An L C R circuit contains resistance of 100 Ω and a supply of 200 V at 300 rad angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by 60°. If, on the other hand, only inductor is taken out, the current leads by 60° with the applied voltage. The current flowing in the circuit is

Question

An L C R circuit contains resistance of 100 Ω and a supply of 200 V at 300 rad angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by 60°. If, on the other hand, only inductor is taken out, the current leads by 60° with the applied voltage. The current flowing in the circuit is (A) 1 A (B) 1.5 A (C) 2 A (D) 2.5 A

Tardigrade Admin · Accepted Answer

According to the given question, tan 60°=(ω L/R) text and tan 60°=(1 / ω C/R) ∴ ω L=(1 / ω C)( case of resonance) Now Z=√R2+(ω L-(1/ω C))2=100 Ω ∴ I rms =(E rms /Z)=(200 V /100 Ω)=2 A

$1 A$

$1.5 A$

$2 A$

$2.5 A$

According to the given question,
$tan 6 0^{\circ} = \frac{ω L}{R} and tan 6 0^{\circ} = \frac{1/ ω C}{R}$
$∴ ω L = (1/ ω C)$ ( case of resonance)
Now $Z = R^{2} + (ω L - \frac{1}{ω C})^{2} = 100 Ω$
$∴$ $I_{r m s} = \frac{E _{r m s}}{Z} = \frac{200 V}{100 Ω} = 2 A$

1A

1.5A

2A

2.5A

According to the given question, tan60∘=RωL​ and tan60∘=R1/ωC​ ∴ωL=(1/ωC)( case of resonance) Now Z=R2+(ωL−ωC1​)2​=100Ω ∴ Irms​=ZErms​​=100Ω200V​=2A

$1 A$

$1.5 A$

$2 A$

$2.5 A$

According to the given question,
$tan 6 0^{\circ} = \frac{ω L}{R} and tan 6 0^{\circ} = \frac{1/ ω C}{R}$
$∴ ω L = (1/ ω C)$ ( case of resonance)
Now $Z = R^{2} + (ω L - \frac{1}{ω C})^{2} = 100 Ω$
$∴$ $I_{r m s} = \frac{E _{r m s}}{Z} = \frac{200 V}{100 Ω} = 2 A$