Question

The circuit diagram shows a 20V battery connected across a parallel combination of two branches. Branch 1 consists of a single 6 Ohm resistor. Branch 2 consists of three resistors in series: a 3 Ohm resistor, a 2 Ohm resistor, and a 1 Ohm resistor. Points A and B are the terminals of the 2 Ohm resistor, which is part of the series branch. What is the potential difference between points A and B?

• A. 20/3 V
• B. 10/3 V
• C. 6 V
• D. 20 V

Answer 1

Answer: (A)

20/3 V

Answer 2

The circuit consists of a 20V battery connected across a parallel combination of a 6 Ohm resistor and a series combination of 3 Ohm, 2 Ohm, and 1 Ohm resistors. The total resistance of the series branch is $3+2+1=6\Omega$. The current through this series branch is $I = V/R_{series} = 20V/6\Omega = 10/3$ A. Points A and B are the terminals of the 2 Ohm resistor. The potential difference across A and B is the voltage drop across the 2 Ohm resistor, which is $V_{AB} = I \times R_{2\Omega} = (10/3 \text{ A}) \times (2\Omega) = 20/3$ V.