JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 8)
A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k$$\Omega$$, an inductor of inductive reactance XL = 250 $$\pi$$$$\Omega$$ and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take $$\pi$$2 = 10)
4 $$\mu$$F
25 $$\mu$$F
400 $$\mu$$F
40 $$\mu$$F
Explanation
From maximum average power
XL = XC
250$$\pi$$ = $${1 \over {2\pi (50)C}}$$
$$ \Rightarrow $$ C = 4 $$\times$$ 10$$-$$6
XL = XC
250$$\pi$$ = $${1 \over {2\pi (50)C}}$$
$$ \Rightarrow $$ C = 4 $$\times$$ 10$$-$$6
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