1

Electrical Technology (3 – 1 – 3) D.C. Networks:

• Kirchoff’s Laws, • Node Voltage and Mesh Current Methods, • Delta – Star and Star – Delta conversations, • Superposition Principle, • Thevenin’s and Norton’s Theorems, • Maximum Power Transfer Theorems.

Books:

1. Electrical Technology – Edward Hughes, ELBS (7th Edition). 2. Theory and Problems of Basic Electrical Engineering – D.P. Kothari and I.J. Nagrath,

Prentice Hall of India (1st Edition). 3. Electrical Engineering Fundamentals – Vincent Del Toro, Prentice Hall of India (2nd

Edition). 4. Problems in Electrical Engineering – S. Parker Smith, CBS Publishers and

Distributors (IXth Edition) Basic Concepts:

• Electric Circuit: - A closed path composed of active and passive elements to which current flow is confined.

• Active element – which supplies energy to the circuit. • Passive element – which receives energy and then either converts it to heat or stores it

in an electric or magnetic field. Linear and Non-Linear Circuits:

Unilateral and Bilateral Elements:

Current

Vol

tage

R is constant

Current

Vol

tage

R is not constant

+ -

R1

- +

R2

R1 ≠ R2

+ -

R1

- +

R2

R1 = R2

2

Sources (D.C.)

• Voltage source Ideal: Maintains a constant voltage Vs across its terminals regardless of the current drawn (its magnitude as well as direction).

So dynamic resistance of the ideal voltage source is Zero. Practical: Voltage decreases some what as the load current is increased. vL = VS – RS iL; RS is internal resistance of the source. PL = vL.iL

PS = VS.iL

vL = VS when iL = 0 VS open circuit voltage of practical voltage source.

• Current Source Ideal: Maintains a constant current IS (the value and the direction of this current at any instant of time is independent of the value or direction of the volts that appears across the terminal of the source).

So dynamic resistance of the ideal current source is Infinite.

A

B

+

-

VS

VS

Slope is zero

Voltage

Current

A

B

+

- VS

VS

Slope is zero

Voltage

Current

vL RS

iL

+

-

VS

RS

VL LOAD

+

-

A

B

IS

IS Slope is infinity

Voltage

Current

3

Practical: Current decreases as vL is increased.

L S LS

1i = I - vR

RS is internal resistance of the source. iL = IS only when vL = 0. IS Short Circuit current of the practical current source. Source Representation and Conversion

L L SS S

1 1I = - V + VR R

⎛ ⎞⎜ ⎟⎝ ⎠

LL S

S

VI = I -R

A

B

IS

IS Voltage

Current

iL

RS

iL

iL

IS RS vL LOAD

+

-

iL

VS

RS

VL

+

-

+

-

Practical voltage source

iL

IS RS VL

+

- Practical current source

VL

IL SSC

S

VI =R

VOC = VS

VOC = Open circuit voltage (IL = 0)

VL

IL

VOC = RSIS

ISC = Short circuit current (VL = 0)

ISC = IS

4

Network Laws and Theorems Kirchoff’s Laws:

1. Current Law – Algebraic sum of all the currents entering or of all the currents leaving any node of an electric circuit is zero. (KCL)

m

KK=1

i = 0∑

i1 – i2 + i3 – i4 + i5 – i6 = 0

Currents entering the node are taken as positive and leaving the node are taken as negative.

2. Voltage Law:- Algebraic sum of the emfs and resistance voltages (product of current

and resistance) in any continuous path of an electric circuit is zero. (KVL)

n

KK=1

v = 0∑

Example: Node m : IA – ID – IF = 0 Node n : ID + IE – IC = 0 KCL Node p : IB + IF – IE = 0 Node q : IC – IA – IB = 0

2.5A 2Ω 5V

2Ω +

-

i1 i2

i3 i4 i5

i6

f g

m n p

a q b

IF RF

IE

IA

IC

IB

ID

RA RB

RC

RD RE EA EB

+

+ +

+

+

+

+

+

-

- -

- -

- - -

5

Loop amnqa: - IARA + EA – IDRD – ICRC = 0

Loop bpnqb:

- IB RB + EB – IERE – ICRC = 0 KVL Loop amfgpbqa:

- IARA + EA – IFRF – EB + IBRB = 0 Mesh Analysis (Mesh Current Method) Procedure for writing Mesh Equations using KVL

1. Select a set of meshes such that at least one mesh passes through each branch 2. Assign and label mesh currents. 3. Apply KVL around each mesh. 4. Solve resulting equations for unknown mesh currents.

Example 1

I1, I2 and I3 are assigned mesh currents for loops I, II and III. Loop I : - I1RA + EA – (I1 – I3)RD – (I1 – I2)RC = 0 Loop II : -(I2 – I1)RC – (I2 – I3)RE – EB – I2RB = 0 Loop III : - I3RF – (I3 – I2)RE – (I3 – I1)RD = 0

Branch currents: IA = I1; IB = - I2; IC = (I1 – I2); ID = (I1 – I3); IE = (I3 – I2); IF = I3.

IF RF

IE

IA

IC

IB

ID

RA RB

RC

RD

RE EA EB + +

- - I II

III

I1 I2

I3

6

Example 2

- 4 (I1 – 1) – 1(I1 – I2) – 17 = 0 - 3 (I2 – 1) – 2I2 + 17 – 1(I2 – I1) = 0 Solving we get I1 = -2A I2 = 3A

Example 3 - 2I1 – 6(I1 – I2) – 10(I1 – I3) +1 = 0 - 4I2 – 7 – 8(I2 – I3) – 6(I2 – I1) = 0 - 10(I3 – I1) – 8(I3 – I2) – 1I3 + 15 = 0

As the resistances are given in kilo ohm, the currents are in milli – amp. Solving we get I1 = 1.5 mA; I2 = 1 mA; I3 = 2 mA.

Example 4

4Ω

1A

-

I1 I2

1Ω

3Ω

2Ω +

17V

4kΩ

-

I1 I2

1kΩ

6kΩ

2kΩ

+

7V

8kΩ 10kΩ

+

-

- +

1V

15V I3

R2 = 1Ω

2A -

I1 I2

+

R3 = 4Ω

R4 = 2Ω

a b c

g

I3

I II

III

v1

R1 = 3Ω

7

Find Iab and Vcg using mesh current method. Loop I : 3 = 3I1 – 2I2 – I3 ……………..(i) Loop II : v1 = - 2I1 + 6I2 – 4I3 …………(ii) Loop III : 0 = - I1 – 4I2 + 8I3 …………....(iii)

Solving equations (i) and (iii) we get 148I = - = 0.6956 A69

and

325I = - = - 1.0869 A23

.

Iab = I1 – I3 = 0.39 A. Ibc = I2 – I3 = - 0.913 A. and Ibg = I1 – I2 = 1.304 A.

From equation (ii) v1 = 6.27 V.

------- × ------

No. of mesh equations: No. of branches – No. of nodes + 1

8

How to write the Mesh Equations?

