Question

# A series LCR circuit with 100 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60∘. When only the inductance is removed, the current leads the voltage by 60∘. The average power dissipated in the LCR circuit is -

A
100 W
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
200 W
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
300 W
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
400 W
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 400 WWhen capacitance is removed, then tanϕ=XLR ⇒tan60∘=XLR ⇒XL=√3R ...(1) When inductance is removed, then tanϕ=XCR ⇒tan60∘=XCR ⇒XC=√3R ...(2) From equations (1) and (2), XL=XC ⇒ LCR circuit is in resonance. So, Z=R Therefore, irms=VrmsZ=200100=2 A Further, power dissipated in the LCR circuit, Pavg=Vrmsirmscosϕ ⇒Pavg=200×2×cos0∘ [At resonance, ϕ=0∘] ⇒Pavg=400 W Hence, option (D) is the correct answer.

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Motional EMF
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program