The current drawn from the battery in the giv | vedclass

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(C) The given circuit can be redrawn as a Wheatstone bridge as shown in the figure.Let the resistors be $R_1 = 6 \ \Omega$,$R_2 = 6 \ \Omega$,$R_3 = 4

Vedclass · Accepted Answer

$2.4$

Vedclass · Answer

(C) The given circuit can be redrawn as a Wheatstone bridge as shown in the figure.Let the resistors be $R_1 = 6 \ \Omega$,$R_2 = 6 \ \Omega$,$R_3 = 4 \ \Omega$,$R_4 = 4 \ \Omega$,and the central resistor $R_5 = 10 \ \Omega$.The condition for a balanced Wheatstone bridge is $\frac{R_1}{R_3} = \frac{R_2}{R_4}$.Here,$\frac{6}{4} = 1.5$ and $\frac{6}{4} = 1.5$.Since $\frac{R_1}{R_3} = \frac{R_2}{R_4}$,the bridge is balanced,and no current flows through the $10 \ \Omega$ resistor.Thus,the circuit simplifies to two parallel branches,each containing two resistors in series.The upper branch has resistance $R_{up} = 6 \ \Omega + 4 \ \Omega = 10 \ \Omega$.The lower branch has resistance $R_{low} = 6 \ \Omega + 4 \ \Omega = 10 \ \Omega$.The equivalent resistance $R_{eq}$ of the circuit is given by $\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$.Therefore,$R_{eq} = 5 \ \Omega$.The current drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{12 \ V}{5 \ \Omega} = 2.4 \ A$.

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

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