Current $i$ coming from the battery and ammeter reading are :

\frac{3}{8} A, \frac{1}{8} A

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\frac{1}{8} A, \frac{1}{8} A

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2 A, \frac{2}{3} A

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2 A, \frac{1}{8} A

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Solution

The correct option is A $\frac{3}{8} A, \frac{1}{8} A$
Current will flow from battery, 4 ohm resistor, the three 4ohm resistors in parallel, and back into the battery.

Current in battery= $\frac{2}{4 + 4 / 3} = \frac{2 \times 3}{4 \times 4} = \frac{3}{8} A$

current in the ammeter = $\frac{1}{3} \frac{3}{8} = \frac{1}{8} A$

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