V,h)between
points B and D. Remembering that ABC (as well as ADC) is a potential divider on which a voltage
drop of 10 V takes place, we get
Potential of B w.r.t. C = 10 x 10/110 = 10/11 =0.909 V
Potential of D w.r.t. C = 10 x 4/54 = 20/27 = 0.741 V
.. p.d. between B and D is Vocor V,h=0.909 - 0.741 =0.168 V
(iil) Now, remove the 10-V battery retaining its internal resistance which, in this case, happens
to be zero. Hence, it amounts to short-circuiting points A and C as shown in Fig. 2.139 (d).

Ideal Constant-Voltage Source

Ideal Constant-Voltage Source
It is that voltage source (or generator) whose output voltage remains absolutely constant whatever
the change in load current. Such a voltage source muat possess zero internal resistance so that
internal voltage drop in the soruce is zero. In that case, output voltage provided by the source would
remain constant irrespective of the amount of current drawn from it. In practice, none such ideal
constant-voltage source can be obtained. However, smaller the internal resistance r of a voltage
source, closer it comes to the ideal sources described above.
Suppose, a 6-V battery has an internal resistance of 0.005 Q [Fig. 2.94 (a)). When it supplies no
current i.e. it is on no-load, Vo=6 V i.e. output voltage provided by it at its output terminals A and B
is 6 V. If load current increases to 100 A, internal drop =100 x 0.005 =0.5 V. Hence, Vo=6 - 0.5
=5.5 V.
Obviously an output voltage of 5.5 - 6 V can be considered constant as compared to wide
variations in load current from 0 A ot 100 A.
2.16. Ideal Constant-Current Source
It is that voltage source whose internal resistance is infinity. In practice, it is approached by a
source which posses very high resistance as compared to that of the external load resistance. As
shown in Fig. 2.94 (b), let the 6-V battery or voltage source have an internal resistance of 1M Q and
let the load resistance vary from 20 K to 200 K. The current supplied by the source varies from
6.l/1.02 =5.9 J.1A to 6/1.2 =5 J.1A. As seen, even when load resistance increases 10 times, current
decreases by 0.9 J.1A.Hence, the source can be considered, for all practical purposes, to be a constant-
current source.According to this theorem, if there are a number of e.m.fs. acting simultaneously in any linear
bilateral network, then each e.m.f. acts independently of the others i.e. as if the other e.m.fs. did not
exist. The value of current in any conductor is the algebraic sum of he currents due to each e.m.f.
Similarly, voltage across any conductor is the algebraic sum of the voltages which each e.m.f would
have produced while acting singly. In other words, current in or voltage across, any conductor of the
network is obtained by superimpsing the currents and voltages due to each e.m.f. in the network. It
is important to keep in mind that this theorem is applicable only to linear networks where current is
linearly related to voltage as per Ohm's law.
Hence, this theorem may be stated as follows :
In a network of linear resistances containing more than one generator (or source of e.m.f.), the
current whichflows at any point is the sum of all the currents which woludflow at that point if each
generator where considered separately and all the other generators replacedfor the time being by
I "4- 1-.'4- resistancesequalto their internalresistacnces
A seen, there are three independent sources are one dependent source. We will find
the value of v produced by each of the three independent sources when acting alone and add the three
values to find v. It should be noted that unlike independent source, a dependent source connot be set
to zero i.e. it cannot be 'killed' or deactivated.
Let us find the value of vI due to 30 V source only. For this purpose we will replace current
source by an open circuit and the 20 V source by a short circuit as shown in Fig. 2.99 (b). Applying
KCL to node 1, we get Solution. The given circuit has been redrawn in Fig. 2.109 (b) with 15 - V battery acting alone
while the other two sources have been killed. The 12 - V battery has been replaced by a short-circuit
and the current source has been replaced by an open-circuit (O.C) (Art. 2.19). Since the output
terminals are open, no current flows through the 4 Q resistor and hence, there is no voltage drop
across it. Obviously VI equals the votIage drop over 10Q resistor which can ,be found by using the
voltage-divider rule.
VI = 15 x 10/(40 + 50) =3 V
Fig. 2.110 (a) shows the circuit when current source acts alone, while two batteries have been
killed. Again, there is no current through 4 Q resistor. The two resistors of values 10Q and 40 Q are
in parallelacrossthe currentsource. Theircombinedresistancesis 10II40 =8 Q

Nodal Analysis with Current Sources

Consider the network of Fig. 2.68 (a) which has two current sources and three nodes out of
which I and 2 are independent ones whereas No.3 is the reference node.
The given circuit has been redrawn for ease of understanding and is shown in Fig. 2.68 (b). The
current directions have been taken on the assumption that
I. both V) and V2are positive with respect to the reference node. That is why their respective
curents flow from nodes I and 2 to node 3.
2. V) is positive with respect to V2because current has been shown flowing from node 1 to
node 2.
A positive result will confirm out assumption whereas a negative one willr . dicate that actual
directionis oppositeto that assumed. 1. product of potential VI and (l/RI + 1- R3) i.e. sum of the reciprocals of the branch resistances
connected to this node.
2. minus the ratio of adjoining potential V2and the interconnecting resistance R3'
3. all the above equated to the current supplied by the current source connected to this node.
This current is taken positive if flowing into the node and negative if flowing out of it (as per
sign convention of Art. 2.3). Same remarks apply to Eq. (ii) where 12has been taken negative
because it flows away from node 2.
In tenns of branch conductances, the above two equations can be put as taken as the reference node or common ground for all other nodes. We will apply KCL to the three
nodes and taken currents coming towards the nodes as positive and those going away from them as
negative. For example, current going away from node No. 1is (VI - V2)/l and hence would be taken
as negative. Since 4 A current is coming towards node No. 1. it would be taken as poisitive but 5 A
current would be taken as negative.
Source Conversion
A given voltage source with a series resistance can be converted into (or replaced by) and
equivalent current source with a parallel resistance. Conversely, a current source with a parallel
resistance can be converted into a vaoltage source with a series resistance. Suppose, we want to
convert the voltage source of Fig. 2.75 (a) into an equivalent current source. First, we will find the
value of current supplied by the source when a 'short' is put across in terrnials A and B as shown in
Fig. 2.75 (b). This current is I = VIR. withit representsthe equivalentsource. It is shownin Fig.2.75(c). Similarly,a currentsourceof I
and a parallel resistance R can be converted into a voltage source of voltage V =IR and a resistance
R in series with it. It should be kept in mind that a voltage source-series resistance combination is
equivalent to (or replaceable by) a current source-parallel resistance combination if, and only if their
1. respective open-circuit voltages are equal, and
2. respective short-circuit currents are equal.
For example, in Fig. 2.75 (a), voltage across tenninals A and B when they are open (i.e. opencircuit
voltage Voc) is V itself because there is no drop across R. Short-circuit current across AB =I
= VIR.
Now, take the circuit of Fig. 2.75 (c). The open:circuit voltage across AB =drop across R =IR
= V. If a short is placed across AB, whole of I passes through it because R is completely shorted out.The two parallel resistances of 3 Q and 6 Q can be combined into a single resistance of 2 Q as
shown in Fig. 2.79. (a)
The two current sources cannot be combined together because of the 2 Q resistance present
between points A and C. To remove this hurdle, we convert the 2 A current source into the equivalent
4 V voltage source as shown in Fig. 2.79 (b). Now, this 4 V voltage source with a series
resistance of (2 + 2) =4 Q can again be converted into the equivalent current source as shown in Fig.
2.80 (a). Now, the two current sources can be combined into a single 4-A source as shown in Fig.
2.80 (b).