1. In each loop, assume a current direction. 2. Indicate the polarity of voltage drops or rise across each resistor in the loop. 3. Determine the algebraic sum of the voltage around the loop and set it equal to zero. 4. If a resistor has two or more loop current flowing through it, the total current through

the resistor is the loop current of interest plus or minus the other currents dependent on direction, plus if in the same direction, minus if opposite.

5. The polarity of all voltage sources is independent of current direction. Example: Mesh Equation Method in Matrix form: A B 1 B 2 a(R + R )I - R I + 0 = V B 1 B C D 2 D 3-R I + (R + R + R )I - R I = 0 D 2 D E 3 b0 - R I + (R + R )I = -V Placing the equations in matrix form,

A B B 1 a

B B C D D 2

D D E 3 b

R + R - R 0 I V- R R + R + R - R I 0

0 - R R + R I -V

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

The elements of the matrixes can be indicated in general form as follows:

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

R R R I VR R R I VR R R I V

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

How to get the elements for a particular problem R11 (Row 1, Column 1): Sum of all resistances through which mesh current I1 passes. In

the above example it is (RA + RB). Similarly, R22 and R33 are the sums of all resistances through which I2 and I3 respectively pass. R12 (Row 1, Column 2): Sum of all resistances through which mesh currents I1 and I2

pass. The sign of R12 is +ve, if the two currents are in the same direction through each resistance and –ve, if they are in opposite directions. (e.g. – RB)

RA +

- Va Vb

I1 I2 I3 RB

RC

RD

RE +

-

9

In a similar manner one can write for R21, R23, R13 and R31. It should be noted that for all i and j, Rij = Rji. As a result, the resistance matrix is symmetric about the principle diagonal. The current vector requires no explanation. Element V1 in the voltage vector is the sum of all source voltages driving mesh current I1. A voltage is counted +ve in the sum if I1 passes from the –ve to the +ve terminal of the source; otherwise it is counted –ve. In the above example mesh 1 has a source Va driving the current in the direction of I1; mesh 2 has no source; and mesh 3 has a source Vb driving opposite to the direction of I3, making V3 negative. Nodal Analysis (Node – Voltage Method) Procedure of writing nodal equations:

1. Select a reference node and ground it (assume zero potential). 2. Label V1, V2, V3 etc. for each unknown node voltage. 3. Assign current directions. 4. Apply KCL at each node. Use Ohm’s Law. 5. Solve resulting equations to find unknown node voltages. 6. Obtain the solution for currents.

Example 1 For node 1 : I1 – I2 – IA = 0 or, I1 = I2 + IA

i.e. A 1 1 2 1

1 2 A

E - V V - V V - 0= +R R R

or 2 A1

A 1 2 2 1

V E1 1 1V - =R R R R R⎡ ⎤

+ +⎢ ⎥⎣ ⎦

For Node 2: I2 + I3 – IB = 0 or I2 + I3 = IB

i.e. 1 2 B 2 2

2 3 B

V - V E - V V - 0+ =R R R

or 1 B2

B 2 3 2 3

V E1 1 1V - =R R R R R⎡ ⎤

+ +⎢ ⎥⎣ ⎦

.

RA

+

- Ea Eb

I1 I2 I3

RB

R1 R2 R3 +

- IA IB

(Zero potential)

1 2

3

10

Number of equations to be solved = 2. Example 2 Assuming the potentials of a, q and b are at zero.

pn Am

A D F D F A

VV E1 1 1V + + - - - = 0R R R R R R⎡ ⎤⎢ ⎥⎣ ⎦

pm n

C D E D E

VV1 1 1V + + - - = 0R R R R R⎡ ⎤⎢ ⎥⎣ ⎦

m n Bp

B E F F E B

V V E1 1 1V + + - - - = 0R R R R R R⎡ ⎤⎢ ⎥⎣ ⎦

Example 3

Node 1 : C B AA

A 1 2 5 2 5 1

V V E1 1 1 1V + + + - - - = 0R R R R R R R⎡ ⎤⎢ ⎥⎣ ⎦

Node 2 : B AC

C 2 3 3 2

V V1 1 1V + + - - = 0R R R R R⎡ ⎤⎢ ⎥⎣ ⎦

Node 3 : C A BB

B 3 4 5 3 5 4

V V E1 1 1 1V + + + - - - = 0R R R R R R R⎡ ⎤⎢ ⎥⎣ ⎦

.

RF

RA RB

RC

RD RE EA EB

+ +

- -

m

I1

a q b

n p

f g

RA

+

- Ea Eb

I1 I2

I3

RB RC

R1 +

-

R2 R3 R4

R5

IA IB IC

I4 I5

2 1 3

0

11

How to write Nodal Equations by Inspection

1. First, write the product of node potential of a particular node and sum of the reciprocal of the branching resistances from that node.

2. Subtract the ratio of adjacent node potentials and the connecting resistance. 3. Subtract the ratio of the adjacent source voltages and the connecting resistances. 4. Lastly, all set equal to zero.

Numerical Example: 1 2I = 4 + I

1 112 - V V - 0= 4 +1 3

Solving V1 = 6V, I1 = 6A, I2 = 2A 12VP = 12v×6A = 72W (Power delivering) 2

1Ω 1P = I ×1Ω = 36W (Power dissipating) 2

3Ω 2P = I ×3Ω = 12W (Power dissipating) 4A 1P = V × 4A = 24W (Power absorbing) 12V 1Ω 3Ω 4AP = P + P + P (Energy is conserved). Node voltage method in matrix form: Example

1 a 1 1 2

A B C

V - V V V - V+ + = 0R R R

2 b2 1 2

C D E

V - VV - V V+ + = 0R R R

I1

12V

V1

+

-

1Ω

3Ω 4A

I2

1

RA +

- Va Vb RB

RC

RD

RE +

-

1 2 4 5

3 (ref)

0

12

In matrix form:

a1

A B C C A

b

2C C D E E

V1 1 1 1 V+ + -R R R R R

=1 1 1 1 V- + + VR R R R R

⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

Note the symmetry of the coefficient matrix. The 1, 1 – element contains the sum of the reciprocals of all resistances connected to node 1; the 2, 2 – element contains sum of the reciprocals of all resistances connected to node 2. The 1, 2 – and 2, 1 – elements are each equal to the negative of the sum of reciprocals of the resistances of all branches joining nodes 1 and 2. (There is just one such branch in the present problem). On the right-hand side, the current matrix contains Va/RA and Vb/RE, the driving currents. Both these terms are taken positive because they both drive a current into a node. Example Find Iab and Vcg using Node Voltage Method Note Va = 3V For Node a:

a b c1 1i = 1+ V - V - V3 3

⎛ ⎞⎜ ⎟⎝ ⎠

……..............(i)

For Node b:

a b c1 1 10 = -V + 1+ + V - V2 4 4

⎛ ⎞⎜ ⎟⎝ ⎠

……….(ii)

For Node c:

a b c1 1 1 12 = - V - V + V3 4 3 4

⎛ ⎞+ ⎜ ⎟⎝ ⎠

………..(iii)

Solving (i), (ii) and (iii) we get Vb = 2.261 V, Vc = 6.26 V.

and a bab

V - V 3- 2.61I = = = 0.39A1 1

.

