An LCR circuit contains resistance of 110 Ω 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° . If on the other hand, only inductor is removed the current leads by 45° with the applied voltage. The rms current flowing in the circuit will be :

Question

An LCR circuit contains resistance of 110 Ω 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° . If on the other hand, only inductor is removed the current leads by 45° with the applied voltage. The rms current flowing in the circuit will be : (A) 1 A (B) 2.5 A (C) 1.5 A (D) 2 A

Tardigrade Admin · Accepted Answer

tan 45°=(1/ω C R)=(ω L/R) ⇒ XL=XC ⇒ resonance i =( V / R ) =(220/110)=2 A

$1 A$

$2.5 A$

$1.5 A$

$2 A$

$tan 4 5^{\circ} = \frac{1}{ω CR} = \frac{ω L}{R}$
$\Rightarrow X_{L} = X_{C}$
$\Rightarrow$ resonance
$i = \frac{V}{R}$
$= \frac{220}{110} = 2 A$

1A

2.5A

1.5A

2A

tan45∘=ωCR1​=RωL​ ⇒XL​=XC​ ⇒ resonance i=RV​ =110220​=2A

$1 A$

$2.5 A$

$1.5 A$

$2 A$

$tan 4 5^{\circ} = \frac{1}{ω CR} = \frac{ω L}{R}$
$\Rightarrow X_{L} = X_{C}$
$\Rightarrow$ resonance
$i = \frac{V}{R}$
$= \frac{220}{110} = 2 A$