BITS PILANI, K.K. BIRLA GOA CAMPUS ELECTRICAL SCIENCES (EEE F111)

Tutorial -2 on 28-01-2015

1) Find value of current i1 in the following circuit using

superposition theorem.

ANS: With only 30V voltage source active

Using mesh analysis: Apply KVL to each mesh,

-30+6I1+VX=0 & -Vx+4 I2+2 I2=0

Vx=30-6I1 & Vx= 6I2

Since LHS are equal,RHS of two mesh equations are equal

30-6I1=6I2.-----------equn 1

Also, I1-I2=8i11 & I2=i11-----Substituting in above equn-1

30=60 i11 Hence i11=0.5A

With only 3A.current source active

6I1+Vy=0 & -Vy+4 (I2-I3)+2 I2=0

Vy=-6I1 & Vy= 6I2 - 4I3

Since LHS are equal, RHS of two mesh equations are equal

6I1=6I2 - 4I3.-----------equn 2

Also, I1-I2=8i12 & I2=i12 & I3=3 A---Substituting in equn-2

12=60 i12 Hence i12=0.2A

Using Superposition theorem, i1=i11+i12=0.5+0.2=0.7A

Resistors 30 ohms in parallel with 15 ohms.(30 || 15 = 10 ohms)

KVL equation around mesh is : 10I+150+10I+[(1/3)vx3]=0 & since Vx3=(-10I).

Solving these two equations, 20I+150-10I=0; 60I+450-10I=0; 50I+450=0; I=-9A;

Vx3=-10I=90V. Hence Vx=15-30+90=75V-------Ans.

Q. In the circuit shown, find the value of R that will absorb maximum power from the circuit. Find also the value of this maximum power.

Solution: ThR R= :

KVL:

3 8( 1) 0 5

4 A 40 V 40 1 A

x x x

tt Th

V V i V iVi V R

+ + + = =

= = = =

VTh:

KVL: 2 21

3 8 04x x x

V V i i V + + = =

1 2 15

5( ) 2 A 104

40 V, 2 80 V

x x x

x Th x

V i i i V V

V V V

= = = +

= = =

2

max

640040 W

4 4 40ThVPR

= = =

+

Vx

5 R8 2 A

+ 3Vx

+

Vx

5 8

+ 3Vx

1 A

+

Vt

i

+

Vx

5 8 2 A

+ 3Vx

+

VTh

i1 i2

Q. The variable resistor oR in the circuit in Fig.3 is adjusted until it absorbs maximum power from the circuit.

o

2

th

a) Find the value of R .

b) Find the maximum power.

10 V3 A2 A Ro

4 : 6 :

Ro

Rth

Vth

When o thR R it will absorb maximum power. Figure 3

Maximum power is max4o

thR

VP

R

4 : 6 :

Rth = 10 :

In order to find thR all independent

sources are killed.

For V : th

KCL at V :th

24

8.............(1)

th

th

V V

V V

10 V3 A2 A

4 : 6 : 10 VVth V

KCL at V :

103

4 6

3 3 2 20 36

3 5 16..........(2)

th

th

th

V V V

V V V

V V

Multiply Eq. (1) by 5 and then sum up Eq(1) and (2)

max

2 24

12 V

10

1443.6

40o

th

th

o

R

V

V

R

P W

:

Tut-2Tut-2tut2

max_power2max_power3