Unit-3

59

DC CircuitsUNIT 3 DC CIRCUITS

Structure 3.1 Introduction

Objectives

3.2 Kirchhoff’s Laws 3.2.1 Kirchhoff’s Voltage Law 3.2.2 Kirchhoff’s Current Law

3.3 Nodal and Mesh Analysis

3.4 Network Theorems 3.4.1 Thevenin’s Theorem 3.4.2 Norton’s Theorem

3.5 Source Transformation

3.6 Summary

3.7 Answers to SAQs

3.1 INTRODUCTION

German physicist Gustav R. Kirchhoff’s is best known for his statement of two more important basic laws governing the performance of a circuit. In his honour these laws are named as Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL). In this unit, you will get understanding of fundamental details and understanding of these laws.

Analysis of circuits using Kirchhoff’s voltage current laws no doubt gives a solution directly, but in the circuits having more loops, it becomes difficult to apply these laws due to large number of variables (voltages and currents) provides large type sets of equations to solves. In such cases Nodal and Mesh analysis provides elegant way of analysing the circuits by reducing the number of unknown variables.

In this unit, we shall also study some important fundamental theorems of circuit theory with illustrative examples to explain the applicability of these theorems in dc circuits.

Objectives After studying this unit, you should be able to

• write the voltage and current equations using Kirchhoff’s voltage and current laws and solve circuit problems,

• calculate the current, voltage, power in circuit using Nodal and Mesh analysis,

• apply Thevenin’s theorem and Norton’s Theorem to reduce complex circuit,

• analyse the current by determining the values of electrical quantities using some important network theorems, and

• describe some transformation techniques to convert a voltage source to a current source and vice-versa.

3.2 KIRCHHOFF’S LAWS

In 1845, Kirchhoff, at the age of 23, published his paper regarding two basic law for solving the electrical circuits. These laws became basis for the development of the network analysis or circuit theory.

60

3.2.1 Kirchhoff’s Voltage Law (KVL) Electrical

This law is based on the law of conservation of energy. According to this, electrical energy supplied to any circuit (closed loop) is equal to the energy consumed by its passive elements. Since this law relates the voltage in a closed circuit of an electrical network it is known as Kirchhoff’s Voltage Law (KVL) or Kirchhoff’s Mesh Law. Kirchhoff’s voltage law states that in a closed electric circuit the algebraic sum of emfs and voltage drops is zero. By convention, the emfs or voltage rises are taken to be positive and voltage drops are taken to be negative.

I1

R1 R2

V1 V2

E2E1

A

B C

D Figure 3.1 : Closed Loop

In the closed circuit ABCDA, shown in Figure 3.1, applying Kirchhoff’s voltage law, we have 1 1 2 2 0E V V E− − − =

1 1 2 2 0E IR IR E− − − =

1 2 1 2E E IR IR− = +

Sum of voltage rises = Sum of voltage drop or Energy supplied = Energy consumed.

3.2.2 Kirchhoff’s Current Law (KCL) Kirchhoff’s current law states that algebraic sum of all the current meeting at a point or junction is zero. Since this law relates the current flowing through the circuit, it is known as Kirchhoff’s Current Law (KCL). In Figure 3.2, applying KCL to the junction, we have

1 3 4 2 5 0I I I I I+ + − − =

or, 1 3 4 2I I I I I5+ + = +

I5I4

I3

I2

I1

O

Figure 3.2 : KCL at Junction

∑ Current entering the junction = ∑ Current leaving the junction.

KCL also indicates the law of conservation of charges and can also be states as at any node of a circuit, at every instant of time, the sum of incoming currents is equal to the sum of outgoing currents. Application of Kirchhoff’s Voltage Law

Kirchhoff’s voltage law is used to determine the value of current in the multi loop circuit. For this purpose, Maxwell’s loop method (Mesh Analysis) is used.

61

DC CircuitsApplication of Kirchhoff’s Current Law

Kirchhoff’s current law is used to determine the voltages at different nodes or junctions of the circuit. For this purpose, Maxwell’s node method (Nodal Analysis) is used. These two methods are being explained in next section with the help of an example.