Nodal Analysis With Sources

The node-equation method is based directly on Kirchhoffs current law unlike loop-current
method which is based on Kirchhoff s voltage law. However, like loop current method, nodal method also has the advantage that a minimum numt
2 2 3 ber of equations need be written to determine
I I"A-/2 fBI-h the unknown quantities. Moreover, it is par-
+ ~ + ticularly suited for networks having many
15 parallel circuits with common ground con-
£2 T nected such as electronic circuits.
For the application of this method, every
juncion in the network where three or more
branches meet is regarded a node. One of
these is regarded as the reference node or
datum node or zero-potential node. Hence
the number of simultaneous equations to be solved becomes (n - 1) where n is the number of independent
nodes. These node equations often become simplified if all voltage sources are converted
into current sources First Case
Consider the circuit of Fig. 2.60 which has three nodes. One of these i.e. node 3 has been take
in as the reference node. VArepresents the potential of node 1 with reference to the datum node 3.
Similarly, VBis the potential difference between node 2 and node 3. Let the current directions which
have been chosen arbitarily be as shown Though the above nodal equations (ii) and (iii) seem to be complicated, they employ a very
simple and systematic arrangement of terms which can be written simply by inspection. Eq. (ii) at
node I is represented by
1. The product of node potential VAand (lIRI + lIR2 + lIR4) i.e. the sum of the reciprocals of
the branch resistance connected to this node. .
2. Minus the ratio of adjacent potential VBand the interconnecting resistance R2.
3. Minus ratio of adjacent battery (or generator) voltage-EI and interconnecting resistacne RI'
4. All the above set to zero.
Second Case
Now, consider the case when a third battery of
e.m.f. E3 is connected between nodes 1 and 2 as
shown in Fig. 2.62.
It must be noted that as we travel from node 1to --£
node 2, we go from the -ve terminal of E3to its +ve T 1
terminal. Hence, according to the sign convention
given in Art. 2.3, E3must be taken aspositive. However,
if we travel from node 2 to node 1, we go from
the +ve to the -ve terminal of E3' Hence, when
viewedfrom node 2, E3 is taken negative.It is exactly the same expression as given under the First Case discussed above except for the
additional tenn involving E3' This additional tenn is taken as +EiR2 (and not as - EiR2) because
this third battery is so connected that when viewed from mode I, it represents a rise in voltage. Had
it been connected the other way around, the additional tenn would have been taken as -EiR2'
~s seen, the additional tenns is -EiR2 (and not + EiR2) because as viewed from this node, E3
represents afall in potential.
It is worth repeating that the additional tenn in the above Eq. (i) and (ii) can be either +EiR2 or
-EiR2 depending on whether it represents a rise or fall of potential when viewed from the node
under consideration.

Maxwell's Loop Curent Method

This method which is particularly well-suited to coupled circuit soluions employs a system of
loop or mesh currents instead of branch currents (as in Kirchhoffs laws). Here, the currents in
different meshes are assigned continuous paths so that they do not split at a juncion into branch
currents. This method eliminates a great deal of tedious work involved in the branch-current method
and is best suited when energy sources are voltage sources rather than current sources. Basically,
this method consists of writing loop voltage equations by Kirchhoff's voltage law in terms of unknown
loop currents. As will be seen later, the number of independent equations to ~
reducesfromb by Kirchhoffs lawsto b - (j - 1)forthe loopcurrentmethodwhereb is the number
of branches andj is the number of junctions in a given network connected in a network consisting of five
resistors. Let the l~op currents for the
three meshes be II' 12and 13' It is obvi-
E ous that current through R4 (when con-
2 sidered as a part of the first loop) is (/) -
12)and that through Rs is (12-13), However,
when R4 is considered part of the
second loop, current through it is (/2 -
H I). Similarly, whe Rs is considered part
of the third loop, current through it is (13
-12), Applying Kirchhoff's voltage law
to the three loops, Consider the network of Fig. 2.52, which contains
resistances and independent voltage sources and has three
meshes. Let the three mesh currents be designated as II' 12
and 13and all the three may be assumbed to flow in the clockwise
direcion for obtaining symmetry in mesh equationsmesh (i) which equals the sum of all resistance in mesh (i). Similarly, the second item in the first
row represents the mutual resistance between meshes (i) and (ii) i.e. the sum of the resistances
common to mesh (i) and (ii). Similarly, the third item in the first row represents the mutual-resistance
of the mesh (i) and mesh (ii).
The item E.. in general, represents the algebraic sum of the voltages of all the voltage sources
acting around mesh (i). Similar is the case with E2 and E3' The sign of the e.m.fs is the same as
discussed in Art. 2.3 fe. while going along the current, if we pass from negetive to the positive
terminal of a battery, then its e.m.f. is taken postive. If it is the other way around, then battery e.mf.
is taken negative.
In general, let
RII =self-resistance of mesh (i)
R22 =self-resistance of mesh (ii) i.e. sum of all resistances in mesh (ii)
R33 =Self-resistance of mesh (iii) i.e. sum of all resistances in mesh (iii)
RI2 =R21 =- [Sum of all the resistances common to meshes (i) and (ii)] *
R23 =R32 =- [Sum of all the resistances common to meshes (ii) and (iii)]*
The above equaitons can be written in a more compact form as [Rm][1m]= [Em]' It is known as
Ohm's law in matrix form.
In the end, it may be pointed out that the directions of mesh currents can be selected arbitrarily.
If we assume each mesh current to flow in the clockwise direction, then
(i) All self-resistances will always be postive and (ii) all mutual resistances will always be
negative. We will adapt this sign convention in the solved examples to follow.
The above main advatage of the generalized form of all mesh equations is that they can be easily
remebered because of their symmetry. Moreover, for any given network, these can be written by
inspection and then solved by the use of determinants. It eliminates the tedium of deriving simultaneous
equations.