No. of nodal equations = No. of nodes – 1.

1Ω

2A i

Iab

4Ω

2Ω

a b c

g

v1 = 3V

3Ω

13

Numerical Example: Find the value of the current through 8 Ω resistor. Application of Kirchoff’s laws:

Current Law: I1 = I2 + I4 at A ………… (i) I2 + I3 = I5 at B ……........ (ii) Voltage Law: Loop ADEFA: - 15I4 + 4 – 5I1 = 0 … ……... (iii) Loop ABCDA: - 10I2 – 8I5 + 15I4 = 0………. (iv) Loop BCGHB: - 8I5 + 6 – 12I3 = 0…………. (v) From (i) and (iii) 4 – 5I2 – 20I4 = 0 …………... (vi) From (ii) and (v) 6 – 20I5 + 12I2 = 0………….. (vii) Arranging (iv), (vi) and (vii) - 10I2 + 15I4 – 8I5 = 0 - 5I2 – 20I4 = - 4 + 12I2 – 20I5 = - 6

5

-10 15 0-5 -20 -4

-12 0 -6I = = 0.32 A

-10 15 -8-5 -20 0

+12 0 +20

15Ω

+

- 4V

6V

I1 I2 I3

+

- I4 I5

F A

8Ω

5Ω 10Ω 12Ω

B H

E D C G

14

Mesh Analysis: Assign three loops or mesh currents I1, I2 and I3 with arbitrary directions.

Loop ADEFA: 15(I1 – I2) + 4 + 5I1 = 0 ………......... (i) Loop ABCDA: 10I2 + 8(I2 – I3) + 15(I2 – I1) = 0…… (ii) Loop BCGHB: - 8(I3 – I2) + 6 – 12I3 = 0…………… (iii)

Rearranging 20 I1 – 15 I2 = - 4 - 15 I1 + 33 I2 – 8 I3 = 0 8 I2 – 20 I3 = - 6 Solving for I2 and I3 the current through 8 Ω resistors can be obtained.

2 3240 2130I = - and I = -

7420 7420

∴ I8Ω = (I3 – I2) in the direction B to C = 0.32 A. Nodal Voltage Method: Assume the voltage of E, D, C or G is zero. Voltage of node A is VA. Voltage of node B is VB. For node A

BA

V1 1 1 4V + + - =15 5 10 10 5⎡ ⎤⎢ ⎥⎣ ⎦

……………………….. (i)

For node B

15Ω

+

- 4V

6V

I1 I2 I3

+

-

F A

8Ω

5Ω

10Ω

12Ω B H

E D C G

15Ω

+

- 4V 6V

+

-

F A

8Ω

5Ω 10Ω 12Ω

B H

E D C G

VA VB

Zero

15

AB

V1 1 1 6V + + - =8 10 12 10 12⎡ ⎤⎢ ⎥⎣ ⎦

……………………… (ii)

Simplifying and rearranging 11 VA – 3 VB = 24 - 12 VA + 37 VB = 60

Solving B79×12V =

371

So current through 8 Ω resistor

BVI = = 0.32 A8

16

Delta (Δ) – Star (Y) Conversion Star (Y) – Delta (Δ) Conversion

Given/Wanted Wanted/Given Equivalence based on the effective resistance between terminals 1, 2 and 3. Resistance between terminals 1 and 2 Star connection: R1 + R2

Delta connection: ( )12 23 31

12 23 31

R R + RR + R + R

For equivalence

R1 + R2 = ( )12 23 31

12 23 31

R R + RR + R + R

………………………….(1)

Similarly for terminals 2 and 3 and 3 and 1

R2 + R3 = ( )23 31 12

12 23 31

R R + RR + R + R

………………………….(2)

R3 + R1 = ( )31 12 23

12 23 31

R R + RR + R + R

………………………….(3)

Adding R1 + R2 + R3 = 12 23 23 31 31 12

12 23 31

R R + R R + R RR + R + R

…………(4)

Subtracting (2), (3) and (1) from (4) in sequence

31 121

12 23 31

R RRR + R + R

= ………………………….(5)

12 232

12 23 31

R RRR + R + R

= ………………………….(6) Δ – Y Conversion

23 313

12 23 31

R RRR + R + R

= ………………………….(7)

Denoting R12 + R23 + R31 = R and forming three possible products from (5), (6) and (7)

3

1

2 3

R1

R2

R3

1

2

R31 R12

R23

Y – Δ

Δ – Y

17

212 23 31

1 2 2

R R RR RR

= ……………………………(8)

2

12 23 312 3 2

R R RR RR

= ……………………………(9)

2

12 23 313 1 2

R R RR RR

= ……………………………(10)

Adding (8), (9) and (10)

12 23 31 12 23 311 2 2 3 3 1 2

R R R (R + R + R )R R + R R + R RR

=

12 23 31R R RR

= …………..(11)

from (7), (5), (6) and (11) we obtain the following relations:

1 2 2 3 3 1 1 212 1 2

3 3

R R + R R + R R R RR R + R +R R

= = ………….(12)

Δ – Y Conversion 1 2 2 3 3 1 2 323 2 3

1 1

R R + R R + R R R RR R + R +R R

= = …………(13)

1 2 2 3 3 1 3 131 3 1

2 2

R R + R R + R R R RR R + R +R R

= = ………….(14)

How to Remember?

i. Δ → Y: To find one leg of the Y, multiply the two adjacent arms of the Δ and divide by sum of the three resistances forming there.

ii. Y → Δ: To find any side of the Δ, and the two resistances (between the extremities of which this arm of Δ exists) and also add the quotient obtained by the resistance of the third leg of the Y.

Example

A

I

a

5Ω

5V

B

A' B'

C'

5Ω

5Ω

3.25Ω

1.5Ω 2.5Ω

1Ω

4.7Ω 4.5Ω

b

c

+ -

18

Find the value of the current I

a1 1.5 1.5R 0.3

1.5 +1+ 2.5 5×

= Ω = Ω = Ω

b1 2.5 2.5R 0.5

5 5×

= Ω = Ω = Ω

c1.5 2.5R 0.75

5×

= Ω = Ω

A OR (4.7 0.3) 5′ = + Ω = Ω B OR (4.5 0.5) 5′ = + Ω = Ω C OR (3.25 0.75) 4′ = + Ω = Ω

a Ra

2.5Ω 1.5Ω

1Ω

Rb

Rc

b

c

A'

0.75Ω 5Ω

0

B'

C'

5Ω

5Ω

3.25Ω

4.7Ω

4.5Ω 0.5Ω 0.3Ω

B'

16.25Ω

13Ω

5Ω

C'

A'

