The reading of the ammeter in amperes for the | vedclass

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(A) The circuit is a balanced Wheatstone bridge. Let the nodes be labeled. The central resistor of $6 \ \Omega$ is connected between the two middle po

Vedclass · Accepted Answer

$0.33$

Vedclass · Answer

(A) The circuit is a balanced Wheatstone bridge. Let the nodes be labeled. The central resistor of $6 \ \Omega$ is connected between the two middle points. Due to the symmetry of the circuit,the potential difference across the central $6 \ \Omega$ resistor is zero,so no current flows through it.Thus,the circuit simplifies to two parallel branches,each consisting of two $6 \ \Omega$ resistors in series.Resistance of the upper branch = $6 \ \Omega + 6 \ \Omega = 12 \ \Omega$.Resistance of the lower branch = $6 \ \Omega + 6 \ \Omega = 12 \ \Omega$.These two branches are in parallel,so the equivalent resistance $R_{eq}$ is:$1/R_{eq} = 1/12 + 1/12 = 2/12 = 1/6 \ \Omega^{-1} \Rightarrow R_{eq} = 6 \ \Omega$.The total current $i$ drawn from the $2 \ V$ battery is:$i = V / R_{eq} = 2 \ V / 6 \ \Omega = 1/3 \ A \approx 0.33 \ A$.The ammeter is placed in the main circuit,so it measures the total current $i = 0.33 \ A$.

The reading of the ammeter in amperes for the following circuit is:

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