Question : In a series LCR circuit, the voltage across R is 100 V, and R = 1KΩ, C=2μF. The resonant angular frequency is 200 rad/s. At resonance, the voltage across L is
a) 40 V
b) 100 V
c) 250 V
d) 500 V
Doubt by Deepanshi
Solution :
VR=100V
R=1KΩ=1000Ω
C=2μF=2×10-6F
ω=200 rad/s
It is given that the circuit is at resonance.
V=VR=100 V
R=1000Ω
I=V/R=100/1000
I=0.1 A
XC=1/ωC
XC=1/[(200)×(2×10^-6)]
XC=1000000/400
XC=10000/4
XC=2500Ω
At resonance
Voltage across Inductor = Voltage across Inductor
VL=VC
VL=IXC
VL=0.1×2500
VL=250 V
Hence, (c) 250V, would be the correct option.
