Question

A battery charger supplies 10 A to charge a storage battery which has an open - circuit voltage of 5.6V. If the voltmeter connected across the charger reads 6.8V, what is the internal resistance of the battery at this time?

a. 13Ω

b. 052Ω

c. 012Ω

d. 21Ω

Answer 1

Solve. A battery charger supplies 10 A to charge a storage battery which has an open - circuit voltage of 5.6V. If the voltmeter connected across the charger reads 6.8V, what is the internal resistance of the battery at this time?

a. 13Ω

b. 052Ω

c. 012Ω

d. 21Ω

Solution.

Ohm's principal discovery was that the amount of electric current through a metal conductor in a circuit is directly proportional to the voltage impressed across it, for any given temperature. Ohm expressed his discovery in the form of a simple equation, describing how voltage, current, and resistance interrelate:

$$U = I r$$

Open-circuit voltage is the difference of electrical potential between two terminals of a device when disconnected from any circuit. There is no external load connected. No external electric current flows between the terminals.

For the series network,

$$U _ {1} + U _ {2} + U _ {3} + \dots = I R _ {1} + I R _ {2} + I R _ {3} + \dots$$

Since the current $I$ is the same in each resistor.

Since the battery is charging

$$E = U + V _ {oc} = I r + V _ {oc}$$

where

$$I = 10\,A$$ $$V _ {oc} = 5.6\,V$$ $$E = 6.8\,V$$

Then

$$r = \frac {E - V _ {oc}}{I} = \frac {6.8 - 5.6}{10} = \frac {1.2}{10} = 0.12 \, \Omega$$

Answer: c. 0.12 Ω