0195136047.pdf

1.3 KIRCHHOFF’S LAWS 45

Power absorbed by resistorR 4 =( 0. 1 )^2 × 50 = 0 .5W

Power delivered by sources= 2. 4 − 1. 2 = 1 .2W

Power absorbed by resistorsR 1 ,R 2 ,R 3 , andR 4 = 0. 1 + 0. 4 + 0. 2 + 0. 5 = 1 .2W

The conservation of power is satisfied by the circuit.

EXAMPLE 1.3.5

Given the network in Figure E1.3.5,

(a) Find the currents through resistorsR 1 ,R 2 , andR 3.

(b) Compute the voltageV 1.

(c) Show that the conservation of power is satisfied by the circuit.

R 1 = R 2 =

I 1

V 1

I 2 I 3

12 A 6 Ω 6 A

Ground node

Node 1

4 Ω R 3 = 12 Ω

Figure E1.3.5

Solution

(a) Applying the KCL at node 1, we have

12 =

V 1

R 1

+

V 1

R 2

+

V 1

R 3

+ 6 =V 1

(

1

6

+

1

4

+

1

12

)

+ 6 =

V 1

2

+ 6

Therefore,V 1 =12 V. Then,

I 1 =

12

6

=2A

I 2 =

12

4

=3A

I 3 =

12

12

=1A

(b)V 1 =12 V

(c) Power delivered by 12-A source= 12 × 12 =144 W

Power delivered by 6-A source=− 6 × 12 =−72 W

Power absorbed by resistorR 1 =I 12 R 1 =( 2 )^26 =24 W