## Mastering Physics Solutions: Measuring the EMF and Internal Resistance of a Battery

Exercise 16.39

Part A = 3.06 V
Part B = 3.636 * 10-2 Ω
Part C = 1.82 Ω

Solutions Below:

When switch S in the figure is open, the voltmeter V of the battery reads 3.06 V. When the switch is closed, the voltmeter reading drops to 3.00 V, and the ammeter A reads 1.65 A. Assume that the two meters are ideal, so they do not affect the circuit.

Part A

Find the emf Ε.

Electromotive force is a synonym for voltage. In this case, it’s 3.06 (the voltage of the battery, ignoring internal resistance).

3.06 V

Part B

Find the internal resistance r of the battery.

You know there’s a difference of 0.06 V between opening and closing the switch (3.06 – 3.00). This is due to the internal resistance of the battery. The problem also says that there’s 1.65 A of current in the circuit. Ohm’s Law gives us the formula V = IR:

V = IR
0.06 = 1.65 * Rinternal
Rinternal = 3.636 * 10-2 Ω

3.636 * 10-2 Ω

Part C

Find the circuit resistance R.