5Ω

5Ω

13Ω

19

A BR 3.83′ ′ = Ω B CR 3.62′ ′ = Ω C AR 3.62′ ′ = Ω

B C5 4R 5 4 13

5′ ′×⎛ ⎞= + + Ω = Ω⎜ ⎟

⎝ ⎠

C A4 5R 4 5 13

5′ ′×⎛ ⎞= + + Ω = Ω⎜ ⎟

⎝ ⎠

A B5 5R 5 5 16.25

4′ ′×⎛ ⎞= + + Ω = Ω⎜ ⎟

⎝ ⎠

A'

5Ω 5Ω

4Ω

B'

C'

1 A +

3.62Ω

-

B

A' B'

C'

3.62Ω

3.83Ω

5V

I A +

7.24 Ω

-

B

A' B' 3.83 Ω

5V

I A + -

B

A' B' 2.5 Ω

5V

20

Example Find the value of the input resistance Rin of the circuit from Fig. (i)

ab36 12R 36 12 64.61

26×

= + + = Ω

ao36 26R 26 36 140

12×

= + + = Ω

bo26 12R 26 12 44.66

36×

= + + = Ω

From Fig. (ii)

ac30 15R 30 15 62.31

26×

= + + = Ω

co15 26R 15 26 54.00

30×

= + + = Ω

ao26 30R 26 30 108

15×

= + + = Ω

a

26Ω

15Ω

30Ω

Rin

b c

26Ω

26Ω

36Ω

12Ω

6Ω 0.6Ω

a

12

0

Rab

b

36

26

Ra0

Rb0

Fig. (i)

a

15

0

Rac

c

30

26

Ra0

Rc0

Fig. (ii)

21

From Fig (iii)

bc6 0.6R 6 0.6 6.738

26×

= + + = Ω

bo6 26R 6 26 292

0.6×

= + + = Ω

co0.6 26R 0.6 26 29.2

6×

= + + = Ω

acR 42.94= Ω bcR 5.85= Ω abR 32.11= Ω inR 19.37= Ω

0.6Ω

0

Rbc

c

6Ω

26Ω Rb0 Rc0

Fig. (iii)

b

a

62.31Ω 64.61Ω

6.74Ω b c

60.96Ω

18.95Ω

19.84Ω

Rin

c

44.95

63.82

64.61

a

b

62.31

6.74

138.14

Rin

22

Superposition Theorem: In a network of linear and bilateral resistors that is energized by two or more sources, (a) The current in any resistor or (b) The voltage across any resistor is equal to (a) the algebraic sum of separate currents in the resistor, or (b) the voltages across the resistor, assuming that each sources acting independently of the others, is applied separately in turn, while the others are replaced by their respective internal values of resistance. Note: In case of ideal sources: (i) voltage source to be replaced by a short circuit and (ii) current source by an open circuit. Illustration:

A A 2 B B 1I = I - I ; I = I - I ; I = I + I′ ′ ′ ′ ′ ′′ Where R1 and R2 are the internal resistances of sources. Example 1 Find I

I = I1 + I2

120I = = 2.5A

(5 + 3)↓ and 2

5I = 5 = 3.125A(3+ 5)

↓

∴ I = (2.5 + 3.125)A = 5.625A ↓ In Fig. (a) Both sources acting. In Fig. (b) Voltage source acting only In Fig (c) Current source acting only

IA EA

R1 I

R

IB EB

R2

′AI

I' R1

R R2

′1I R1

I''

R R2

′BI

′2I

(a)

+

- 20V

5Ω

3Ω 5A

10Ω

I

(b)

20V

5Ω

3Ω

10Ω

I1

(c)

5Ω

3Ω 5A

10Ω

I2

23

Example 2

Compute the current in 23 Ω resistor When the voltage source (i.e. 200 V source) acting alone:

eq27(4 + 23)R 47 60.5

54= + = Ω

T200I 3.31A60.5

= =

2327I 3.31 1.65A.54Ω′ = × = ↓

When the current source (i.e. 20-A source) acting alone

eq(27)(47)R 4 21.15

74= + = Ω

2321.15I (20) 9.58A.

(21.15 + 23)Ω′′ = =

Therefore, total current through 23 Ω resistor 23 23 23I I + I (1.65 9.58)A 11.23A.Ω Ω Ω′ ′′= = + = In the downward direction.

4Ω

20A +

-

27Ω 200V

23Ω 47Ω

a.c. +

-

27Ω 200V

′23ΩI 4Ω 47Ω

23Ω

20A 27Ω S.C.

23Ω

4Ω 47Ω

′′23ΩI

24

Example Find Iab and Vcg using Superposition Theorem. First consider the voltage source only:

eq ac cb ab bg7 23R (R R ) R R 28 8

⎛ ⎞= + + = + Ω = Ω⎜ ⎟⎝ ⎠

3I = = 1.043 A238

ab7I = 1.043× = 0.913 A8

acbI = 1.043 - 0.913 = 0.13 A

cg cb bgV = V + V = (4 × 0.13 + 2 × 1.043)V = 2.61 V. Note: ‘c’ is at a higher potential than ‘g’. Next consider only the current source:

Current through 3 Ω resistor = 14/3 2A = 1.217A(14/3) + 3

.

1Ω

2A 3V

4Ω

2Ω

a b c

g

3Ω

+

-

g

1Ω

3V

4Ω

2Ω

a b c

3Ω

+

-

I

25

Through 4 Ω resistor = (2 – 1.217)A = 0.783 A.

Through 1 Ω resistor (b to a) = 2 ×0.783A = 0.522A3

.

Voltage across 3 Ω resistor c-g terminal) Vcg = 1.217 × 3 V = 3.651 V. Note: ‘c’ is at a higher potential than ‘g’. Total current (due to both sources) flowing through 1 Ω resistor (from a to b) i.e. Iab = (0.913 – 0.522) A = 0.391 A. Total voltage (due to both sources) across 3 Ω resistor i.e. Vcg = (2.61 + 3.651)V = 6.261 V. Note: ‘c’ is at higher potential. Limitations of application of Superposition Theorem:

• Not applicable for power calculation, as power is the product of voltage and current or proportional to the square of current or voltages which are not linear operations.

• Not applicable to circuits containing non linear elements such as Diodes, Transistors, Thermistors etc.

Remark: For a network containing a large number of independent sources the method

becomes tedious. Thevenin’s Theorem: The current through a resistance RL connected across any two terminals A and B of an active network is obtained by dividing the potential difference between A and B with RL disconnected (VTH), by (RTH + RL), where RTH is the resistance of the network measured between points A and B with RL disconnected and all sources are replaced by their respective internal resistances. Note: Ideal voltage source is replaced by S.C. and Ideal current source is replaced by O.C.

1Ω

2A

4Ω

2Ω

a b c

g

3Ω

1Ω 2A

4Ω

2Ω

(a, g)

b c

g

3Ω

Iba

26

Explanation:

A linear, active network which contains one or more voltage or current sources can be replaced by a single voltage source can be replaced by a single voltage source and a series resistance. The voltage of the equivalent source is called Thevenin equivalent voltage and the series resistance is called Thevenin equivalent resistance. Illustration: To find VTH:

• Disconnect RL • Find voltage between A and B

2TH

1 2

RV = ER + R

To find RTH:

• Replace the source by its internal resistance and look back into the network from A and B with RL disconnected.

