Reading of ammeter for the following circuit is:-

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\frac{1}{2} A

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\frac{2}{3} A

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Solution

The correct option is C $\frac{2}{3} A$
$2 V = 2 * I_{1} + 2 * (I_{1} + I_{2})$ ---eq.1
$2 V = 2 * (I_{1} + I_{2}) + 2 I_{2}$ ---eq.2
eq1 - eq.2:
$0 = 2 I_{1} - 2 I_{2}$
$I_{1} = I_{2}$
eq1 + eq.2:
$4 = 2 I_{1} + 4 (I_{1} + I_{2}) + 2 I_{2}$
$4 = 6 I_{1} + 6 I_{2}$
$6 (I_{1} + I_{2}) = 4$
$I_{1} + I_{2} = \frac{4}{6}$
$I_{1} + I_{2} = \frac{2}{3}$
Reading of A is $= I_{1} + I_{2} = \frac{2}{3} A$
Hence option 'C' is correct.

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