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Question

(A) The circuit can be redrawn as a Wheatstone bridge. The resistances are $R_{FC} = 2R$,$R_{FD} = 2R$,$R_{CE} = 2R$,and $R_{DE} = 2R$. The resistance

Vedclass · Accepted Answer

$\frac{V}{4R}$

Vedclass · Answer

(A) The circuit can be redrawn as a Wheatstone bridge. The resistances are $R_{FC} = 2R$,$R_{FD} = 2R$,$R_{CE} = 2R$,and $R_{DE} = 2R$. The resistance between $C$ and $D$ is $2R$.Since $\frac{R_{FC}}{R_{FD}} = \frac{2R}{2R} = 1$ and $\frac{R_{CE}}{R_{DE}} = \frac{2R}{2R} = 1$,the bridge is balanced.Therefore,no current flows through the central resistance $CD$.The circuit simplifies to two parallel branches,each containing two resistances of $2R$ in series.The equivalent resistance of each branch is $2R + 2R = 4R$.The current through the branch $FC$ is $I_{FC} = \frac{V}{4R}$.

Five equal resistances each of $2R$ are connected as shown in the figure. $A$ battery of $V$ volts is connected between $A$ and $B$. Then,the current through $FC$ is:

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