Two identical cells each of emf 1.5 V are connected in series across a 10 Ω resistance. An ideal voltmeter connected across 10 Ω resistance reads 1.5 V. The internal resistance of each cell is __________ Ω.

Solution

Physics sir · 4 steps

Half the 3 V is lost inside the cells, so their internal resistance must equal the external 10 Ω.

1. Current from voltmeter. The ideal voltmeter reads the drop across the 10 Ω resistor, so the current is I = 1.5/10 = 0.15 A.
2. Series cells. In series the emfs add to 2 × 1.5 = 3 V, and the internal resistances add to 2r. The same current flows through everything.

   I = 2E/(R + 2r) ⇒ 0.15 = 3/(10 + 2r)

3. Solve for r. 10 + 2r = 3/0.15 = 20, so 2r = 10 Ω.
4. So. Each cell has internal resistance r = 5 Ω.

The common slip: 10. Stopping at 2r = 10 Ω gives 10, but that is the combined internal resistance of both cells. Each cell has half of it.

The idea to keep. Find the current from the reading across the external resistor first. Then total emf divided by that current gives the total resistance, internal plus external.

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