A cell E₁ of emf 6V and internal resistance 2Ω is connected with another cell E₂ of emf 4V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is :

Solution

Physics sir · 4 steps

Check the plates first: the two cells push against each other, so the 6 V cell charges the 4 V cell.

1. Read the polarities. The long thin plate is the positive terminal. The negative terminals of E₁ and E₂ both face X, and the wire joins their positive ends at P and Y. So in the loop the two emfs oppose each other.
2. Find the current. The net emf is 6 - 4 = 2 V and the total resistance is 2 + 8 = 10 Ω. So I = 2/10 = 0.2 A. The stronger cell E₁ sets the direction, so the current enters E₂ at its positive terminal and charges it.

   V_Y - V_X = E₂ + I r₂ = 4 + 0.2 × 8 = 5.6 V

3. Cross-check with E1. E₁ is discharging, so its terminal voltage is E₁ - I r₁ = 6 - 0.2 × 2 = 5.6 V. P and Y are joined by a plain wire, so this is the same voltage across X and Y.
4. So. The potential difference across X and Y is 5.6 V, option (c).

The idea to keep. Terminal voltage is E - Ir for a cell that is discharging and E + Ir for a cell being charged. First find which way the current flows through each cell.

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