Question

A battery of emf 10 V is connected to resistance as shown in figure. The potential difference $VA-VB$ between the points A and B is

• A. -2 V
• B. 2 V
• C. 5 V
• D. $\frac{20}{11} V$

Hint:

A battery of emf 10 V is connected to resistance as shown in figure. The potential difference $VA-VB$ between the points A and B is \includegraphics[width=0.5\linewidth]7phy.png \labelfig:placeholder

Answer 1

The Correct Option is B

To find the potential difference \( V_A - V_B \) between the points A and B, let's analyze the circuit given in the image.

1. The circuit forms a bridge with a battery of 10 V. The resistances in the bridge are 1 Ω, 3 Ω, 3 Ω, and 1 Ω.
2. The resistances between A and B in each arm of the bridge are parallel: \(R_1 = 1 \, \Omega + 3 \, \Omega = 4 \, \Omega\) and \(R_2 = 3 \, \Omega + 1 \, \Omega = 4 \, \Omega\).
3. The total resistance in the parallel branches between A and B is: \(\frac{1}{R_{AB}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \, \Omega\)
4. Hence, \(R_{AB} = 2 \, \Omega\).
5. This setup is a balanced Wheatstone bridge, so the current through the bridge (between points A and B) is zero.

Now, using the battery and external 3 Ω resistor:

1. The total resistance of the circuit is: \(R_{\text{total}} = 3 \, \Omega + R_{AB} = 3 \, \Omega + 2 \, \Omega = 5 \, \Omega\)
2. The current through the circuit is: \(I = \frac{V}{R_{\text{total}}} = \frac{10 \, V}{5 \, \Omega} = 2 \, A\)
3. The potential difference across each parallel path of 4 Ω resistor is: V_A - V_B is 2 V.