Five resistors,each of resistance R,are arran | vedclass

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(B) The circuit can be redrawn as a Wheatstone bridge. By observing the symmetry,the potential at the top node $C$ and the bottom node $D$ are equal r

Vedclass · Accepted Answer

$V/2R$

Vedclass · Answer

(B) The circuit can be redrawn as a Wheatstone bridge. By observing the symmetry,the potential at the top node $C$ and the bottom node $D$ are equal relative to the input terminals $A$ and $B$,or more simply,the bridge is balanced. In a balanced Wheatstone bridge,no current flows through the central resistor connected between nodes $C$ and $D$. Thus,the circuit simplifies to two parallel branches,each consisting of two resistors of resistance $R$ in series. The resistance of each branch is $R + R = 2R$. Since these two branches are in parallel,the total equivalent resistance $R_{eq}$ is given by $1/R_{eq} = 1/(2R) + 1/(2R) = 2/(2R) = 1/R$,so $R_{eq} = R$. The total current drawn from the battery is $I = V/R_{eq} = V/R$. Since the two branches are identical (each having resistance $2R$),the current $I$ splits equally between them. Therefore,the current flowing through the branch $AFCEB$ is $I/2 = (V/R) / 2 = V/(2R)$.

Five resistors,each of resistance $R$,are arranged as shown in the figure. $A$ battery of voltage $V$ is connected between points $A$ and $B$. Find the current flowing through the branch $AFCEB$.

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