In the figure shown,the power generated in y | vedclass

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(D) The current $i$ in the circuit is given by:$i = \frac{10}{2 + R + y}$The power $P$ generated in the variable resistor $y$ is:$P = i^2 y = \left( \

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(D) The current $i$ in the circuit is given by:$i = \frac{10}{2 + R + y}$The power $P$ generated in the variable resistor $y$ is:$P = i^2 y = \left( \frac{10}{2 + R + y} \right)^2 y = \frac{100y}{(2 + R + y)^2}$For the power to be maximum,the derivative of $P$ with respect to $y$ must be zero:$\frac{dP}{dy} = 100 \left[ \frac{(2 + R + y)^2(1) - y(2)(2 + R + y)}{(2 + R + y)^4} \right] = 0$This simplifies to:$(2 + R + y) - 2y = 0$$2 + R - y = 0$$y = 2 + R$Given that the power is maximum when $y = 5\,\Omega$,we substitute this value:$5 = 2 + R$$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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