Kirchhoff's Mesh Law or Volta2e Law (KVU

It srares as follows :
the algebraic sum of the products of currents and resistances in each of the conductors in any
closed path (or mesh) in a network plus the algebraic sum of the e.mjs. in that path is zero.
In other words, 1:IR + 1:e.mj. = 0 ...round a mesh
It should be noted that algebraic sum is the sum which takes into account the polarities of the
voltage drops.
The basis of this law is this : If we.start from a particular junction and go round the mesh till we
come back to the starting point, then we must be at the same potential with which we started. Hence,
it means that all the sources of e.m.f. met on the way must necessarily be equal to the voltage drops
in the resistances, every voltage being given its proper sign, plus or minus.
2.3. Determination of Voltage Sign
In applying Kirchhoff's laws to specific problems, particular attention should be paid to the
algebraic signs of voltage drops and e.m.fs., otherwise results will come out to be wrong. Following
sign conventions is suggested :
(a) Sign of Battery E.M.F.
A rise in voltage should be given a + ve sign and afall in voltage a -ve sign. Keeping this in
mind, it is clear that as we go from the -ve terminal of a battery to its +ve terminal (Fig. 2.3), there
is a rise in potential, hence this voltage should be given a + ve sign. If, on the other hand, we go from
+ve tenninal to -ve tenninal, then there is afall in potential, hence this voltage should be preceded by a -ve sign. It is important to note that the sign of the battery e.m.! is independent of the direction
of the current through that branch.
(b) Sign of IR Drop
Now, take the case of a resistor (Fig. 2.4). If we go through a resistor in the same direction as the
current, then ther is a fall in potential because current flows from a higher to a lower potential..
Hence, this voltage fall should be taken -ve. However, if we go in a direction opposite to that of the
current, then there is a rise in voltage. Hence, this voltage rise should be given a positive sign.
It is clear that the sign of voltage drop across a resistor depends on the direction of current
through that resistor but is independent of the polarity of any other source of e.m.! in the circuit
under consideration. Assumed Direction of Current Fig. 2.5
In applying Kirchhoff's laws to electrical networks, the question of assuming proper direction
of current usually arises. The direction of current flow may be assumed either clockwise or
anticlockwise. If the assumed direction of current is not the actual direction, then on solving the
quesiton, this current will be found to have a minus sign. If the answer is positive, then assumed
direction is the same as actual direction (Example 2.10). However, the important point is that once
a particular direction has been assumed. the same should be used throughout the solution of the
question.
Note. It should be noted that Kirchhoff's laws are applicable both to d.c. and a.c. voltages and
currents. However, in the case of alternating currents and voltages, any e.m.f. of self-inductance or
that existing across a capacitor should be also taken into account.Solving Simultaneous Equations
Electric circuit analysis with the help of Kirchhoff's laws usually involves solution of two or
three simultaneous equations. These equations can be solved by a systematic elimination of the
varia~les but the procedure is often lengthy and laborious and hence more liable to error. Determinants
and Cramer's rule provide a simple and straight method for solving network equations through
manipulation of their coefficients. Of course, if the number of simultaneous equaitons happens to be
very large, use of a digital computer can make the task easy.
2.6. Determinants
The symbol I ~ ~ I is called a determinant of the second order (or 2 x 2 ~eterminant) because
it contains two rows (ab and cd) and two columns (ac and bd). The numbers a, b, c and d are called
the elements or constituents of the determinant. Their number in the present case is 22=4.
The evaluation of such a determinant is accomplished by cross-multiplicaiton is illustrated

Electric Circuits and Network Theorems

There are certain theorems, which when applied to the solutions of electric networks, wither
simplify the network itself or render their analytical solution very easy. These theorems can also be
applied to an a.c. system, with the only difference that impedances replace the ohmic resistance of
d.c. system. Different electric circuits (according to their properties) are difined below:
I. Circuit. A circuit is a closed conducting path through which an electric current either
flows or is inteneded flow.
2. Parameters. The various elements of an electric circuit are called its parameters like resistance,
inductance and capacitance. These parameters may be lumped or distributed.
3. Liner Circuit. A linear circuit is one whose parameters are constant i.e. they do not change
with voltage or current.
4. Non-linear Circuit. It is that circuit whose parameters change with voltage or current.
5. Bilateral Circuit. A bilateral circuit is one whose properties or characteristics are the same
in either direction. The usual transmission line is bilateral, because it can be made to perform
its function equally well in either direction.
6. Unilateral Circuit. It is that circuit whose properties or characteristics change with the
direction of its operation. A diode rectifier is a unilateral circuit, because it cannot perform
rectification in both directions.
7. Electric Network. A combination of various electric elements, connected in any manner
whatsoever, is called an electric network.
8. Passive Network is one which contains no source of e.m.f. in it.
9. Active Network is one which contains one or more than one source of e.m.f.
10. Node is a junction in a circuit where two or more circuit elements are connected together.
II. Branch is that part of a network which lies between two juncions.
12 Loop. It is a close path in a circuit in which no element or node is epcountered more than
once.
13. Mesh. It is a loop that contains no other loop within it. For example, the circuit of Fig. 2.1
(a) has even branches, six nodes, three loops and two meshes whereas the circuit of Fig.
2.1 (b) has four branches, two nodes, six loops and three meshes.
It should be noted that, unless stated otherwise, an electric network would be assumed passive
in the following treatment.
We will now discuss the various network theorems which are of great help in solving complicated
networks. Incidentally, a network is said to be completely solved or analyzed when all volt-.
ages and all currents in its different elements are determined.
There are two general approaches to network analysis :
(i) Direct Method
Here, the network is left in its original form while determining its different voltages and currents.
Such methods are usually restricted to fairly simple circuits and include Kirchhoff slaws,
Loop analysis, Nodal analysis, superposition theorem, Compensation theorem and Reciprocity theorem
etc.
(ii) Network Reduction Method
Here, the original network is converted into a much simpler equivalent circuit for rapid calculation
of different quantities. This method can be applied to simple as well as complicated networks.
Examples of this method are: Delta/Star and StarlDelta conversions. Thevenin's theorem and
Norton's Theorem etc.
2.2. KirchhoWs Laws *
These laws are more comprehensive than Ohm's law and are used for solving electrical networks
which may not be redily solved by the latter. Kirchhoffs laws, two in number, are particularly
useful (a) in determining the equivalent resistance of a complicated network of conductors and
(b) for calculating the currents flowing in the various conductors. The two-laws are :
1. KirchhoWs Point Law or Current Law (KCL)
It states as follows:
in any electrical network, the algebraic sum of the currents meeting at a point (or junction) is
zero.
Put in another way, it simply means that the total current leaving a juncion is equal to the total
current entering that junction. It is obviously true because there is no accumulation of charge at the
junction of the network. '
Consider the case of a few conductors meeting at a point A as in Fig. 2.2 (a). Some conductors
have currents leading to point A, whereas some have currents leading away from point A. Assuming
the incoming currents to be positive and the outgoing currents negative, we have

