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 address the problem, we first establish the circuit's configuration and apply Ohm's Law. Initially, the total resistance is the sum of the internal resistance of the first cell and the wire: 0.5 Ω + 0.5 Ω = 1 Ω. The electromotive force (EMF) of the first cell is 1.1 V. Consequently, the current, \( I \), is 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 an identical EMF is connected in series, increasing the total EMF to 1.1 V + 1.1 V = 2.2 V. Assuming the current remains constant at 1.1 A, Ohm's Law is reapplied to solve for the new total resistance, \( R_t \), using the combined EMF:

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

Solving for \( R_t \) yields:

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

This total resistance comprises the internal resistance of both cells and the wire:

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

Rearranging to solve for \( r \):

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

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

Thus, the internal resistance \( r \) of the second cell is determined to be 1 Ω.