In the given circuit, the terminal potential difference of the cell is :

Solution

Physics sir · 3 steps

The cell's own 1 Ω eats part of the 3 V; find the current first, then subtract that drop.

1. External resistance. The two 4 Ω resistors are connected across the same two wires, so they are in parallel: R = (4 × 4)/(4 + 4) = 2 Ω.
2. Find the current. The current flows through the internal resistance r = 1 Ω and the external 2 Ω in series: I = E/(R + r) = 3/(2 + 1) = 1 A.

   V = E - Ir = 3 - (1)(1) = 2 V

3. So. The terminal potential difference is 2 V. As a check, IR = 1 × 2 = 2 V across the external circuit gives the same value, so the answer is (c).

The idea to keep. Terminal voltage is the EMF minus the drop inside the cell, V = E - Ir. It equals the drop across the external circuit, IR, so you can use either one as a check.

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