Duality Between Series and Parallel Circuits

There is a certain peculiar pattern of relationship between series and parallel circuits. For example,
in a series circuit. current is the same whereas in a parallel circuit. voltage is the same. Also, in a
series circuit, individual voltages are added and in a parallel circuit, individual currents are added. It
is seen that while comparing series and parallel circuits, voltage takes the place of current and current
takes tha place of voltage. Such a pattern is known as "duality" and the two circuits are said to be
duals of each other.
As arranged in Table 1.4 the equations involving voltage, current and resistance in a series
circuit have a corresponding dual counterparts in terms of current, voltage and conductance for a
parallel circuit. It is the voltage of one point in a circuit with respect to that of another point (usually called the
reference or common point).
Consider the circuit of Fig. 1.83 (a) where the most negative end-point C has been taken as the
reference. With respect to point C, both points A and B are positive thoughA is more positive than B.
The voltage of point B with respect to that of C i.e. VBC=+ 30 V.
Similarly, VAC=+ (20 + 30) =+ 50 V.
This point is often called ground or earth because originally it meant a point in a
circuit which was actually connected to earth either for safety in power
systems or for efficient radio reception and transmission. Although, this
meaning still exists, yet it has become usual today for 'ground' to mean
any point in the circuit which is connected to a large metallic object such
as the metal chassis of a transmitter, the aluminium chassis of a receiver, a
wide strip of copper plating on a printed circuit board, frame or cabinet
which supports the whose equipment. Sometimes, reference point is also
called common point. The main advantage of using a ground system is to
simplify our circuitry by saving on the amount of wiring because ground is used as the return path
for may circuits. The three commonly-used symbols for grounnd are shownVE = V0 - voltage fall across 4 Q restors =20 - 8 =+ 12 V
Also VA = VE - fall across 6 Q resistor =12 (2 x 6) =0 V
(ii) In Fig. 1.87, point D has been taken as the ground. Starting from point D, as we go to E
there is a fall of 8 V. Hence, VE = - 8 V. Similarly, VA=- (8 + 12) =- 20 V.
As we go from A to B, there is a sudden increase of 34 V because we are going from negative
terminal of the battery to its positive tenninal.
:. VB = - 20 + 34 =+ 14 V
Ve = VB - voltage fall across 2 Q resistor = \4 - 4 = + 10 V.
It should be so because C is connected directly to the positive terminal of the 10 V battery.
Choice of a reference point does not in any way affect the operation of a circuit. Moreover, it
also does not change the voltage across any resistor or between any pair of points (as shown below)
because the ground current ig=O.

Shorts' in Parallel Circuits

Suppose a 'short' is placed across RJ (Fig. 1.43). ft becomes directly connected across the
ba~ery and draws almost infinite current because not only its own resistance but that of the connecting
wires AC and BD is negligible. Due to this excessive current, the wires may get hot enough to
burn out unless the circuit is protected by a fuse.Following points about the circuit of Fig. 1.43 (a) are worth noting.
\. not only is RJ short-circuited but both Rl and R2 are also shorted out i.e. short across one
branch means short across all branches.
2. there is no current is shorted resistors. If these were three bulbs, they will not glow.
3. the shorted components are not damaged, For example, if we had three bulbs in Fig. 1.43
(a), they would glow again when circuit is restored to normal conditions by removing the
short-circuited.
It may, however, be noted from Fig. 1.43 (b) that a short-circuit across RJ may short out R2but
not R, since it is protected by R4'
1.25. Division of Current in Parallel Circuits
In Fig. 1.44, two resistances are joined in parallel across a voltage V.
branch, as given in Ohm's law, is Hence, the division of current in the branches of a parallel
circuit is directly proportional to the conductance of the branches
or inversely proportional to their resistances. We may also
express the branch currents in terms of the total circuit current thus Equivalent Resistance ,
The equivalent resistance of a circuit (or network) between its any two points (or terminals) is
given by that single resistance which can replace the entire given circuit between these twopoints. It
should be noted that resistance is always between two given points of a circuit and can have different
values for different point-pairs as illustrated by Example 1.42. it can usually be found by using
series and parallel laws of resistances. Concept of equivalent resistance is essential for understanding
network theorems like Thevenin's theorem and Norton's theorem etc. discussed.

Varistor (Nonlinear Resistor)

