In the circuit shown in the figure., the equivalent resistance between point $A$ and $B$ is

\frac{3 R}{2}

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\frac{3 R}{4}

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\frac{3 R}{10}

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\frac{3 R}{12}

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Solution

The correct option is A $\frac{3 R}{2}$
In parallel
$\frac{1}{r e q} = \frac{1}{3 R} + \frac{1}{3 R} = \frac{2}{3 R}$
$R e q . = \frac{3 R}{2}$

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In the circuit shown in the figure., the equivalent resistance between point A and B is

The correct option is A 3R2In parallel1req=13R+13R=23RReq.=3R2

In the circuit shown in the figure., the equivalent resistance between point $A$ and $B$ is

The correct option is A $\frac{3 R}{2}$
In parallel
$\frac{1}{r e q} = \frac{1}{3 R} + \frac{1}{3 R} = \frac{2}{3 R}$
$R e q . = \frac{3 R}{2}$