Two cells of emf 2E and E with internal resistance r₁ and r₂ respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

Solution

Physics sir · 4 steps

Terminal voltage zero means the drop Ir₁ inside the first cell eats up its whole emf 2E.

1. Find the current. In the figure both cells face the same way, so their emfs add to 3E. Everything is in series, so the current is I = 3E/(R + r₁ + r₂).
2. Zero terminal voltage. The potential difference across the first cell is 2E - Ir₁. Setting this to zero gives Ir₁ = 2E.

   (3E r₁)/(R + r₁ + r₂) = 2E

3. Solve for R. Cross-multiply: 3r₁ = 2R + 2r₁ + 2r₂. So 2R = r₁ - 2r₂, which gives R = r₁/2 - r₂.
4. So. The terminals of the first cell are at the same potential when R = r₁/2 - r₂, option (b).

The idea to keep. Terminal voltage of a cell is E - Ir. Find I from the whole circuit first, then set that one cell's E - Ir to zero. The emf in the condition is that cell's own emf.

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