It is a voltage-dependent metal-oxide material whose resistance decreases sharply with increasing
voltage. The relationship between the current flowing through a varistor and the voltage applied
across it is given by the relation : i = kenwhere i = instantaneouscurrent,e is the instantaneous
voltage and 11is a constant whose value depends on the metal oxides used. The value of 11for
silicon-carbide-based varistors lies between 2 and 6 whereas zinc-oxide-based varistors have a value
ranging from 25 to 50.
The zinc-oxide-based varistors are primarily used for protecting solid-state power supplies from
low and medium surge voltage in the supply line. Silicon-carbide varistors provide protection against
high-voltage surges caused by lightning and by the discharge of electromagnetic energy stored in the
magnetic fields of large coils.
1.20. Short and Open Circuits
When two points of circuit are connected togetherby a thick metallic wire (Fig. 1.38),they are said
to be short-circuited. Since 'short' has practically zero resistance,it gives rise to two important facts :
(i) no voltage can exist across it because V =IR =I x 0 =0
(ii) currentthroughit (calledshort-circuitcurrent)is very large(theoretically,infinity)
Two points are said to be open-circuitedwhen there is no direct connectionbetweenthem
(Fig. 1.39), Obviously,an 'open' representsa break in the continuityof the circuit. Due to this break .
(i) resistance between the two points is infinite.
(ii) there is no flow of current between the two points.
1.21. 'Shorts' in a Series Circuit
Since a dead (or solid) short has almost zero resistance, it causes the problem of excessivecurrent
which, in turn, causes power dissipation to increase many times and circuit componentsto bum out.In Fig. 1.40 (b), 3-Q resistor has been shorted out by a resistanceless copper wire so that RCD=O.
Now, total circuit resistance R= 1 + 2 + 0 = 3 Q. Hence, / = 12/3 = 4 A and P =:42X3 = 48 W.
Fig. 1.40 (c) shows the situation where both 2 Q and 3 Q resistors have been shorted out of the
circuit. In this case, 2
R = 1 Q, / =12/1 = 12 A and P =12 xl = 144 W
Because of this excessive current (6 times the nonnal value), connecting wires and other circuit
components can become hot enough to ignite and bum out.
1.22. 'Opens' in a Series Circuit
In a nonnal series circuit like the one shown in Fig. 1.41 (a), there exists a current flow and the
voltage drops across different resistors are proportional to their resistances. If the circuit becomes
'open' anywhere, following two effects are produced:
(i) since 'open' offers infinite resistance, circuit current becomes zero. Consequently, there is
no voltage drop across RI and R2.
(ii) wholeof the appliedvoltage(i.e. /00 V in this case) isfelt across the 'open' i.e. across terminalsA and B [Fig. 1.41 (b)]. The reason for this is that RI and R2 V}become negligible as compared to the infinite resistance of the 'open' which has 100V practicallly whole of the applied voltage dropped across it (as per Voltage Divider
Rule of art. 1.15). Hence, voltmeter in Fig. 1.41 (b) will read nearly 100 V i.e. the supply voltage. 'Opens' in a Parallel Circuit .
Since an 'open' offers infinite resistance, there would be no current in that part of the circuit
where it occurs. In a parallelcircuit, an 'open' can occur either in the main line or in any parallelbranch.
As shown in Fig. 1.42 (a), an open in the main line prevents flow of current to all branches.
Hence, neither of the two bulbs glows. However, full applied voltage (i.e. 220 V in this case) is
available across the open.

Nonlinear Resistors

Those elements whose V - I curves are not straight lines are called nonlinear elements because
their resistances are nonlinear resistances. Their V - I characteristics can be represented by an
equation of the form 1= kV = b where n is usually not equal to one and the constant b mayor may not
be equal to zero.
Examples of nonlinear elements are filaments of incandescent lamps, diodes, thermistors and
varistors. A varistor is a special resistor made of carborundum crystals held together by a binder.
Fig. 1.37(a) shows how current through a varistor increase rapidly when the applied voltage increases
beyond a certain amount (nearly 100V in the present case).There is a corresponding rapid decrease in resistance when the current increases. Hence, varistors
are generally used to provide over-voltage protection in certain circuits.
A thermistor is made of metallic oxides in a suitable binder and has a large negative coefficient
of resistance i.e. its resistance decreases with increase in temperature as shown in Fig. 1.30 (b). Fig.
1.30 (c) shows how the resistance of an incandescent lamp increases with voltage whereas Fig. 1.30
(d) shows the V-I characteristics of a typical silicon diode. For a germanium diode, current is related
to its voltage by the relation.

Types of Resistors

Carbon Composition
It is a combination of carbon particles and a binding resin with different proportions for providing
desired resistance. Attached to the ends of the resistive element are metal caps which have axial
leads of tinned copper wire for soldering the resistor into a circuit. The resistor is enclosed in a
plastic case to prevent the entry of moisture and other harmful elements from outside. Billions of
carbon composition resistors are used in the electronic industry every year. They are available in
power ratings of lI8, lI4, 1/2, I and 2 W, in voltage ratings of 250,350 and 500 V. They have low
failure rates when properly used.
Such resistors have a tendency to produce electric noise due to the current passing from one
carbon particle to another. This noise appears in the form of a hiss in a loudspeaker connected to a
hi-fi system and can overcome very weak signals. That is why carbon composition resistors are used
where performance requirements are not demanding and where low cost in the main consideration.
Hence, they are extensively used in entertainment electronics although better resistors are used in
critical circuits.
(b) Deposited Carbon
Deposited carbon resistors consist of ceramic rods which have a carbon film deposited on them.
They are made by placing a ceramic rod in a methane-filled flask and heating it until, by a gascracking
process, a carbon film is deposited on them. A helix-grinding process forms the resistive
path. As compared to carbon composition resistors, these resistors offer a major improvement in
lower current noise and in closer tolerance. These resistors are being replaced by metal film and
metal glaze resistors. '
(c) High-Voltage Ink Film
These resistors consist of a ceramic base on which a special resistive ink is laid down in a helical
band. These resistors are capable of withstanding high voltages and find extensive use in cathoderay
circuits, in radar and in medical electronics. Their resistances range from I ill to 100,000Mil
with voltage range upto 1000 kV.
(d) Metal Film
Metal film resistors are made by depositing vaporized metal in vacuum on a ceramic-core rod.
The resistive path is helix-ground as in the case of deposited carbon resistors. Metal film resistors
have excellent tolerance and temperature coefficient and are extrememly reliable. Hence, they are
very suitable for numerous high grade applications as in low-level stages of certain instruments
although they are much more costlier.
(e) Metal Glaze
A metal glaze resistor consists of a metal glass mixture which is appied as a thick film to a
ceramic substrate and then fired to form a film. The value of resistance depends on the amount of
metal in the mixture. With helix-grinding, the resistance can be made to vary from I Q to many
megohms.
Another category of metal glaze resistors consists of a tinned oxide film on a glass substrate.
if) Wire-wound
Wire-wound resistors are different from all other types in the sense that no film or resistive
coating is used in their construction. They consist of a ceramic-core wound with a drawn wire
having accurately-contralled characteristics. Different wire alloys are used for providing different
resistance ranges. These resistors have highest stability and highest power rating.
Because of their bulk, high-power ratings and high cost, they are not suitable for low-cost or
high-density, limited-space applications. The completed wire-wound resistor is coated with an insulating
material such as baked enamel.
(g) Cermet (Ceramic Metal)
The cermet resistors are made by firing certain metals blended with ceramics on a ceramic
substrate. The value of resistance depends on the type of mix and its thickness. These resistors have
very accurate resistance values and show high stability even under extreme temperatures. Usually,
they are produced as small rectangles having leads for being attached to printed circuit boards (PCB).

