A series L-C-R circuit is designed to resonate at an angular frequency $\omega$0 = 105 rad/s. The circuit draws 16W power from 120V source at resonance. The value of resistance 'R' in the circuit is _________ $\Omega$.
Answer (integer)
900
Solution
Given, angular frequency at resonance, $\omega$<sub>0</sub> = 10<sup>5</sup> rads<sup>$-$1</sup><br/><br/>Power drawn from circuit, P = 16 W<br/><br/>and supply voltage, V = 120 V<br/><br/>Let resistance of circuit = R.<br/><br/>As, $P = {V^2}/R$<br/><br/>$\Rightarrow R = {V^2}/P = {{120 \times 100} \over {16}}$<br/><br/>$= 30 \times 30 = 900\,\Omega$
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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