JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 25)
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
900
Explanation
Given, angular frequency at resonance, $$\omega$$0 = 105 rads$$-$$1
Power drawn from circuit, P = 16 W
and supply voltage, V = 120 V
Let resistance of circuit = R.
As, $$P = {V^2}/R$$
$$ \Rightarrow R = {V^2}/P = {{120 \times 100} \over {16}}$$
$$ = 30 \times 30 = 900\,\Omega $$
Power drawn from circuit, P = 16 W
and supply voltage, V = 120 V
Let resistance of circuit = R.
As, $$P = {V^2}/R$$
$$ \Rightarrow R = {V^2}/P = {{120 \times 100} \over {16}}$$
$$ = 30 \times 30 = 900\,\Omega $$
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