JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Current Electricity

Two identical cells each of emf 1.5 V are connected in series across a

10Ω

resistance. An ideal voltmeter connected across

10Ω

resistance reads 1.5 V . The internal resistance of each cell is

____

Ω

Answer: 5

Solution

Two cells in series give total emf =

2 \times 1.5 = 3

V. Let the internal resistance of each cell be

r

. Total internal resistance =

2 r

. The current in the circuit is

I = \frac{3}{10 + 2 r}

. The voltage across the

10Ω

resistor is

V = I \times 10 = 1.5

V. Therefore,

\frac{3 \times 10}{10 + 2 r} = 1.5

, which gives

30 = 1.5 (10 + 2 r) = 15 + 3 r

. Solving:

3 r = 15

, so

r = 5Ω

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer