JEE MAIN - Physics (2006 - No. 12)
In a series resonant $$LCR$$ circuit, the voltage across $$R$$ is $$100$$ volts and $$R = 1\,k\Omega $$ with $$C = 2\mu F.$$ The resonant frequency $$\omega $$ is $$200$$ $$rad/s$$. At resonance the voltage across $$L$$ is
$$2.5 \times {10^{ - 2}}V$$
$$40$$ $$V$$
$$250$$ $$V$$
$$4 \times {10^{ - 3}}V$$
Explanation
Across resistor, $$I = {V \over R} = {{100} \over {1000}} = 0.1A$$
At resonance,
$${X_L} = {X_C} = {1 \over {\omega C}}$$
$$ = {1 \over {200 \times 2 \times {{10}^{ - 6}}}} = 2500$$
Voltage across $$L$$ is
$$I{X_L} = 0.1 \times 2500 = 250V$$
At resonance,
$${X_L} = {X_C} = {1 \over {\omega C}}$$
$$ = {1 \over {200 \times 2 \times {{10}^{ - 6}}}} = 2500$$
Voltage across $$L$$ is
$$I{X_L} = 0.1 \times 2500 = 250V$$
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