A-level Physics (Advancing Physics)/Potential Dividers/Worked Solutions

'''1. A 12 kΩ resistor and a 20 kΩ resistor are connected to a 9V battery. A voltmeter is connected across the 12 kΩ resistor. What is the reading on the voltmeter? (Assume negligible internal resistance.)'''

$$V_{12 k\Omega} = 9 \times \frac{12}{12 + 20} = 3.375 V$$

'''2. A potential divider consists of 100 5Ω resistors, with a wiper which moves on one resistor for every 3.6° a handle connected to it turns. The wiper is connected to a voltmeter, and the circuit is powered by a 120V power source with negligible internal resistance. What is the reading on the voltmeter when the handle turns 120°?'''

First, work out the number (n) of resistors between the terminals of the voltmeter:

$$n = \frac{120}{3.6} = 33.\bar{3}$$

The handle, then, has not turned enough to reach the 34th resistor, so the voltmeter is connected just after the 33rd resistor.

$$V = 120 \times \frac{33 \times 5}{100 \times 5} = 39.6\mbox{ V}$$

'''3. A 9V battery with internal resistance 0.8Ω is connected to 3 resistors with conductances of 3, 2 and 1 Siemens. A voltmeter is connected across the 3 and 2 Siemens resistors. An ammeter is placed in the circuit, between the battery and the first terminal of the voltmeter, and reads 2A. What is the reading on the voltmeter?'''

First, work out the resistances of the resistors:

$$R = \frac{1}{G} = \frac{1}{3}, \frac{1}{2} , 1 \mbox{ }\Omega$$

Then, work out the external potential difference (i.e. excluding the potential difference lost due to the battery's internal resistance):

Vexternal = E - IRinternal = 9 - (2 x 0.8) = 7.4V

$$V_{3\Omega, 2\Omega} = 7.4 \times \frac{\frac{1}{3} + \frac{1}{2}}{\frac{1}{3} + \frac{1}{2} + 1} \approx 3.36\mbox{ V}$$