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JEE PYQ: Alternating Current Question 12

Question 12 - 2021 (24 Feb Shift 2)

A series LCR circuit is designed to resonate at an angular frequency $\omega_0 = 10^5$ rad/s. The circuit draws 16 W power from 120 V source at resonance. The value of resistance ‘$R$’ in the circuit is ____ $\Omega$.

Show Answer

Answer: 900

Solution

$P = \frac{V^2}{R}$. $16 = \frac{120^2}{R} \Rightarrow R = \frac{14400}{16} = 900\Omega$.


Learning Progress: Step 12 of 38 in this series