A resistance of `1 Omega` is connected to a battery of emf 2 V and internal resistance `0.4 Omega` what current is flowing through the circuit ? If another battery of double the emf is connected in series with the earlier one then current flowing in the circuit becomes double . what should be the internal resistance of battery ?

Explanation

To solve the problem step by step, let's break it down into two parts: finding the current flowing through the circuit initially and then finding the required internal resistance when another battery is added.

### Step 1: Calculate the initial current flowing through the circuit.

Given:
- Resistance \( R = 1 \, \Omega \)
- EMF of the battery \( E = 2 \, V \)
- Internal resistance \( r = 0.4 \, \Omega \)
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