Page 310 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 25] DIRECT-CURRENTCIRCUITS 295

it provides a current of I is therefore

V = V e − Ir
Terminal voltage = emf − potential drop due to internal resistance

When a battery or generator of emf V e is connected to an external resistance R, the total resistance in the
circuit is R + r, and the current that flows is

V e
I =
R + r
emf
Current =
external resistance + internal resistance

BATTERIES
The emf of a set of batteries connected in series is the sum of the emf’s of the individual cells. The internal
resistance of the set is the sum of the individual internal resistances. Thus

V e, series = V e,1 + V e,2 + V e,3 + ···
r series = r 1 + r 2 + r 3 + ···

Batteries that are intended to be connected in parallel should always have the same emf; otherwise currents
would circulate among them, wasting energy. If V e is the emf of each battery, the emf of a set of batteries
connected in parallel is also V e . The internal resistance of the set is

1 1 1 1
= + + +· · ·
r parallel r 1 r 2 r 3

SOLVED PROBLEM 25.13
A dry cell of emf 1.5 V and internal resistance 0.05 is connected to a flashlight bulb whose resistance
is 0.4 . Find the current in the circuit.

V e 1.5V
I = = = 3.33 A
R + r 0.4 + 0.05
SOLVED PROBLEM 25.14
A battery whose emf is 45 V is connected to a 20- resistance, and a current of 2.1 A flows. (a) Find the
internal resistance of the battery. (b) Find the terminal voltage of the battery.
(a) From I = V e /(R + r) we obtain

V e 45 V
r = − R = − 20 = 21.4 − 20 = 1.4
I 2.1A
(b) V = V e − Ir = 45 V − (2.1A)(1.4) = 42 V

SOLVED PROBLEM 25.15

A generator has an emf of 120 V and an internal resistance of 0.2 .(a) How much current does the
generator supply when the terminal voltage is 115 V? (b) How much power does it supply? (c)How
much power is dissipated in the generator itself?