The current drawn from the battery in the giv | vedclass

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(A) The given circuit is a balanced Wheatstone bridge because the ratio of resistances in the arms is $\frac{3}{2} = \frac{3}{2}$.Since the bridge is

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$2.4$

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(A) The given circuit is a balanced Wheatstone bridge because the ratio of resistances in the arms is $\frac{3}{2} = \frac{3}{2}$.Since the bridge is balanced,no current flows through the central $5 \Omega$ resistor.Therefore,the $5 \Omega$ resistor can be removed from the circuit.Now,the upper branch consists of a $3 \Omega$ and $2 \Omega$ resistor in series,giving a total resistance of $3 + 2 = 5 \Omega$.The lower branch also consists of a $3 \Omega$ and $2 \Omega$ resistor in series,giving a total resistance of $3 + 2 = 5 \Omega$.These two branches are connected in parallel across the $6 	ext{ V}$ battery.The equivalent resistance $R_{eq}$ of the two $5 \Omega$ resistors in parallel is $R_{eq} = \frac{5 	imes 5}{5 + 5} = 2.5 \Omega$.The current drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{6}{2.5} = 2.4 	ext{ A}$.

The current drawn from the battery in the given network is (Internal resistance of the battery is negligible). (in $A$)

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