LINEAR ACTIVE

NETWORK

A

B

IL RL IL = ?

LINEAR ACTIVE

NETWORK

A

B

VTH V

ORIGINAL CIRCUIT

LINEAR NETWORK WITHOUT SOURCES

A

B

RTH

RESISTANCE SEEN

THROUGH A & B

VTH RL

+

-

RTH

IL

A

B

+

-

A

R1

E

R3

R2 RL

B

27

1 2TH 3

1 2

R RR = RR + R

+

Note: Here source of emf is shorted.

To obtain load current IL

• Draw Thevenin’s equivalent circuit between A and B i.e. an equivalent source VTH and an equivalent resistance RTH in series.

• Connect RL across A and B.

• Find current THL

TH L

VR =R + R

Example

To get VTH

( ) ( )TH

8V = 412 + 4

= 2V

A

R1 R3

R2

B

A

RTH

B

VTH

+

-

+

-

A 12Ω

8V RL

B

4Ω

4Ω 2Ω

6Ω

IL

IL = ?

+

-

A 12Ω

8V

B

4Ω

4Ω

6Ω

VTH

28

To get RTH

( )( )( )TH

12 4R = 4 + 6 13

12 + 4+ = Ω

Thevenin’s equivalent circuit →

L2I = A

13 + 2

= 0.133A Proof of Thevenin’s Theorem:

A

12Ω 4Ω

4Ω

B

A

13Ω

2V

B

2Ω

Active Linear Bilateral Network

A

B

IL

RL

Reading of Ammeter is IL

A

Active Linear Bilateral Network

A

B

Reading of Voltmeter is VTH (say) V

Active Linear Bilateral Network

A

B

IL = 0

RL

Reading of Ammeter is zero

A

VTH

29

Now apply superposition principle:

So the equivalent resistance of the network is Req which is nothing but RTH. So we can replace the complicated active linear bilateral network by an equivalent simple network containing a simple voltage source (VTH) and a series resistance RTH to produce the same current through RL when connects to the points A and B.

THL

TH L

VI =R + R

Ex. Find Iab using Thevenin’s theorem and hence calculate Vcg. To find VTh: Step 1: Disconnect the load i.e, 1Ω and circuit is as shown in Fig. 2 Step 2: Apply any method (say Node Voltage Method) at Node C: 1 22 I I 0− − =

c cV 3 V 02 03 6− −

− − =

c 1 2V 6v, I = 1A & I = 1A⇒ = ( )Th ag bgV V V 3 2 v 1v.= − = − =

Linear Bilateral Network with all sources replaced by their internal resistances

IL

RL

VTH

Active Linear Network

A

B

IL

RL

RTH

IL

RL VTH

1Ω

2A 3V

4Ω

2Ω

a b c

g

3Ω

+

-

Fig. 1

30

To find RTh: Step 3: Replace all sources by their internal resistances and look into the network from a and b. (Fig. 3).

Th2 7R 1.5552 7×

= = Ω+

To find IL i.e, Iab: Step 4: Draw Thevenin’s equivalent circuit as in Fig. 5

ab1I A 0.39A

1.555 1= =

+

VTH

2A 3V

4Ω

2Ω

a b c

g (zero potential)

3Ω

+

-

Fig. 2

I1

I2

4Ω

2Ω

a b c

3Ω

Fig. 3

4Ω a b

c

3Ω

Fig. 4

2Ω

1.555Ω

a

b

Iab 1Ω

Fig. 5

RTH

VTH 1V

+

-

31

To compute Vcg ( )bg ag abV V V 3 1 0.39 v 2.61v= − = − × =

( )bgI 2.61 2 A 1.305A;= =

( )cbI 1.305 0.39 0.915A= − =

( )cgV 4 0.915 2 1.305 v 6.27v= × + × =

Alternative Statement of Thevenin’s Theorem: In any two terminal network of linear resistors and independent sources, the current in a load resistor connected to the output terminals is equal to the current that would exist in the same resistor if it were connected in series with

(a) A simple source of EMF whose voltage is measured at the open circuited network terminals and

(b) A simple resistance whose magnitude is that of the network looking back from the two terminals into the network with all sources replaced by their internal resistances.

Norton’s Theorem It states that any two terminals A & B of a network composed of linear bilateral resistances and independent sources may be replaced by an equivalent current source and a parallel resistance. The current magnitude of the source is the current measured in the short circuit placed across the terminals A & B. the parallel resistance is the equivalent resistance looking back into the network through the poins A & B with all independent sources have been replaced by their internal resistances.

N

L scN L

RI IR R

= ⋅+

1Ω

2A 3V

4Ω

2Ω

a b c

g

3Ω

+

-

Fig. 6

LINEAR ACTIVE

NETWORK

A

B

IL RL

ISC RL

IL

A

B

RN

32

It is clear that the equivalent parallel resistance RN has got the same value as that of RTh i.e, RN = RTh. Norton’s equivalent circuit is dual to thevenin’s equivalent circuit.

The current magnitude of current source will be obtained by short circuiting A & B terminals.

So THsc

TH

VIR

= . Therefore,

N TH THL sc

N L TH TH L

R V RI IR R R R R

= ⋅ = ⋅+ +

TH

TH L

VR R

=+

The above result is obtained directly from thevenin’s theorem.

THTH N

sc

VR RI

= =

Illustration: To find Isc:

• Disconnect RL • Short circuit A & B terminals • Find current through short circuited path

2sc

2 3 2 31

2 3

REI R R R RRR R

= ⋅++

+

LINEAR ACTIVE

NETWORK

A

B

ISC

A

B

RTH = RN

VTH

A

B

RTH +

-

LINEAR ACTIVE

NETWORK

HIGH RESISTANCE VOLTMETER GIVES VTH

V

LINEAR ACTIVE

NETWORK

LOW RESISTANCE AMMETER GIVES ISC

A

+

- E

R1 R3

RL IL

R2

A

B

33

2

1 2 2 3 3 1

ERR R R R R R

=+ +

To find RN (or RTN):

1 2N TH 3

1 2

R RR R RR R

= = ++

1 2 2 3 3 1

1 2

R R R R R RR R+ +

=+

Note: Here source of EMFE is shorted. To obtain load current IL:

THL sc

TH L

RI IR R

= ⋅+

Example

+

- E

R1 R3 ISC R2

A

B

R1 R3 RN (or RTh) R2

A

B

RTh IL

RL

A

B

ISC

A

B

IL 4Ω

4Ω

12Ω

2Ω RL

6Ω

+

- 8V

34

To get Isc:

sc8 4I 4 10 1412

14

⎛ ⎞= ⎜ ⎟× ⎝ ⎠+

2 A.13

=

To get RN or RTH:

TH12 4R 4 6 13 .12 4

×= + + = Ω

+

To get IL:

L2 13I A

13 13 2= ⋅

+

0.133A.=

Ex: Find Iab and Vcg using Norton’s Theorem.

