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In a series L - C - R circuit frequency of the source is constant and power dissipated in the circuit is maximum when L = 80 mH and $C = 100 μ F .$ If inductance of the inductor is changed to 100 mH, what will be the capacitance of the capacitor for which the power dissipated in the circuit will be again maximum ?

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$80 μ F$

$120 μ F$

$60 μ F$

$140 μ F$

answer is A.

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Detailed Solution

(1)

Power dissipated in the circuit is maximum when

$\begin{array}{l} ω L = \frac{1}{ω C} \\ i . e . w h e n ω^{2} = \frac{1}{L C} \\ ∴ \frac{1}{L^{'} C^{'}} = \frac{1}{L C} \\ \Rightarrow {C^{'}}^{'} = \frac{L}{L^{'}} . C = \frac{80}{100} \times 100 μ F = 80 μ F \end{array}$

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