JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 7)
For a series LCR circuit with R = 100 $$\Omega$$, L = 0.5 mH and C = 0.1 pF connected across 220V$$-$$50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :
0$$^\circ$$, resistive circuit
$$ \approx $$ 90$$^\circ$$, predominantly inductive circuit
0$$^\circ$$, resonance circuit
$$ \approx $$ 90$$^\circ$$, predominantly capacitive circuit
Explanation
R = 100$$\Omega$$
$${X_L} = \omega L = 50\pi \times {10^{ - 3}}$$
$${X_C} = {1 \over {\omega C}} = {{{{10}^{11}}} \over {100\pi }}$$
$${X_C} > > {X_L}$$
& $$\left| {{X_C} - {X_L}} \right| > > R$$
$${X_L} = \omega L = 50\pi \times {10^{ - 3}}$$
$${X_C} = {1 \over {\omega C}} = {{{{10}^{11}}} \over {100\pi }}$$
$${X_C} > > {X_L}$$
& $$\left| {{X_C} - {X_L}} \right| > > R$$
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