To find the Norton’s source current. Step 1: Remove the 1Ω resistor connection between a and b and short the terminals a & b. Find IN by any method. Let us apply Nodal voltage method taking g is at zero potential.

A

B

ISC 4Ω

4Ω

12Ω

6Ω

8V

A

B

4Ω 4Ω

12Ω

6Ω

IL

RL 2Ω 13Ω

213

A

35

At Node C: bc ca2 I I+ =

i.e, b c c aV V V V24 3− −

+ =

Note here b aV V 3V= = Which gives c cgV V 6.428V= =

At Node b:

( )N bc bg

3 6.428Vb Vc 3 3I I I4 2 4 2

−−= + = + = +

0.643A= − This indicates that Norton’s current direction is opposite to the assigned short circuit current (i.e, from b to a). The equivalent resistance between a and b which all sources are replaced by their equivalent internal resistance is RN = RTh = 1.555Ω. Therefore Norton’s equivalent circuit becomes

( )L

1.555I 0.643 A1.555 1

= ⋅+

( )0.39A. a to b=

1Ω

2A 3V

4Ω

2Ω

a b c

g

3Ω

+

-

Fig. 1

2A

3V

4Ω

2Ω

a b c

g

3Ω

Ica

Ibc

Ibg

IN

0.643A 1.555Ω 1Ω

IL

a

b

36

Ex. Obtain Thevenin’s & Norton’s Equivalent Circuits between Points A & B.

TH6V 12 9V

2 6= ⋅ = =

+ Open circuit voltage across A & B.

Isc = Sort circuit current when A & B shorted

12 6 0.67mA6 12 6 1226 12

= − =× +++

TH N2 6R R 12 K2 6×⎛ ⎞= = + Ω⎜ ⎟+⎝ ⎠

13.5K= Ω

Thevenin’s Equivalent Circuit

Norton’s Equivalent Circuit

A

2KΩ

12KΩ

6KΩ

B

12V

+

-

A 2KΩ

12KΩ

6KΩ

B

RTH = RN

9V

A

B

13.5 KΩ +

-

0.67 mA

A

B

13.5 KΩ

37

Find Norton’s Thevenin’s Equivalent Circuit Ex. For RTH:

( )( )TH

3 5 1R 2

3 5 1× +

= Ω = Ω+ +

For Isc:

Norton’s Equivalent Circuit

Thevenin’s Equivalent Circuit

RTH

A

B

5Ω 3Ω

1Ω

9A

A

B

5Ω

9V

3Ω

1Ω

+ -

45V

A

B

5Ω

9V

3Ω

1Ω

+ -

+

-

6A

A

B

2Ω

12V

A

B

2Ω +

-

38

Converting the Current Source to A Voltage Source

Short circuiting A & B source voltage = (45-9)v =36v total resistance = (5+1)Ω = 6Ω. Note: 3Ω resistance will be ineffective. scI 36 6 A = 6A=

Practical Verification of (i) Superposition theorem (ii) Thvenin’s theorem

Ex.

36V

A

B

6Ω +

-

ISC

220V D.C.

I1

A

R1

+

-

A

A

R2

R3 I2

I3

220V D.C.

+

-

220V D.C.

R1

+

-

A

R2

R3

RL

110V D.C.

+

-

V

+

-

12Ω

4V

5Ω 10Ω

8Ω 15Ω

+

- 6V

39

Find current through 8Ω resistor I8Ω by applying

(i) Super position theorem (ii) Thevenin’s theorem (iii) Norton’s theorem

i) Superposition Theorem: (a) Replacing 6V battery by its internal resistance

112 8 24R12 8 5

×= = Ω

+

2 174R 10 R5

= + = Ω

23

2

R 15R 7.45R 15

×= = Ω

+

'1

4I 0.321A;7.45 5

= =+

' '2 1

2

15I I ;15 R

= ⋅+

' '3 2

8I I ;20

=

' ' 24 1

2

RI I ;15 R

= ⋅+

' '5 2

12I I20

=

' ' ' '2 1 5 2

12I 0.503 I hence I I 0.097A20

= = = in the

A B

C D

I1' I2' I3'

I4' I5'

A B

CD

I1' I2'

I4'

R1

A

D

I1'

I2'I4'

R2 = 10 + R1

A

D

I1'R3

40

Direction From B to C. b) Replacing 4V battery by its internal resistance

415.5R 3.75

15 5= = Ω

+

'' '' ''1 2 2

15I I 0.75 I15 5

= =+

'' ''4 2I 0.25 I=

5 5R R 10 13.75= + = Ω

'' '' ''2 3 3

5

8I I 0.365 IR 8

= =+

'' '' ''5 3 3

13.75I I 0.635 I21.75

= =

56

5

R 8 13.75 8R 5.06R 8 21.75

⋅ ×= = = Ω

+

'' ''3 5

6

6I 0.353A, I 0.223AR 12

= = =+

in the direction from B to C.

Hence current through 8Ω resistance ( )' ''5 5 5I I I 0.097 0.223 A= = + = +

0.32A= The direction of current is from B to C. ii) Thevenin’s Theorem: Disconnecting 8Ω resistance from the network ( )1 1 24 -5I -15 I - I = 0

I1'' I2'' I3''

I4'' I5''

R4

I2'' I3''

I5''

R5

I2''

I3''

I5''

R6

I3''

41

( )2 2 1-22 I - 6 -15 I - I = 0 Solving 2 1I = -0.116A & I = 0.113A

TH 2V = 6 +12 I ( )= 6 -12×0.116 v = 4.608v

TH

5×1510 + 1220R = = 6.42Ω

5×1510 + +1220

⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠

The current through 8Ω resistor

TH8Ω

TH

V 4.608I = = AR +8 6.42 +8

= 0.32A From B to C.

iii) Norton’s Theorem: ( )1 1 24 -5 I -15 I - I = 0 ............ (1)

( )2 2 1-10 I -15 I - I = 0 ............. (2) 3-12 I - 6 = 0 .......................... (3)

15Ω 4V 6V

I1 I2

A

5Ω 10Ω 12Ω

B

D C

15Ω

I2

A

5Ω 10Ω

12Ω B

D C

15Ω

5Ω 10Ω

12Ω

B C 4.608V

B

C

12Ω

+

-

8Ω

6.42A

42

SC 2 6I = I - I

From(3) 3I = -0.5A

From (2) 2 1 1 215 5I = I I = I25 3

⇒

From (1) 2 254 - 20. I 15 I = 03

+

24×3I = A = 0.218A55

( )SCI = 0.218 + 0.5 A = 0.718A

8Ω6.42I = 0.718× = 0.32A

6.42 + 8

From B to C.

