A battery of internal resistance $4$ $Ω$ is connected tot he network of resistance as shown. In order to give the maximum power to the network, the value of $R$ in $Ω$ should be:

\frac{4}{9}

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\frac{8}{9}

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2

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18

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Solution

The correct option is C $2$
Effective resistance = $\frac{1}{3} R + \frac{1}{6} R + \frac{3}{6} R = \frac{6}{3}$

$= 2 R$

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A battery of internal resistance $4 Ω$ is connected to the network of resistance as shown. In order to deliver the maximum power to the network, the value of R in ohm should be:

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A battery of internal resistance 4 Ω is connected tot he network of resistance as shown. In order to give the maximum power to the network, the value of R in Ω should be:

The correct option is C 2Effective resistance =13R+16R+36R=63=2R

A battery of internal resistance $4$ $Ω$ is connected tot he network of resistance as shown. In order to give the maximum power to the network, the value of $R$ in $Ω$ should be:

The correct option is C $2$
Effective resistance = $\frac{1}{3} R + \frac{1}{6} R + \frac{3}{6} R = \frac{6}{3}$

$= 2 R$