JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 1)
An LCR circuit contains resistance of 110$$\Omega$$ and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45$$^\circ$$. If on the other hand, only inductor is removed the current leads by 45$$^\circ$$ with the applied voltage. The rms current flowing in the circuit will be :
1 A
2.5 A
2 A
1.5 A
Explanation
Since $$\phi $$ remain same, circuit is in resonance
$$ \therefore $$ $${I_{rms}} = {{{v_{rms}}} \over z}$$ = $${{{v_{rms}}} \over R}$$
$$ = {{220} \over {110}}$$
$$ \Rightarrow $$ $${I_{rms}} = 2A$$
$$ \therefore $$ $${I_{rms}} = {{{v_{rms}}} \over z}$$ = $${{{v_{rms}}} \over R}$$
$$ = {{220} \over {110}}$$
$$ \Rightarrow $$ $${I_{rms}} = 2A$$
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