Question

When a battery voltage is measured by placing a high-resistance voltmeter across the terminals of the battery, the voltmeter reads 6.3V. With the voltmeter leads still attached, the battery terminals are then connected to a resistor in series with an ammeter. In this arrangement, the ammeter reads 150 mA, and the voltmeter reads 5.9V. What are the emf and the internal resistance of the battery?

Answer 1

(a) In the first case, the voltmeter is of high-resistance so we can say that it is an ideal voltmeter.

Potential across the terminal measured by voltmeter, E = 6.3 V

So, the emf of the battery = 6.3 V

(b) Now the current measured by the ammeter, I = 150 mA = 0.150 A

New voltage = 5.9 V

So, the voltage drop, $\Delta V$ = 6.3 - 5.9 = 0.40 V

This voltage drop is due to the internal resistance of the battery.

Therefore, the internal resistance of the battery, r = $\Delta V$ / I = 0.40 / 0.150 = 2.67 $\Omega$

Hope you understand the solution !