Resistance in Series

When some conductors having resistances R1,R2 and R3etc. are joined end-on-end as in Fig.
1.12, they are said to be connected in series. It can be proved that the equivalent resistance or total.
resistance between points A and D is equal to the sum of the three individual resistances. Being a
series circuit, it should be remembered that (i) current is the same through all the three conductors
(ii) but voltage drop across each is different due to its different resistance and is given by Ohm's Law
and (iii) sum of the three voltage drops is equal to the voltage applied across the three conductors.
There is a progressive fall in potential as we go from point A to D as shown
Also -I = -+1-+- I I
G G. G2 GJ
As seenfrom above, the main characteristics of a series circuit are :
I. samecurrent flows through all parts of the circuit.
2. different resistors have their individual voltage drops.
3. voltage drops are additive.
4. applied voltage equals the sum of different voltage drops.
5. resistancesare additive.
6. powers are additive.
Fig. 1.13
17
D
-Ohm's Law
1.15. Voltage Divider Rule
Since in a series circuit, same current flows through each of the
given resistors, voltage drop varies directly with its resistance. In Fig.
1.14 is shown a 24- V battery connected across a series combination
of three resistors.
Total resistance R = R I + R2 + RJ = 12 Q
According to Voltage Divider Rule, various voltage drops are :
R 2
VI = V. --R-1= 24x -12 = 4 V
R2 4
V2 = V. - = 24 x _2
= 8V
R 1
V RJ 6 J = V.- =24x-=12 V
R 12
1.16. Resistances in Parallel
Three resistances,asjoined in Fig. 1.15aresaid to be connected
in parallel. In this case (i) p.d. across all resistancesis the same
(il) current in each resistor is different and is given by Ohm's Law
and(iii) the total current i.. the sum of the three separatecurrents.
V V V
I - 1+1 +1 --+-+-
- I 2 J- R R R . I 2 3
, = ~ where V is the applied voltage.
R = equivalent resistance of the parallel combination.
~ _ -Y.+ y +-Y. or 1- = -1-+ -.l +-.L
R - RI R2 RJ R RI R2 RJ
G = GI + G2+ G]
Fig. 1.15
Now,
..
Also
R
.
12 R
.
IJ R,J
I I
V
R2
24V
RJ
-
Fig. 1.14
II RI
A
2 VI
B
4 V2
C
6 VJ
DThe main characteristics of a parallel circuit are :
I. same voltage acts across all parts of the circuit
2. different resistors have their individual current.
3. branch currents are additive.
4. conductances are additive.
5. powers are additive.

Conductance and Conductivity

Conductance (G) is reciprocal of resistance*. Whereas resistance of a conductor measures the
opposition which it offers to the flow of current, the conductance measures the inducement which it
offers to its flow.
I I A aA
From Eq. (i) of Art. 1.6, R = P A or G =P.T =I
where a is called the conductivity or specific conductance of a conductor. The unit of conductance
is siemens(S). Earlier,this unitwascalledmho.
It is seenfromthe aboveequationthat the conductivityof a materialis givenby
G IG siemens x I metre G I. a = - = 2 = - sIemens /metre A A metre A
Hence, the unit of conductivity is siemens/metre (S/m).
1.9. Effect of Temperature on Resistance
The effect of rise in temperature is :
(i) to increase the resistance of pure metals. The increase is large and fairly regular for normal
ranges of temperatur~. The temperature/resistance graph is a straight line (Fig. 1.6). As
would be presently clarified, metals have a ~ temperature co-efficient at resist~e.
(ii) to increase the resistance of alloys, though in their case, the increase is relatively small and
irregular. For some high-resistance alloys like Eureka (60% Cu and40% Ni) and manganin,
the increase in resistance is (or can be made) negligible over a considerable range of temperature.
'
(iii) to decrease the resistance of electrolytes, insulators (such as paper, rubber, glass, mica etc.)
and partial conductors such as carbon. Hence, insulators are said to possess a negative
temperature-coefficient of resistance. .
1.10. Temperature Coefficient of Resistance
Let a metallic conductor having a resistance of Roat O°Cbe heated of tOCand let its resistance
at this temperature be R,. Then, considering normal ranges of temperature, it is found that the
increase in resistance ~ R =Rt - ~depends
(i) directly on its initial resistance
(ii) directly on the rise in temperature
(iii) on the nature of the material of the conductor.
* In a.c. circuits, it has a slightly different meaning.
R,- Ro oc R x t or R, - Ro=a Rot ...(i)
where a (alpha) is a constant and is known as the temperature coefficient of resistance of the conductor.
" R,-Ro /:1R
Rearrangmg Eq. (I), we get a = Rox t Rox t
If Ro = I Q,t = 1°C, then a =/:1R= R,- Ro
Hence, the temperature-coefficient of a material may be defined as :
the increase in resistance per ohm original resistance per °C rise in temperature.
From Eq. (i), we find that R, = Ro(l + at)It should be remembered that the above equation holds good for both rise as well as fall in temperature.
As temperature of a conductor is decreased. its resistance is also decreased. In Fig. 1.6 is shown the
temperature/resistance graph for copper and is practically a straight line. If this line is extended backwards,
it would cut the temperature axis at a point where temperature is - 234.5°C (a number quite easy
to remember). It means that theoretically, the resistance of copper conductor will become zero at this
point though as shown by solid line, in practice, the curve departs from a straight line at very low
temperatures. From the two similar triangles of Fig. 1.6 it is seen that :
!l t + 234.5 _(I+ L- Ro = 234.5 234.5)
R, = Ro(I + 23~.5) or R, =Ro (l + a t) where a = lJ234.5 for copper.
..
1.11. Value of a at Different Temperatures
So far we did not make any distinction between values of a at different temperatures. But it is
found that value of a itself is not constant but depends on the initial temperature on which the
increment in resistance is based. When the increment is based on the resistance measured at O°C,
then a has the value of <Xo.At any other initial temperature tOC,value of a is a, and so on. It should
be remembered that, for any conductor, <Xohas the maximum value.
Suppose a conductor of resistance Roat O°C(point A in Fig. 1.7) is heated to tOC(point B). Its
resistance R, after heating is given by
R, = Ro (\ + <Xot) ...(i)
where <Xois the temperature-coefficient at O°C.
Now, suppose that we have a conductor of resistance R, at temperature tOe. Let this conductor
be cooled from tOCto O°e. Obviously, now the initial point is B and the final point is A. The final
resistance Ro is given in tenns of the initial resistance by the following equation
Ro = R,[1+ a,(- t)] = R, (1 - a/. t) ...(ii)
B R-Ro
From Eq. (ii) above, we have a, = -R
'
I Xt
Sub"tuting the value of R, from Eq. (i), we get
O/j
.5
(3oU
O/j
c
.~~
::r:
a = Ro(1+ aot) - Ro= ao :. a = ao ...(iii)
I Ro(1+ ao t) x t I + ao t I I + ao t
In general, let a( =tempt. coeff. at t (oC ; ~ =tempt. coeff. at t2°C.
Then from Eq. (iii) above, we get
-ao or --1 I + ao t)
I + ao t( a) ao
Fig. 1.7
= I + ao t2
a2 ao
Subtracting one from the other, we get
Similarly,
I I I 1 1
- -- = (t2-t()or - = - +(t2-t)or~=
a2 a) a2 a) 1/al + (t2 - t)
Values of a for copper at different temperatures are given in Table No. 1.3.
Table 1.3. Different values of a for copper
In view of the dependence of a on the initial temperature, we may define the temperature
coefficient of resistance at a given temperature as the charge in resistance per ohm per degree
centrigrade change in temperaturefrom the given temperature.
In case Ro is not given, the relation between the known resistance RI at tlOCapd the unknown
resistance R2at t2°C can be found as follows :
R2 = Ro (1 + ao t2) and R) =Ro (1 + <Xotl)
R2 _ I + aot2
. . Rt - I + aot)
The above expression can be simplified by a little approximation as follows:
R2 -I
R! = (I + <Xot2)(I + au II)
...(iv)
= (1 + <Xot2)(I - (loti) [Using Binomial Theorem for expansion and
. = 1 + <X(ot2- tl) neglecting squares and high.er powers of (<2Xtol)]
.. !?2 = RI [I + <Xo(t2 - t\)] [Neglectmg product (<Xot(t2)]
For more accurate calculations, Eq. (iv) should, however, be used.
1.12. Variations of Resistivity with Temperature
Not only resistance but specific resistance or resistivity of metallic conductors also increases
with rise in temperature and vice versa.
Tempt. in °C 0 5 to 20 30 40 50
a 0.00427 0.00418 0.00409