Maximum Power Transfer Theorem

I1 I2

A B

D C

I3

ISC

0.718 A

B

C

8Ω 6.42Ω

r

E RL = r RL

3Ω

4A

14V RL

2Ω

5Ω

5V +

-

+ -

A

B

43

Maximum Power Transfer Theorem In D.C. Circuit: The electrical power transferred to an adjustable resistor RL is maximum, when RL is equal to the thevenin’s (or Norton’s) resistance RTH (or RN) between the two terminals across which RL is connected. Proof: According to thevenin’s theorem the equivalent circuit of any active linear circuit across any two points A & B can be represented as below:

THL

TH L

VI ................................... (1)R R

=+

2

2 THL L L L

TH L

VP I R R ............(2)R R

⎛ ⎞= = ⋅⎜ ⎟+⎝ ⎠

for maximum or minimum value of PL for adjustable RL

dPL ...................................... (3)dRL

( )( )TH L2

L TH L

d 1i.e. V 2 R 0dR R R

⎧ ⎫⎡ ⎤⎪ ⎪⋅ ⋅ =⎢ ⎥⎨ ⎬+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

RL = RTh

RTh

RL VTh

B

A

+

-

RTh = RN RL Ise RL = RN = RTh

A

RTH

RL

B

IL

VTh

+

-

44

( ) ( )( )

2TH L TH L L

TH L L4TH L

R +R 2 R +R Ri.e. 0 R R 2R

R +R−

= ⇒ + =

therefore, L THR R .......................... (4)=

Now 2 2

L TH2L TH

d P V 0 dR 8R

= − < ⇐ condition for maximum power

L THR R ............................... (5)= So when RL = RTH, the power transferred to RL is maximum.

max

2 2 22 TH TH TH

L L L TH2TH TH L

V V V1 1P I R R ........... (6)4R 4 R 4 R

= = ⋅ = ⋅ = ⋅

Total power delivered by the source

( )2 2T L TH L L LP I R R 2I R ........................................ (7)= + =

So the maximum power delivered to load is 50% of the total power drawn from the source. Remark:-

1) The power transferred to load resistance vs the load resistance plot is shown in Fig. 1.

2) Operating at the maximum power transfer condition is meaningful only in low-power electronic circuits, such as communication and instrumentation systems, where it is desirable for the signal source to transmit as much power as possible to the receiver or load.

3) In power utility systems, however, this theorem has limited applications as operating the network at maximum power transfer condition has only 50% transmission efficiency and a large voltage drop and also large power loss in the transmission lines.

PL max

RTh RL

PL

Fig. 1

45

Analysis of Electric Circuits in presence of one Non-linear (N.L) resistance

Problem: (i) A known non-linear resistance (NLR) is connected across a variable voltage source, we want to know the current drawn by the resistance. Solution:

Method 1 (Analytical) Method 2 (Graphical)

Method 1: If the expression governing the v-i characteristic of NRL is given, then the problem may be solved analytically. It is straight forward. Method 2: If the v-i characteristic of the non-linear element is known, then the problem could also be solved graphically.

If v = v1, obtain i from graph as i1 if it is required to find voltage required for current i3 (say), this could also be obtained from the graph as v = v3. Problem:

Suppose a given NLR is connected in series with a known linear resistance (LR). If a voltage source of E volts is connected across the combination, how to find out the current? Solution:

The solution of this problem may again be obtained either analytically (if the expression for v-i characteristic of NLR is given) or graphically (if the graphical representation of v-i characteristic of NLR is provided). But the solution is not very straight forward lime problem (i), since the voltages across each of the resistors are not known before hand.

NL R

i

v

i

v

i1

i2

i3

v3 v2 v1

v – i characteristic of NLR

46

Ex.

0, for 1.5v= ≤i v

( )23

0.4 1.5 1.5v10

= − ≥i v v

Using KVL: 38 2.4 10= + × ×v i ( )22.4 0.4 1.5 assuming 1.5= + × − ≥v v v

2or, 0.96 1.88 5.84 0− − =v v or, = 3.633V or -1.675Vv (The second solution is not possible since 1.5≥v ) Therefore, i = 1.82mA For graphical solution of such types of problems are as follows: To begin with note

i. LR & NLR are in series, as I will be same

ii. Sum of the voltage drops in LR & NLR must be equal to E

Step 1: Draw the i-v characteristics of NLR & LR in the first and 2nd quadrant respectively of the same graph paper as shown.

Step 2: One rather length procedure could be: Assume the current to be any arbitrary value OC. Draw a horizontal line through which cuts LR & NLR characteristics at A and B. then voltage drop across LR is AC & voltage drop across NLR is CB. Now check whether AC+CB = AB = E or not. If not, repeat this procedure with position of C shifted elsewhere, till the above condition is fulfilled. If it AB = E, the current will be i = OC.

v

NL R

8V

i

v 1.5 V

A B C

O

LR NLR i

C v v

47

Step 3: Although step (2) is apparently lengthy, the correct location of C can be fixed in one stroke as follows: Cut out a length XY = E and imagine XY be moved parallel to v-axis. Try to place the segment XY (parallel to v-axis) in such a way that the point X lies on LR characteristic and point Y lies on NLR characteristic. Then we can say with certainly that OC′ will be current.

Step 4: Step (3) is certainly better than step (2), but same parallel shifting of XY is necessary. Now in this step we will see that this shifting is also not necessary. How? Let us draw a line parallel to XO through the point Y such that it meets the v-axis at O′. Obviously OXYO′ is a parallesgram. So OO′ = E. Therefore, the alternative way of getting OC′ can be:

i. Draw NLR characteristic ii. On v-axis cut out OO′ such that OO′ = E.

iii. At O′ sketch LR characteristic iv. The point of intersection of NLR characteristic and this new shifted LR characteristic

will fix up C′. Then C′M = i OM = voltage drop across NLR MO′ = voltage drop across LR Finally, For a complicated network having only one non-linear resistance can be solved as follows:

X Y C'

O

LR NLR

i

E

v v

O'

C'

O

LR NLR i

v O' M

48

1) Obtain the Thevenin’s (or Norton’s) equivalent of the linear part of the circuit looking NLR terminals and represent the network as

2) If NLR is represented as an analytical expression the problem may be solved analytically.

3) If NLR characteristic is given then adopt the graphical method. In this method, we may conceptually break up the circuit into a pure linear and a pure non linear part as follows:

The v-I characteristics of the two parts are:

Since the v-i in the two circuits are identical, the solution is obtained by the intersection of two curves.

A

NL R

VTh

B

RTh

A

VTh

NL R

B

RTh

v

i VTh i

RTh +

-

v i

+

- v

NLR

VTh

VTh / RTh

LR characteristics or load line

v

i i

v

NLR (given) characteristic

49

Transients in D.C. Circuits

A. Resistance – Inductance (R – L) circuit. a) Growth of Current: Let i(t) be the current through the circuit at any instant of time t after closing the switch. Let us also suppose that the current increases by ‘di’ in time ‘dt’ secs. Then se can write using KVL.

