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(A) The circuit consists of a $3 \ V$ battery connected in parallel with a resistor $R_1 = 2 \ \Omega$ and a series combination of resistors $R_2, R_3

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$2$

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(A) The circuit consists of a $3 \ V$ battery connected in parallel with a resistor $R_1 = 2 \ \Omega$ and a series combination of resistors $R_2, R_3, R_4$.First,calculate the equivalent resistance of the series branch containing $R_2, R_3, R_4$:$R_s = R_2 + R_3 + R_4 = 2 \ \Omega + 2 \ \Omega + 2 \ \Omega = 6 \ \Omega$.Now,this $R_s$ is in parallel with $R_1 = 2 \ \Omega$. The equivalent resistance $R_{eq}$ of the circuit is:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_s} = \frac{1}{2} + \frac{1}{6} = \frac{3+1}{6} = \frac{4}{6} = \frac{2}{3} \ \Omega^{-1}$.Therefore,$R_{eq} = 1.5 \ \Omega$ or $\frac{3}{2} \ \Omega$.The total current $i$ drawn from the battery is given by Ohm's law:$i = \frac{V}{R_{eq}} = \frac{3 \ V}{1.5 \ \Omega} = 2 \ A$.

What is the current $i$ in the circuit as shown in the figure (in $A$)?

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