An alternating voltage source of peak voltage $V_{0}$ is connected across a series combination of a capacitance C and a variable resistance R as shown in figure. The value of R is adjusted such that maximum power is dissipated in it. Under such a condition an AC voltmeter connected across “C” would read

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$\frac{\sqrt{3} V_{0}}{\sqrt{2}}$

$\frac{V_{0}}{\sqrt{2}}$

$\frac{V_{0}}{2}$

$\frac{\sqrt{3} V_{0}}{2}$

answer is A.

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Detailed Solution

$i = \frac{ε ω C}{\sqrt{{(R ω C)}^{2} + 1}} (w h e r e ε = \frac{V_{0}}{\sqrt{2}})$

$P = i^{2} R = \frac{ε^{2} ω^{2} C^{2} R}{{(R ω C)}^{2} + 1} \frac{d P}{d R} = 0 \Rightarrow \frac{d}{d R} {\frac{R}{{(R ω C)}^{2} + 1}} = 0 \Rightarrow {{(R ω C)}^{2} + 1} - R [2 R {(ω C)}^{2}] = 0 {(R ω C)}^{2} = 1 \Rightarrow R = \frac{1}{ω C} L_{1} = \frac{m L_{1}}{m L_{2}} = \frac{2260}{333}$