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(C) The given circuit is a Wheatstone bridge. In the balanced condition, the galvanometer branch $BD$ has no current flowing through it.Therefore, the

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$0.2$

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(C) The given circuit is a Wheatstone bridge. In the balanced condition, the galvanometer branch $BD$ has no current flowing through it.Therefore, the two upper resistors ($10 \ \Omega$ and $10 \ \Omega$) are in series, and the two lower resistors ($30 \ \Omega$ and $30 \ \Omega$) are in series.The equivalent resistance of the upper branch is $R_1 = 10 \ \Omega + 10 \ \Omega = 20 \ \Omega$.The equivalent resistance of the lower branch is $R_2 = 30 \ \Omega + 30 \ \Omega = 60 \ \Omega$.These two branches are in parallel with each other.The equivalent resistance $R_{eq}$ of the circuit is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{20} + \frac{1}{60} = \frac{3+1}{60} = \frac{4}{60} = \frac{1}{15}$.Thus, $R_{eq} = 15 \ \Omega$.The current $I$ drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{3 	ext{ V}}{15 \ \Omega} = 0.2 	ext{ A}$.

Resistances are joined as shown in the figure. In the balanced condition, the current $I$ drawn from the battery is (in $\text{ A}$)

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