( ) ( )di tV R i t L

dt= +

or, ( ) ( )di tV Li t ;R R dt= + V I Steady state current i.e. current at t =

R= = ∞

or, ( ) ( )di tLI i t R dt

− =

or, ( )( )

di tR dtL I i t

=−

integrating both sides w.r.t. t, i(t )

t

oo

R di(t)dtL I i(t)

=−∫ ∫

( ) ( )i( t )

o

R It ln I i(t) lnL I i t

= − − =⎡ ⎤⎣ ⎦ −

( ) ( )RL ti t I 1 e ........................ (1)−= −

The voltage across R : ( ) ( )R

L tRv t V 1 e−= −

& the voltage across L: ( ) ( ) ( )RL t

R

di tv t L Ve ............... 2

dt−= =

R

V L

Closed at t = 0 +

-

i(t)

i

v

RL- ti = I(1- e )

0.632I

T

V v

RL- t

Lv = Ve

T

50

Note: At t = 0, the current through the inductor is zero i.e. it is open circuited Time constant of an R-L circuit: It may be defined as follows:

i. It is the time by which the current through the inductor grows and reaches the final steady state value (I) if the initial rate of growth of current is maintained.

Now from the figure (a) initial rate of increase of current IT

= and from eqn (1) the initial rate

of increase of current

( )t 0

t 0

di t R RI t I .dt L L=

=

= ⋅ = ⋅

therefore, I R LI T secsT L R= ⋅ ⇒ =

ii. In eqn (2) let us put t = T

( ) ( ) ( ) ( )R R L

L L RT 1i T I 1 e I 1 e 1 1 e 0.632I− − ⋅ −= − = − = − =

So the time constant may also be defined as the time during which the current through the inductance grows to 63.2% of its final steady state value. iii. It may also be defined from figure (b), as the time required for the voltage across the

inductor to reach from V to O b) Decay of current: Let initially i.e. at t = 0 (when the switch is thrown to 2), the current through the inductor is I. Now after time ‘t’ secs of closing the switch in position (2), i(t) be the current flowing in the circuit as shown and let ‘di’ be the decay of current through the inductor in time ‘dt’ sec. So we may write

( ) ( )di tR i t L 0

dt+ =

or ( )( )

di tR dtL i t

= −

Integrating, ( )( )

t i

o I

di tR dtL i t

= −∫ ∫

leading to ( ) RL ti t I.e−= (Note at t = 0, i(a) = I)

R V

L

2 (for decay)

i(t)

1 (for growth)

51

Here again, the time constant may be defined as before & we got LTR

= sec or its is the time

by which the current falls to 36.8% of its initial value.

B. Resistance – Capacitance circuit

i. Charging of a capacitor At t = 0, the switch S is closed. After time t secs of closing the switch let the voltage across the capacitor be v(t) and also let us suppose that the voltage changes by dv in dt secs. Therefore, the charging current

( ) ( ) ( )dv tdq di t cv t cdt dt dt

= = =⎡ ⎤⎣ ⎦

Applying Kirchoff’s law

( ) ( ) ( ) ( )dv tV R i t v t RC v t

dt= + = +

or ( ) ( )dv tV v t RC

dt− =

or ( )( )

dv tdtdc V v t

=−

Integrating ( )( )

( )t v t

o o

dv tdtRC V v t

=−∫ ∫

Which results in ( ) ( )tRCv t V 1 e−= −

( ) ( ) ( )tRC

dv t di t C CV 1 edt dt

−= = −

tRC

V eR

−=

I

T

i (t)

i (t)

t

0.368I

R

V C

S

+

-

i(t)

+ + +

- - -

52

At the instant of switching on t = 0, ( ) Vi o IR

= = and so ( ) tRCi t I e−=

The charge accumulated in the capacitor at time t ( ) ( ) ( ) ( )t t

RC RCq t Cv t CV 1 e Q 1 e− −= = − = −

Where Q is the final charge accumulated in the capacitor at t = ∞. Initially at the instant of switching on, the capacitor behaves as short circuit and whole of the applied voltage will be across the resistance as there is no charge on C at that time. Time Constant: i) It is the time required for the potential difference across C or the charge accumulated in C from zero to its final value if it continues to increase at its initial rate.

Initial rate of increase of p.d. = VT

volts/sec. from graph.

Again ( )t 0

dv t Vdt RC

=

= . So T = RC sec.

ii) At t = T, v(T) = 0.632V & q(T) = 0.632Q also i(T) = 0.368 I. So the time constant may also be defined as that time by which voltage (or charge) across the capacitor plates is 63.2% of their final values or the charging current decreases to 36.8% of its initial value.

i (t)

t T

I

v (t)

T

V

t

i (t)

0.368 I

0.632 V

v (t)

1 S

R

C + -

i 2

V v

53

ii) Discharging of a capacitor: Let initially the capacitor is fully charged to a p.d. of V volts and stored charge Q coulombs. Let at t = 0, the switch S is thrown to 2. Let the voltage across the capacitor fall to v(t) after t secs of closing the switch and the corresponding discharging current be i(t) amps.

Therefore, ( ) ( )v t dv tC

R dt− = since the direction of current is opposite

( )( )

dv tdtRC v t

= − . Integrating we get

( ) tRCv t V e−=

( ) ( ) tRC-dv t

i t C I edt

= =

( ) tRC& q t cv CVe−= =

Time Constant: i) It is the time by which the voltage across the capacitor (or the charge stored in the capacitor) reaches the final value (i.e. zero for discharge) if the initial rate of decrease of voltage (or charge) is maintained constant. It gives T = RC. ii) It is also defined as defined as the time in secs during which the voltage (or charge) across the capacitor falls by 36.8% of its initial value.

v(t)

v(t)

T

T t

t i(t)

V

i(t)

-0.632 I

- I

54

D.C. Transients in R-L-C circuit :

( ) ( ) ( )i t dtdi tRi t L V

dt C+ + =

Differentiating

( ) ( ) ( )2

2

di t d i t i tR L 0

dt dt C+ + =

The solution of this equation 1 2m t m ti A e B e= + where m1 & m2 are roots of

2 1Lm Rm 0C

+ + = which gives 2 4L

C1

R Rm

2L− + −

= & 2 4L

C2

R Rm

2L− − −

= .

Case 1. 2 4LRC

> ; m1 & m2 are both real & unequal

Soln. 2 4L2 4L R R CC t

2 LR R

t2L

2 4LC

Vi e eR

− − −− + −⎧ ⎫⎪ ⎪= −⎨ ⎬− ⎪ ⎪⎩ ⎭

Case 2. 2 4LRC

= 1 2Rm m2L

= = −

( ) R2L tVi t t e

L−=

R L C

t = 0 V

i

i(t)

t

i(t)

t

55

Case 3. 2 4LRC

< m1 & m2 are both complex

( ) R2L

2

2t

2L RC 4

V 1 Ri t e sin tLC 4L

−⎛ ⎞

= −⎜ ⎟⎜ ⎟− ⎝ ⎠

t

i(t)