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Dear student, 
Given R1 = R2 = 2Ω

Internal resistance = r

In parallel current flows = 1.2 A

In series current flows = 0.4A

Assuming that potential of the source remains constant in both cases.

Equivalent Resistance for Parallel connection

Thus, sum total of potential drops at individual resistance = total potential difference applied

--------------1.

Now, for series connection

Equivalent resistance together with the internal resistance in the series it becomes

Thus, sum total of potential drops at individual resistance = total potential difference applied

----------------2.

From eqn. 1 and eqn. 2

(ii) E = I(R+r) => 1.2(1+0.5) = 1.8 V Regards