In the figure shown the power generated in y | vedclass

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Question

The current $i$ is$:$

$i=\frac{10}{2+R+y}$

Power, $P=i^{2} R$

$P=\frac{100}{(2+R+y)^{2}} y$

The power will be maximum where $d P / d y=0$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

The current $i$ is$:$

$i=\frac{10}{2+R+y}$

Power, $P=i^{2} R$

$P=\frac{100}{(2+R+y)^{2}} y$

The power will be maximum where $d P / d y=0$

Differentiating by parts we will get $(2+y+R)^{2}(1)-y(2)(2+R+y)=0$

Or $2+R+y=2 y$

Or $2+R=y$

For $y=5 \Omega$

$R=3 \Omega$

In the figure shown the power generated in $y$ is maximum when $y = 5\,\Omega$ . Then $R$ is .............. $\Omega$

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