Answer to Question #132300 in Electricity and Magnetism for Picksie

Question

A battery with emf 1.5V is used to power a small lightbulb, which appears to be burning rather faintly. Using a voltmeter and ammeter you determine that the current in the circuit is 355mA, and the potential difference between the terminals of the battery is 0.85V. What is the internal resistance of the battery?

Answer 1

If the potential difference between the terminals of the battery is $0.85V$, then the potential difference across the internal resistance of the battery is $1.5V - 0.85V = 0.65V$. As far, as the current in the circuit is $355mA$, according to the Ohm's law, obtain the value of the internal resistance:

$$r = \dfrac{V}{I} = \dfrac{0.65V}{355\times 10^{-3}A}\approx 1.83\Omega$$

Answer. 1.83 Ohm.