3.3 NODAL AND MESH ANALYSIS

Kirchhoff’s voltage law is used to determine the value of current in the multi loop circuit. For this purpose, Maxwell’s loop method is used as explained by an example given below.

Figure 3.3 shows the circuit having two loops and one voltage source in first loop.

I1

R4

I2

V

R1 R3

R2

Figure 3.3 : Maxwell’s Loop Method

Assume, two loop currents flowing in the directions shown in Figure 3.3. Now we can write the KVL equations for each loop. KVL Equation for First Loop

∑ Voltage rise = Voltage drops ∑

1 1 2 1 2( )V R I R I I= + −

1 2 1 2( ) 2R R I R I= + − . . . (i)

KVL Equation in Second Loop

2 2 1 3 2 4 20 ( )R I I R I R I= − + +

2 1 2 3 4 2( )R I R R R I= − + + + . . . (ii) After solving equations, we can find two loop currents I1 and I2. Voltage division rule is also one of the applications of KVL.

Application of Kirchhoff’s Current Law Kirchhoff’s current law is used to determine the voltages at different nodes or junctions of the circuit. For this purpose, Maxwell’s node method is used. This method is being explained with the help of an example given below. Again consider the same circuit of Figure 3.3.

For Applying Maxwell’s Node Method Assume voltage V1 at node as shown in Figure 3.4 and one reference node with zero voltage.

I1

R4

Reference Node (Voltage = 0)

I3

V

R1 R3V1

I2

R2

62

Figure 3.4 : Maxwell’s Node Method Electrical

Assume three currents I1, I2 and I3 leaving the node V1, then apply KCL :

Algebraic of currents at any node is zero.

1 2 3 0I I I+ + =

1 1 1

1 2 3 40V V V V

R R R R−

+ + =+

11 2 3 4

1 1 1 VV1R R R R R

⎡ ⎤+ + =⎢ ⎥+⎣ ⎦

From this equation node, voltage V1 can be obtained and then current passing through different elements can be determined.

For example, current through R1 is

11

1

V VIR−

=

If value of I1 comes negative, this means actual direction of current I1 will be opposite to the assumed direction as shown in Figure 3.4.

Current division rule is also one of the applications of KCL.

Example 3.1

Find currents I1 and I2 in the given circuits by applying KVL.

Figure 3.5

Solution

We apply KVL for first loop :

1 110 1 1 ( )I I I2= + −

110 2 I I2= − . . . (i)

KVL for second loop :

2 1 21 ( ) 4 0I I I− + =

2 15 0I I− = . . . (ii)

∴ 12 0.2

5II = = 1I

1I

. . . (iii)

Put the value of I2 in Eq. (i)

1 110 2 0.2 1.8I I= − =

⇒ 1 5.55 AmpI =

63

DC CircuitsFrom Eq. (iii), . 2 1.11 AmpI =

Example 3.2

Find the node voltage V and then current passing through each element by using KCL.

Figure 3.6

Solution

Apply KCL at the node whose voltage is V.

Algebraic summation of all the current meeting at node is zero.

10 01 1 4

V V V−+ + =

11 1 104

V ⎡ ⎤+ + =⎢ ⎥⎣ ⎦

2.25 V = 10

V = 4.44 volt.

Now we have to find current through each element.

Figure 3.7

10 10 4.44 5.56 Amp1

VI −= = − =

1 4.44 Amp1VI = =

2 1.11 Amp4VI = = .

SAQ 1 (a) In the given DC circuit, determine the currents I1 and I2 by using KVL.

64

Figure for SAQ 1(a) Electrical

(b) In the following circuit, find the node voltage V and then determine the current passing through each element (use KCL).

Figure for SAQ 1(b)

(c) Find the current passing through 2Ω resistor by using Maxwell loop method.

