JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Current Electricity

Two cells of emf 1 V and 2 V and internal resistance

2 Ω

and

1 Ω

, respectively, are connected in series with an external resistance of

6 Ω

. The total current in the circuit is

I_{1}

. Now the same two cells in parallel configuration are connected to same external resistance. In this case, the total current drawn is

I_{2}

. The value of

(\frac{I _{1}}{I _{2}})

\frac{x}{3}

. The value of x is ____.

Answer: 4

Solution

In series the two cells aid each other, giving an emf of

1 + 2 = 3 V

against a total resistance

2 + 1 + 6 = 9 Ω

, so

I_{1} = \frac{3}{9} = \frac{1}{3} A

. In parallel the equivalent emf is

\frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + 1} = \frac{5}{3} V

and the equivalent internal resistance is

\frac{2 \times 1}{2 + 1} = \frac{2}{3} Ω

, giving

I_{2} = \frac{5/3}{2/3 + 6} = \frac{1}{4} A

. The ratio

\frac{I _{1}}{I _{2}} = \frac{1/3}{1/4} = \frac{4}{3}

, so that

x = 4

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Integer answer