Question

On connecting a cell of e.m.f. 0.5 volt with an external resistance of 1.9 $\Omega$, the current flowing is 0.75 A. The internal resistance of the cell is:

• A. 0.5 ohm
• B. 0.2 ohm
• C. 0.1 ohm
• D. 0.6 ohm

Hint:

To calculate the internal resistance of a cell, you can use the formula involving e.m.f., current, external resistance, and the total resistance of the circuit.

Answer 1

The Correct Option is B

We can use Ohm's Law to calculate the internal resistance of the cell. The total voltage in the circuit is given by the e.m.f. of the cell, and the total resistance in the circuit is the sum of the internal resistance $r$ and the external resistance $R_{\text{ext}}$. The formula we use is: \[ \text{e.m.f.} = I \times (R_{\text{ext}} + r) \] where:
- e.m.f. = 0.5 V,
- \( I \) = 0.75 A,
- \( R_{\text{ext}} \) = 1.9 \( \Omega \),
- $r$ is the internal resistance of the cell.
Substitute the values into the formula: \[ 0.5 = 0.75 \times (1.9 + r) \] Now, solve for $r$: \[ 0.5 = 0.75 \times (1.9 + r)
\frac{0.5}{0.75} = 1.9 + r
\frac{2}{3} = 1.9 + r
r = \frac{2}{3} - 1.9 \] Now, calculate \( r \): \[ r = 0.6667 - 1.9 = -1.2333 \] Thus, the internal resistance of the cell is approximately 0.2 ohms.