10V

5Ω

25V 5Ω

2Ω 1Ω

Figure for SAQ 1(c)

(d) Find the current passing through 1Ω resistor by using KVL (by Maxwell loop method)

Figure for SAQ 1(d)

(e) Solve the above problem (SAQ 1(d)) by using KCL (apply node method).

3.4 NETWORK THEOREMS

We have already studied Kirchhoff’s laws, Nodal and Mesh analysis in the previous sections of this unit, these general methods of network analysis tend to become time consuming and laborious in the case of large and complex networks. In such cases some special techniques are used to solve complicated networks to reduce the quantum of labour and time involved in solution of circuits. Such special techniques are called as network theorems and these techniques are applicable to a useful and fairly wide class of networks, and their conclusions are simple.

65

DC Circuits3.4.1 Thevenin’s Theorem

This is the most extensively used theorem in circuit theory. Any of the circuit we have been working with can be viewed as a two terminal network thevenin’s theorem provides us with an easy way to develop an equivalent circuit of a two terminal network.

Applications Sometimes it is required to study the variation of current or voltage in a particular branch by varying the resistance of that branch while remaining network remains the same, e.g. designing of electronic circuit. At such places thevenin’s theorem is quite suitable.

Statement This theorem states that any two terminal network containing energy sources and resistors can be replaced by on equivalent network consisting of a single source of emf and a series resistor, Rth. This emf Eth, is equal to potential difference between the terminals of the network, when the resistor, R, is removed. The resistance of the series resistor, Rth, is equal to the equivalent resistance of the network measured between the terminals, with the resistor, R, removed and all energy sources eliminated (but not their internal resistances) here, ideal independent voltage sources is replaced by short circuit and ideal independent current sources by open circuit. Equivalent circuits developed by thevenin’s theorem are often called as thevenin’s equivalent circuits.

We can explain thevenin’s theorem by applying it to a two-terminal network using following steps :

Figure 3.8 : A Two Terminal Network

Here by applying thevenin’s theorem you will learn to determine the thevenin’s equivalent circuit across the terminals where load RL is connected. Step 1

Create open circuit between the terminals for this remove RL in which current is to be determined.

Step 2 Determine the open circuit voltage (Vth) between the terminals. So, voltage across R3 = i R3

(Voltage across R2 will be zero as no current is following through it due to open circuit)

1 3

Eir R R

=+ +

Vth = Voltage across R3 only 31 3

E Rr R R

⎛ ⎞= ⎜ ⎟+ +⎝ ⎠

66

Electrical

Figure 3.9 : Network to Find Open Circuit Voltage

Step 3

Determine the internal resistance of the equivalent circuit (Rth) after replacing all sources in the original circuit with resistances equal to their internal resistances. When the original sources are viewed as ideal sources, voltage sources are replaced by short circuit and current sources are replaced by open circuit. After this total resistance between the terminals is calculate or measured.

Figure 3.10 : Network to Find Rth

1 3th 2

1 3

( ) .r R RR R

r R R+

= ++ +

Step 4

Replace the entire network by the Vth and Rth across the terminals (in which Rth is in series with Vth) or make thevenin’s equivalent circuit across the terminals.

Figure 3.11 : Thevenin’s Equivalent Network

Step 5

Connect load resistance RL back to its terminals from where it was removed.

Figure 3.12 : Thevenin’s Equivalent after Connecting RL

Step 6

Determine the current flowing through the load RL.

th

th L

VI

R R=

+

Example 3.3

67

DC Circuits

i

Use thevenin’s theorem to find the following for the rework shown in figure below

(a) the equivalent emf of the network when view from terminals a and b,

(b) the equivalent resistance of the network when looked into from terminal a and b, and

(c) current in the load resistance RL of 30 Ω.

Figure 3.13

Solution

(a) For finding Vth or open circuit voltage (Voc) across the terminals remove RL = 30 Ω.

Figure 3.14

Now, voltage across terminals ab = voltage across 24 Ω

th 24ocV V= = ×

i = current through 24 Ω 48 1.5 Amp.

24 8 2= =

+ +

∴ th 24 1.5 36V V= × = (Ans.)

(b) For equivalent resistance Rth

Figure 3.15

There are two parallel paths between terminal a and b

∴ 24 (6 2) 624 (6 2)thR × +

= = Ω+ +

(Ans.)

(c) For finding current in load resistance RL of 30 Ω.

Thevenin’s equivalent circuit is

68

th

th

36 1 Amp6 30L

VI

R R= = =

+ + (Ans.)

Electrical

Figure 3.16

Example 3.4

Find current through the 5 Ω resistor.

Figure 3.17

Solution

Remove 5 Ω resistor through which current is to be determined (say terminal a and b).

Figure 3.18

Open circuit voltage across ab is

th 10ocV V= = Ω

To obtain thevenin’s resistance across the open circuited terminal, the sources are eliminated as shown in figure below.

Figure 3.19

It is evident from figure that Rth = 0.

Thus, current through the 5 Ω resistor is given by

69

DC Circuits

th

th

10 2 Amp5 5

VI

R= = =

+ (Ans.)

Figure 3.20

SAQ 2 Find current through the 0.2 Ω resistor in the figure using thevenin’s theorem.

1.7 1.7 A0.9 0.1

THL

TH L

VIR R

= = =+ +

A

2 A

B

0.1 Ω

0.4 Ω0.5 Ω

5 A

Figure for SAQ 2

3.4.2 Norton’s Theorem

Like Thevenin’s theorem, Norton’s theorem is also frequently used in electronic circuits. As discussed earlier Thevenin’s theorem is used to simplify a network into a constant voltage source and a series resistance, Norton’s theorem can be used to simplify a network into a constant current source and a parallel resistance. In circuit analysis interchange of voltage sources and current sources by use of Thevenin’s and Norton’s theorem is sometimes very useful. (This source transformation will be discussed in next section of this unit). Statement

This theorem states that any two terminal linear network containing energy sources and resistors and can be replaced by an equivalent network consisting of a single current source (Isc or IN) and a equivalent parallel resistor (RN). Where IN or Isc is the short circuit current between the two terminals when these terminals are short circuited and RN is the equivalent resistance as seen from the terminals. This parallel resistance is the same as the Thevenin’s equivalent resistance as defined earlier and we can explain Norton’s theorem by applying it to a two terminal network using following steps :

70

Electrical

Figure 3.21 : A Two Terminal Network

Here by applying Norton’s theorem, you will learn to determine the Norton’s equivalent circuit across the terminal where load RL is connected.

Step 1

Short circuit the terminals across which the load resistor is connected. Now calculate the current which would flow between them. This current is called short circuit current (Isc) or Norton’s current (IN).

Figure 3.22 : Network to Find Isc or IN

Here, 1

N scEI IR

= =

Step 2

Determine RN as seen from the terminal. For calculating RN you have to redraw the network by replacing each voltage source by short circuit in series with its internal resistance if any and each current source by open circuit in parallel with its internal resistance

Method for calculating RN in same as for calculating Rth is Thevenin’s theorem.

So, after redrawing the network

1 2 3

1 2

( )N

3

R R RR

R R R+

=+ +

Figure 3.23 : Network to Find Equivalent Resistance

Step 3 Draw Norton’s equivalent circuit across the load terminals by replacing the entire network by IN or Isc and RN in parallel with it.

Figure 3.24 : Norton’s Equivalent Circuit

Step 4

Connect the load back to its terminal and find current (IL) through it

71

DC Circuits

NL N

N L

RI I

R R= ×

×

Figure 3.25 : Norton’s Equivalent Circuit after Connecting RL

Example 3.5

Find the current through 15 Ω resistance of the circuit shown in figure using Norton’s theorem.

Figure 3.26

Solution

Remove 15 Ω resistance through which current is to be determined and short circuit the terminals from where it has removed (say a, b)

30 3.758N scI I= = = Ω

Figure 3.27

Now to find RN, replace battery of 30 V by short circuit resistance. Now network as seen from the terminals a and b is

8 (10 14) 68 (10 14)NR +

= = Ω+

72

Figure 3.28 Electrical

Norton’s equivalent circuit is shown in figure is

Figure 3.29

Now, current through 12 Ω resistance using current division rule is

63.75 1.0714 Amp6 15

I = × =+

SAQ 3 Determine the voltage across 100 Ω resistor in the circuit given below by Norton’s theorem.

