Page 307 - Schaum's Outline of Theory and Problems of Applied Physics

292 DIRECT-CURRENTCIRCUITS [CHAP. 25

SOLVED PROBLEM 25.8
Two 240- light bulbs are to be connected to a 120-V power source. To determine whether they will
be brighter when connected (a) in series or (b) in parallel, calculate the power they dissipate in each
arrangement.

(a) The equivalent resistance of the two bulbs in series is
R = R 1 + R 2 = 240 + 240 = 480

The current in the circuit is therefore

V 120 V
I = = = 0.25 A
R 480
Since the bulbs are in series, this current passes through each of them. The power each bulb dissipates is
2
2
P = I R = (0.25 A) (240 ) = 15 W
(b) When the bulbs are in parallel, the potential difference across each is 120 V. Hence the power each bulb
dissipates is

V 2 (120 V) 2
P = = = 60 W
R 240
The bulbs will be brighter when they are connected in parallel.

SOLVED PROBLEM 25.9
A circuit has a resistance of 50 . How can it be reduced to 20 ?

To obtain an equivalent resistance of R = 20 , a resistor R 2 must be connected in parallel with the circuit of
R 1 = 50 .Tofind R 2 we proceed as follows:

1 1 1 1 1 1 R 1 − R
= + = − =
R R 1 R 2 R 2 R R 1 R 1 R
R 1 R (50 )(20 )
R 2 = = = 33.3
R 1 − R 50 − 20

SOLVED PROBLEM 25.10
Find the equivalent resistance of the circuit shown in Fig. 25-4(a).
Figure 25-4(b) shows how the original circuit is decomposed into its series and parallel parts, each of which is
treated in turn. The equivalent resistance of R 1 and R 2 is

R 1 R 2 (10 )(10 )

R = = = 5
R 1 + R 2 10 + 10
This equivalent resistance is in series with R 3 , and so

R = R + R 3 = 5 + 3 = 8
Finally R is in parallel with R 4 ; hence the equivalent resistance of the entire circuit is

R R 4 (8 )(12 )

R = = = 4.8
8 + 12
R + R 4