Question

A cell of emf 2 V is connected to an external resistor. If the current through the resistor is 200 mA and the terminal voltage of the cell is 87.5% of the emf of the cell, then the internal resistance of the cell is:

• A. \( 1.50 \, \Omega \)
• B. \( 1.25 \, \Omega \)
• C. \( 2 \, \Omega \)
• D. \( 2.25 \, \Omega \)

Hint:

To find the internal resistance, use the formula \( V = E - I \times r \), where \( V \) is the terminal voltage, \( E \) is the emf, \( I \) is the current, and \( r \) is the internal resistance.

Answer 1

The Correct Option is B

Given: - \( E = 2 \, \text{V} \) (emf of the cell),
- Current \( I = 200 \, \text{mA} = 0.2 \, \text{A} \),
- Terminal voltage \( V = 87.5\% \) of the emf, so \( V = 0.875 \times E = 0.875 \times 2 = 1.75 \, \text{V} \).
Using the formula for terminal voltage: \[ V = E - I \times r \] Substitute the given values: \[ 1.75 = 2 - 0.2 \times r \] Solve for \( r \): \[ 0.2 \times r = 2 - 1.75 = 0.25 \quad \Rightarrow \quad r = \frac{0.25}{0.2} = 1.25 \, \Omega \] Thus, the internal resistance of the cell is \( 1.25 \, \Omega \).