Page 311 - Schaum's Outline of Theory and Problems of Applied Physics

296 DIRECT-CURRENTCIRCUITS [CHAP. 25

(a) From V = V e − Ir we obtain
V e − V 120 V − 115 V
I = = = 25 A
r 0.2
(b) P = IV = (25 A)(115 V) = 2875 W
2
2
(c) P = I r = (25 A) (0.2 ) = 125 W
SOLVED PROBLEM 25.16
A source of what potential difference is required to charge a battery of V e = 6 V and r = 0.1 at a rate
of 10 A?

The required potential difference must equal the emf of the battery plus the Ir drop in its internal resistance.
Hence

V applied = V e + Ir = 6V + (10 A)(0.1 ) = 7V

SOLVED PROBLEM 25.17

What are the advantages of connecting a set of batteries in parallel?
(1) The reduced internal resistance of the set enables more current to be drawn by a load. (2) The ampere-hour
capacity of the set is the sum of the capacities of the individual batteries and so is greater than that of any of them.

SOLVED PROBLEM 25.18
Three batteries, each of emf 12 V and internal resistance 0.1 , are connected in series with a load of
2 . What is the current in the load?
The emf of the set of batteries is

V e = 12 V + 12 V + 12 V = 36 V
and its internal resistance is

r = 0.1 + 0.1 + 0.1 = 0.3
Hence the current in the load is

V e 36 V
I = = = 15.7A
R + r (2 + 0.3)
SOLVED PROBLEM 25.19
The batteries of Prob. 25.18 are connected in parallel with the same load. What is the current in the load
now?
The emf of the set of batteries is 12 V, and its internal resistance is found as follows:
1 1 1 1 3
= + + =
r 0.1 0.1 0.1 0.1
0.1
r = = 0.033
3
The current in the load is
12 V
V e
I = = = 5.9A
R + r (2 + 0.33)