A network of four resistances is connected to 9 V battery, as shown in figure. The magnitude of voltage difference between the points A and B is __________ V.

Solution

Physics sir · 3 steps

No wire joins A and B, so each 6 Ω branch takes its own current. Find the potential at A and at B.

1. Two parallel branches. The upper branch is 2 Ω then 4 Ω through A. The lower branch is 4 Ω then 2 Ω through B. Both run across the same 9 V, and each has 6 Ω.
2. Branch current. Each branch carries I = 9/6 = 1.5 A. Take the left junction at 9 V and the right junction at 0 V.

   V_A = 9 - 1.5 × 2 = 6 V, V_B = 9 - 1.5 × 4 = 3 V

3. So. |V_A - V_B| = 6 - 3 = 3 V.

The common slip: 0. Treating this as a balanced Wheatstone bridge gives 0. The ratios are 2:4 on top and 4:2 below, so the bridge is not balanced.

The idea to keep. To find the voltage between two points in separate branches, fix one junction as your reference. Then walk from it to each point, subtracting IR across every resistor on the way.

Answer from the NTA final answer key. Solution by PrepWiser.