Figure for SAQ 3

SAQ 4 Fill in the blanks :

(a) The circuit where parameters change with voltage or current is called _________________ circuit.

(b) The circuit where parameters are constant is called a __________ circuit.

(c) An ideal constant voltage source has _____________ internal resistance whereas a constant current source has ______________ internal resistance.

(d) Total resistance of a parallel circuit is _____________ the smallest branch resistance.

(e) Maxwell’s loop (or Mesh) current method is best suited when energy sources are ________________ sources rather than ________ sources.

73

DC Circuits3.5 SOURCE TRANSFORMATIONS

The voltage and current sources are mutually transferrable. Any practical voltage source (or simply, a voltage source) consists of an ideal voltage source in series with an internal resistance (for ideal source, this impedance is being zero and the output becomes independent of the load current). Any practical current source is demonstrated by ideal current source in parallel with their internal resistance. Let Figures 3.30(a) and (b) demonstrates their equivalence.

(a) A Practical Voltage Source (b) A Practical Current Source Figure 3.30 : Source Transformation

Assume some load resistance rL connected at terminal a-b of both voltage and current source.

aL

a L

VIr r

=+

b

bL

b L

rI I

r r=

+

Two sources to become identical they should deliver same current to load

b

a L b L

rV Ir r r r

=+ +

However, for the current source, the terminal voltage at a-b would be I rb, a-b being open

i.e. bV I r=

Now, we finally get

a L br r r r+ = + L

i.e. a br r=

So, for any practical voltage source, if ideal voltage be V and internal resistance be ra, the voltage source can be replaced by a current source I with the internal resistance in parallel to the current source.

Example 3.6

Convert the following current source to equivalent voltage source.

Figure 3.31

74

Electrical

r

Solution For equivalent voltage source V is given by V I= ×

10 10A= × Ω

100 V=

Series resistance for voltage source will be same as parallel resistance in case of current source (that is 10 Ω as given). So, equivalent voltage source is

Figure 3.32

Example 3.7 Convert voltage source shown in figure to a current source.

Figure 3.33

Solution Short circuit the voltage source to find the constant current

24 V 4 A6

ocSC

s

VI

R= = =

Ω

Figure 3.34

The Rs of the two sources will be the same. So, the constant current is 4 A and current source is as shown in figure below.

Figure 3.35

75

DC Circuits3.6 SUMMARY

In this unit, you have studied about methods of network analysis which are general in nature such as KCL, KVL, Nodal and Mesh analysis. These general methods may be used for any network and enables us to find current and voltages in all branches of the network. However in some network problems it is required only restricted analysis such as to find the current, voltage and power in a load resistance across given two terminals of a network. For such restricted analysis the desired results may be obtained more conveniently using certain network theorems. In this unit, you have treated with some of the important network theorems like Thevenin’s and Norton’s theorem.

Here you have also learnt source transformation techniques.

Solved examples along with SAQs related to each topic have been given for better understanding.

3.7 ANSWERS TO SAQs

SAQ 1

(a)

Figure for Answers to SAQ 1(a)

By Maxwell loop method

Apply KVL for loop 1 :

1 15 3 (I I I= + − 2 )

2

2

or, . . . (i) 1 25 4 3I I= −

KVL for loop 2 :

2 1 20 3 ( ) 2 4I I I I= − + +

1 20 3 9I I= − + . . . (ii)

1 3I I= . . . (iii)

Put Eq. (iii) in Eq. (i) to find value of I2

2 25 4 (3 ) 3I I= −

2 212 3I I= −

29I=

So, 25 Amp 0.555 Amp9

I = =

From Eq. (iii)

15 53 A9 3

I ⎛ ⎞= =⎜ ⎟⎝ ⎠

mp

or, 1 1.667 AmpI =

76

(b) Electrical

Figure for Answers to SAQ 1(b)

