A cell supplies a current of 1.2 A through two resistors each of 2 `Omega` connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate :
the internal resistance,

Explanation

To solve the problem of finding the internal resistance of a cell that supplies a current of 1.2 A through two resistors of 2 ohms each connected in parallel, and a current of 0.4 A when the same resistors are connected in series, we can follow these steps:

### Step 1: Analyze the Parallel Circuit
1. **Calculate the equivalent resistance (R_parallel)** of the two resistors in parallel.
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1 \, \Omega \]
Thus, \( R_{\text{parallel}} = 1 \, \Omega \).
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