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(A) The given circuit can be redrawn as a Wheatstone bridge. The resistors are arranged such that the bridge is balanced because the ratio of resistan

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(A) The given circuit can be redrawn as a Wheatstone bridge. The resistors are arranged such that the bridge is balanced because the ratio of resistances in the arms is $\frac{3}{3} = \frac{2}{2} = 1$. In a balanced Wheatstone bridge,no current flows through the central $5 \ \Omega$ resistor. Thus,the circuit simplifies to two parallel branches,each consisting of two resistors in series. The upper branch has a resistance of $3 \ \Omega + 2 \ \Omega = 5 \ \Omega$. The lower branch has a resistance of $3 \ \Omega + 2 \ \Omega = 5 \ \Omega$. The equivalent resistance $R_{eq}$ of these two parallel branches is given by $\frac{1}{R_{eq}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$,which gives $R_{eq} = 2.5 \ \Omega$. The current $I$ drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{6 \ V}{2.5 \ \Omega} = 2.4 \ A$.

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

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