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Question

Physics Question on LCR Circuit

An AC generator producing 10V(rms) at 200 rad/s is connected in series with a 50Ω50 \Omega resistor, a 400 mH inductor and a 200μF200 \mu F capacitor. The rms voltage across the inductor is

A

2.5 V

B

3.4 V

C

6.7 V

D

10.8 V

Answer

10.8 V

Explanation

Solution

Given E=10VE =10 V
ω=200rad/s\omega =200 \,rad / s
R=50ΩR =50 \,\Omega
L=400mHL =400 \,mH
C=200μFC =200 \,\mu F
We know that,
Z=R2+(XLXC)2Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}
=(50)2+(8025)2=(50)2(55)2=\sqrt{(50)^{2}+(80-25)^{2}}=\sqrt{(50)^{2}-(55)^{2}}
=2500+3025=\sqrt{2500+3025}
Z=5525=74.3ΩZ =\sqrt{5525}=74.3 \,\Omega
I=EZ=1074.3=0.13459AI =\frac{E}{Z}=\frac{10}{74.3}=0.13459 A
EL=IXL=0.13459×80E_{L} =I X_{L}=0.13459 \times 80
=10.76V or 10.8V=10.76 V \text { or } 10.8 V