KCL at node :

5 01 1 6

V V V−+ + =

11 1 56

V ⎛ ⎞+ + =⎜ ⎟⎝ ⎠

V = 3.33 volt

(c)

Figure for Answers to SAQ 1(c)

For loop 1 1 1 1 2 125 5 2 5( ) 25 12 5I I I I I 2I= + + − ⇒ = − . . . (i)

For loop 2 2 2 1 110 5( ) 10 5 6I I I I 2I− = + − ⇒ − = − + . . . (ii)

Solving Eqs. (i) and (ii)

2

25 510 6

2.128 Amp12 5

5 6

I

−−

= =−

−

(d) Assume loop currents I1, I2, I3 as shown in figure.

Figure for Answers to SAQ 1(d)

Here current passing through 1 Ω is I1

KVL equations :

Loop 1 : 1 215 6( )I I I1= + + . . . (i)

Loop 2 : 1 2 2 315 6( ) 10 6( )I I I I= + + + − . . . (ii)

77

DC Circuits

3Loop 3 : . . . (iii) 3 25 6( ) 2I I I− = − +

After solving Eqs. (i), (ii) and (iii), we get

I1 = 6.364 Amp (current passing through 1 Ω resistor).

(e)

(a) (b)

Figure for Answers to SAQ 1(e)

Apply source transformation

Now, network after assuming node voltages V1, V2, V3

Here we have to use super node analysis V2 and V3 from the super node. Apply KCL at node 1 :

1 1 2 53 3

V V V−+ =

1 2 1 21 1 1 5 2 13 3 3

V V V V⎛ ⎞+ − = ⇒ − =⎜ ⎟⎝ ⎠

5 . . . (i)

KCL at node 2 and 3 (super node)

3 32 1 2 5 03 1 6 2

V VV V V −−+ + + =

1 2 30.333 1.333 0.666 2.5V V V− + + = . . . (ii)

And the voltage source is equal to V2 – V3

V2 – V3 = 10 V

V3 = V2 – 10 . . . (iii) After solving Eqs. (i), (ii) and (iii), we get

V1 = 10.682 V

V2 = 6.3648 V

V3 = − 3.635 V

Current through 1 Ω, 2 6.3648 Amp1

VI = =

SAQ 2 Step 1 : The Thevenin’s Equivalent Circuit is,

Figure for Answers to SAQ 2 (Step 1)

78

Electrical

Step 2 : From the Given Circuit, Disconnect RL = 0.1 Ω and find VTH

Figure for Answers to SAQ 2 (Step 2)

Circuit to find VTH

0.4 2 0.5 5 0ABV− − × + × =

or 0.5 5 0.4 2 1.7 VABV = × − × = B is negative polarity and A is positive polarity.

Step 3 : To Calculate RTH

From the given circuit, open the current source. The resultant circuit is,

Figure for Answers to SAQ 2 (Step 3)

0.5 0.4 0.9TH ABR R= = + = Ω

Step 4 : Now Thevenin’s Equivalent is

Figure for Answers to SAQ 2 (Step 4)

Step 5 : Connect Load Resistance of 0.1 Ω Across AB

Figure for Answers to SAQ 2 (Step 5)

1.7 1.7 A0.9 0.1

THL

TH L

VIR R

= = =+ +

79

DC CircuitsSAQ 3

To find Isc, short circuit the terminal A and B and find current through it

0.2 1 1.2 ASCI = + =

To find RN (= RTH)

Disconnect RL = 100 Ω between A and B.

Figure for Answers to SAQ 3 Draw Norton’s equivalent circuit and connect the 100 Ω across A-B terminal

Figure for Answers to SAQ 3

SC NL

N L

I RI

R R=

+

1.2 50 0.4 A50 100

×= =

+

L L LV I R=

0.4 100 40 V= × =

SAQ 4 (a) Non-linear (b) Linear (c) Zero, infinite (d) Less than (e) Voltage, current