0195136047.pdf

44 CIRCUIT CONCEPTS

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

−

+

−

R 1 = 10 Ω +

I

R 2 = 40 Ω

R 4 = 50 Ω

R 3 = 20 Ω

12 V 24 V

a bcd

f e

(a)

Figure E1.3.4

−

+

+− −

+

12 V

5 V

−+1 V −+4 V −+2 V

24 V

a bcd

f e

(b)

Solution

(a) From Ohm’s law, the currentIis given by

I=

24 − 12

10 + 40 + 20 + 50

=

12

120

= 0 .1A

Therefore, voltage drops across the resistors are calculated as follows:

Vba=IR 1 = 0. 1 × 10 =1V

Vcb=IR 2 = 0. 1 × 40 =4V

Vdc=IR 3 = 0. 1 × 20 =2V

Vfe=IR 4 = 0. 1 × 50 =5V

These are shown in Figure E1.3.4(b) with their polarities. Note that capital letters are

used here for dc voltages and currents.

(b) Applying the KVL for the closed pathedcbafe,weget

− 24 + 2 + 4 + 1 + 12 + 5 = 0

which confirms that the KVL is satisfied,

Vbf=Vba+Vaf= 1 + 12 =13 V

Vec=Ved+Vdc=− 24 + 2 =−22 V

(c) Power delivered by 24-V source= 24 × 0. 1 = 2 .4W

Power delivered by 12-V source=− 12 × 0. 1 =− 1 .2W

Power absorbed by resistorR 1 =( 0. 1 )^2 × 10 = 0 .1W

Power absorbed by resistorR 2 =( 0. 1 )^2 × 40 = 0 .4W

Power absorbed by resistorR 3 =( 0. 1 )^2 × 20 = 0 .2W