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

To solve the problem, we need to understand the circuit configuration and apply Ohm's Law. Initially, the total resistance in the circuit is the sum of the internal resistance of the first cell and the wire, which is 0.5 Ω + 0.5 Ω = 1 Ω. The electromotive force (EMF) of the first cell is 1.1 V. Therefore, the current, \( I \), through the circuit can be calculated using Ohm's Law, \( I = \frac{E}{R} \):

\( I = \frac{1.1 \, \text{V}}{1 \, \text{Ω}} = 1.1 \, \text{A} \)

A second cell with the same EMF is added in series. The total EMF of the circuit becomes 1.1 V + 1.1 V = 2.2 V. If the current remains the same at 1.1 A, we apply Ohm's Law again with the total new EMF and solve for the required total resistance, \( R_t \):

\( 1.1 = \frac{2.2}{R_t} \)

Solving for \( R_t \):

\( R_t = \frac{2.2}{1.1} = 2 \, \text{Ω} \)

The total resistance now includes the internal resistance of both cells and the wire, so:

\( R_{\text{total}} = 0.5 \, \text{Ω (wire)} + 0.5 \, \text{Ω (first cell)} + r \, \text{(second cell)} = 2 \, \text{Ω} \)

Solving for \( r \):

\( r + 1 \, \text{Ω} = 2 \, \text{Ω} \)

\( r = 2 \, \text{Ω} - 1 \, \text{Ω} = 1 \, \text{Ω} \)

Therefore, the internal resistance \( r \) of the second cell is 1 Ω.