Five equal resistance each of resistance $R$ are conneted a shown in figure. A battery of $V$ volts is connected between $A$ and $B$ . The current flowing in $A F C E B$ will be

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\frac{V}{R}

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\frac{V}{2 R}

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\frac{2 V}{R}

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\frac{3 V}{R}

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Solution

The correct option is B $\frac{V}{2 R}$
Rearranging the circuit

It is a balanced wheat stone bridge $∴$ No current will pass through $C D$

Equivalent resistance of circuit, $R_{e q} = R$
$∴$ Current $I = \frac{V}{R}$
Now, $i_{1} = i_{2} = \frac{I}{2} = \frac{V}{2 R}$
So, current flowing in $A F C E B = i_{1} = \frac{V}{2 R}$
$∴$ option $(b)$ is correct.

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Five equal resistance each of resistance R are conneted a shown in figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be ​​​​​​​[1 Mark]

The correct option is B V2RRearranging the circuit It is a balanced wheat stone bridge ∴ No current will pass through CD Equivalent resistance of circuit, Req=R ∴ Current I=VR Now, i1=i2=I2=V2R So, current flowing in AFCEB=i1=V2R ∴ option (b) is correct.

Five equal resistance each of resistance $R$ are conneted a shown in figure. A battery of $V$ volts is connected between $A$ and $B$ . The current flowing in $A F C E B$ will be

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