Question

A cell of emf 1.1 V and internal resistance 0.5 Ω is connected to a wire of resistance 0.5 Ω. Another cell of the same emf is now connected in series with the intention of increasing the current but the current in the wire remains the same. The internal resistance of the second cell is________.
Fill in the blank with the correct answer from the options given below.

• A. 1.5 Ω
• B. 2.5 Ω
• C. 1Ω
• D. 2 Ω

Answer 1

The Correct Option is C

Let's analyze the circuit and determine the internal resistance of the second cell.

1. First Circuit:

The first circuit has:

• Emf (E₁) = 1.1 V
• Internal resistance (r₁) = 0.5 Ω
• External resistance (R) = 0.5 Ω

The current (I₁) in the first circuit is:

I₁ = E₁ / (R + r₁)

I₁ = 1.1 V / (0.5 Ω + 0.5 Ω)

I₁ = 1.1 V / 1 Ω

I₁ = 1.1 A

2. Second Circuit:

The second circuit has:

• Emf (E₁) = 1.1 V
• Internal resistance (r₁) = 0.5 Ω
• Emf (E₂) = 1.1 V
• Internal resistance (r₂) = ?
• External resistance (R) = 0.5 Ω

The current (I₂) in the second circuit is:

I₂ = (E₁ + E₂) / (R + r₁ + r₂)

I₂ = (1.1 V + 1.1 V) / (0.5 Ω + 0.5 Ω + r₂)

I₂ = 2.2 V / (1 Ω + r₂)

We are given that I₁ = I₂:

1.1 A = 2.2 V / (1 Ω + r₂)

1.1 (1 + r₂) = 2.2

1 + r₂ = 2.2 / 1.1

1 + r₂ = 2

r₂ = 2 - 1

r₂ = 1 Ω

Therefore, the internal resistance of the second cell is 1 Ω.

The correct answer is:

Option 3: 1 Ω