Electric Current and Ohm's Law

It should be particularly noted that the direction of electronic current is from zinc to copper in
the external circuit. However, the direction of conventional current (which is given by the direction
of flow of positive charge) is from Cu to zinc. In the present case, there is no flow of positive charge
as such from one electrode to another. But we can look upon the arrival of electrons on copper plate
(with subsequent decrease in its positive charge) as equivalent to an actual departure of positive
charge from it.
When zinc is negatively charged, it is said to be at negative potential with respect to the electrolyte,
whereas anode is said to be at positive potential relative to the electrolyte. Between themselves, Cu
plateis assumed to be at a higher potential than the zinc rod. The difference in potential is continuously
maintainedby the chemical action going on in the cell which supplies energy to establish this potential
difference.
1.4. Resistance
It may be defmed as the property of a substance due to which it opposes (or restricts) the flow of
electricity (i.e., electrons) through it.
Metals (as a class), acids and salts solutions are good conductors of electricity. Amongst pure
metals,silver, copper and aluminium are very good conductors in the given order.* This, as discussed
earlier, is due to the presence of a large number of free or loosely-attached electrons in their atoms.
Thesevagrant electrons assume a directed motion on the application of an electric potential difference.
These electrons while flowing pass through the molecules or the atoms of the conductor, collide and
other atoms and electrons, thereby producing heat.
Those substances which offer relatively greater difficulty or hindrance to the passage of these
electrons are said to be relatively poor conductors of electricity like bakelite, mica, glass, rubber,
p.v.c. (polyvinyl chloride) and dry wood etc. Amongst good insulators can be included fibrous
substances such as paper and cotton when dry, mineral oils free from acids and water, ceramics like
hard porcelain and asbestos and many other plastics besides p.v.c. It is helpful to remember that
electric friction is similar to friction in Mechanics.
1.5. The Unit of Resistance
The practical unit of resistance is ohm.** A conductor is said to have a resistance of one ohm if
it permits one ampere current to flow through it when one volt is impressed across its terminals.
For insulators whose resistances are very high, a much bigger unit is used i.e. megaohm =106
ohm (the prefix 'mega' or mego meaning a million) or kilohm = 103ohm (kilo means thousand). In
the case of very small resistances, smaller units like milli-ohm = 10-3 ohm or microhm = 10-6 ohm
are used. The symbol for ohm is Q.
Table 1.1. Multiples and Sub-multiples of Ohm
However. for the same resistance per unit length, cross-sectional area of aluminium conductor has to be
1.6 times that of the copper conductor but it weighs only half as much. Hence, it is used where economy
of weight is more important than economy of space.
** After George Simon Ohm (1787-1854), a German mathematician who in about 1827 formulated the law
of known after his name as Ohm's Law.
Prefix Its meaning Abbreviation Equal to
Mega- One million MO 1060
Kilo- One thousand kO 1030
Centi- One hundredth - -
MiIIi- One thousandth mO 10-30
Micro- One millionth flO 10-<>
Laws of Resistance
The resistance R offered by a conductor depends on the following factors :
(i) It varies directly as its length, I.
(ii) It varies inversely as the cross-section A of the conductor.
(iii) It depends on the nature of the material.
(iv) It also depends on the temperature of the conductor.
Smaller I
Larger A
Low R
Larger I
Smaller A
Greater R
Fig. 1.3. Fig. 1.4
Neglecting the last factor for the time being, we can say that
R oc ~ or R = P ~ ...(i)
where p is a constant depending on the nature of the material of the conductor and is known as its
specific resistance or resistivity.
If in Eq. (i), we put
I = I metre and A =I metre2,then R =P (Fig. 1.4)
Hence, specific resistance of a material may be defined as
the resistance between the opposite faces of a metre cube of that material.
1.7. Units of Resistivity
From Eq. (i), we have p = AR I
In the S.I. system of units,
A metre2 x R ohm = AR ohm-metre
p = I metre 1
Hence, the unit of resistivity is ohm-metre (Q-m).
It may, however, be noted that resistivity is sometimes expressed as so many ohm per m3.
Although, it is incorrect to say so but it means the same thing as ohm-metre.
If I is in centimetres and A in cm2,then p is in ohm-centimetre (Q-cm).
Values of resistivity and temperature coefficients for various materials are given in Table 1.2.
The resistivities of commercial materials may differ by several per cent due to impurities etc.Resistivities and Temperature Coefficients
Example 1.3. A coil consists of 2000 turns of copper wire having a cross-sectional area of 0.8 J
mm-. The mean length per turn is 80 em and the resistivity of copper is 0.02 J1Q~m. Find the
resistance of the coil and power absorbed by the coil when connected across 110 V d.c. supply.
(F.Y. Engg. Pone Univ. May 1990)
. Solution. Length of the coil, I =0.8 x 2000 =1600 m ; A =0.8 mm2=0.8 x 10-6m2.
R = P~ =0.02X 10-6x 1600/0.8X 10-6=40 Q
Power absorbed = V/ R =1102/40=302.5 W
Example 1.4. An aluminium wire 7.5 m long is connected in a parallel with a copper wire 6 m
long. When a current of 5 A is passed through the combination, it isfound that the current in the
aluminium wire is 3 A. The diameter of the aluminium wire is 1 mm. Determine the diameter of the
copper wire. Resistivity of copper is 0.017 J1Q-m..that of the aluminium is 0.028 J1Q-m.
(F.Y. Engg. Pone Univ. May 1991)
Solution. Let the subscript I represent aluminium and subscript 2 represent copper.
I I R P I a
R I = pi and R2=P2 --1.. :. ...1.=...1.. .1. . -L
al a2 RI PI II a2
RI P2. 12 .
.. a2 = al.R.2 -P'IT 1 ...(1)
Now /1 = 3 A ; /2 =5 - 3=2 A.
If Vis thecommonvoltageacrosstheparallelcombinationof aluminiumandcopperwires,then
Material Resistivity in ohm-metre Temperature coefficient at
at 20.C (x 10-JI) 20.C (x 10-4)
Aluminium, commercial 2.8 40.3
Brass 6-8 20
Carbon 3000 - 7000 -5
Constantan or Eureka 49 +0.1 to-o.4
Copper (annealed) 1.72 39.3
German Silver 20.2 2.7
(84% Cu; 12%Ni; 4% Zn)
Gold 2.44 36.5
Iron 9.8 65
Manganin 44 - 48 0.15
(84% Cu; 12%Mn; 4% Ni)
Mercury 95.8 8.9
Nichrome 108.5 1.5
(60% Cu ; 25% Fe ; 15%Cr)
Nickel 7.8 54
Platinum 9 - 15.5 36.7
Silver 1.64 38
Tungsten 5.5 47
Amber 5 x 1014
Bakelite 1010
Glass 1010 _ 1012
Mica 1015
Rubber 1016
Shellac 1014
Sulphur 1015

ELECTRIC CURRENT AND OHM'S LAW

Electron Drift Velocity
Suppose that in a conductor, the number of free electrons available per m3 of the conductor
material is n and let their axial drift velocity be v metres/second. In time dt, distance travelled would
be v x dt. If A.is area of cross-section of the conductor, then the volume is vAdt and the number of
electrons contained in this volume is vA dt. Obviously, all these electrons will cross the conductor
cross-section in time dt. If e is the charge of each electron, then total charge which crosses the
section in time dt is dq =nAev dt.
Since current is the rate of flow of charge, it is given as
= -dq = nAev dt :. I.=nAev
dt dt
Current density J = ilA =ne v amperelmetre2
Assuming a normal current density J =1.55 X 106 Alm2, n = 1029for a copper conductor
6 -19
and e =1. x 10 coulomb, we get
1.55 x 106 = 1029x 1.6X 10-19x v :. v = 9.7 X 10-5m/s = 0.58 cm/min
It is seen that contrary to the common but mistaken view, the electron drift velocity is rather
very slow and is independent of the current flowing and the area of the conductor.
N.H.Current density i.e., the current per unit area. is a vector quantity. It is denoted by the symbol J .
-->
Therefore. in vector notation, the relationship between current I and J is:
-->--> -->
I = J. a [where a is the vector notation for area 'a']
For extending the scope of the above relationship. so that it becomes applicable for area of any shape, we
write:
f
--> -->
J = J .d a
Themagnitudeof the currentdensitycan. therefore.be writtenas J.a.
Example 1.1. A conductor material has a free-electron density of J{j4 electrons per metrl.
When a voltage is applied, a constant drift velocity of 1.5 X ]0-2 metre/second is attained by the
electrons. If the cross-sectional area of the material is 1 cm2,calculate the magnitude of the current.
Electronic charge is 1.6 x ]0-19coulomb. (Electrical Engg. Aligarh Muslim University 1981)
Solution. The magnitude of the current is
= nAev amperes
n = 1024;A = 1cm2 = 10-4m2
e = 1.6 x 10-19C ; v = 1.5X 10-2m/s
= 1024X 10-4x 1.6 X 10-19x 1.5X 10-2=0.24 A
Here,
..
1.2. Charge Velocity and Velocity of Field Propagation
The speed with which charge drifts in a conductor is called the velocity of charge. As seen fromabove. its value is quite low, typically fraction of a metre per second.
However. the speed with which the effect of e.m.f. is experienced at all parts of the conductor
resulting in the flow of current is called the velocit.v of propagation of electri8calfield. It is independent
of current and voltage and has high but constant value of nearly 3 x 10 mls.
Example 1.2. Find the velocity of charge leading to 1 A current which flows in a copper
conductor of cross-section 1 em2and length 10 km. Free electron dellSityof copper =8.5.x 1028per
m3. How long will it take the electric charge to travelfrom one end of the conductor to the other.
Solution. i =neAv or v =ih,('A
.. v = 1/(85 y I02X) x 1.6 x 10 1'1X (I X 10-4) = 7.35 X 10-7 mls =0.735 Jlmls
Time taken by the charge 10lravd conductor length of 10 kIn is
t = distance _ 10x 103 =1.36 x 1010S
velocity 7.35 x 10-7
Now. I year::: 365 x 24 x 3600 = 31,536.000 s
10
t = 1.36 x 10 /31.536.000 = 431 years
1.3. The Idea of Electric Potential
In Fig. 1.1is shown a simple voltaic cell. It consists of copper plate (known as anode) and a zinc
rod (I.e. cathode) immersed in dilute sulphuric acid (H2S04) contained in a suitable vessel. The
chemical action taking place within the cell causes the electrons to be removed from Cu plate and to
be deposited on the zinc rod at the same time. This transfer of electrons is accomplished through the
agency of the diluted H2S04 which is known as the electrolyte. The result is that zinc rod becomes
negative due to the deposition of electrons on it and the Cu plate becomes positive due to the removal
of electrons from it. The large number of electrons collected on the zinc rod is being attracted by
anode but is prevented from returning to it by the force set up by the chemical action within the cell.
Conventional Direction
of Current
--0 --0 --0. Directionof '"
E-l-e-ctronFlow0- .'"
1 1
Water
,'-:' -- - --- --- -- - -- - - -- - - - - :-,
r
Direction
of Flow
1
Cu Zn
Fig. 1.1. Fig. 1.2
But if the two electrodes are joined by a wire externally. then electrons rush to the anode thereby
equalizing the charges of the two electrodes. However. due to the continuity of chemical action. a
continuous difference in the number of electrons on the two electrodes is maintained which keeps up
a continuous flow of current through the external circuit. The action of an electric cell is similar to
that of a water pump which. while working. maintains a continuous flow of water i.e. water current
through the pipe (Fig. 1.2).