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(B) In the given circuit,there are three resistors of $2 \, \Omega$ each connected in parallel. Let these be $R_1, R_2,$ and $R_3$.The equivalent resi

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$\frac{3}{4} \, A$

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(B) In the given circuit,there are three resistors of $2 \, \Omega$ each connected in parallel. Let these be $R_1, R_2,$ and $R_3$.The equivalent resistance of these three parallel resistors is given by $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \, \Omega^{-1}$,so $R_p = \frac{2}{3} \, \Omega$.This parallel combination is in series with another resistor of $2 \, \Omega$ and a battery of $2 \, V$.The total equivalent resistance of the circuit is $R_{eq} = R_p + 2 \, \Omega = \frac{2}{3} + 2 = \frac{8}{3} \, \Omega$.The total current $i$ flowing through the circuit is given by Ohm's law: $i = \frac{V}{R_{eq}} = \frac{2}{8/3} = \frac{6}{8} = \frac{3}{4} \, A$.

The reading of the ammeter as shown in the figure is:

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