wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C μF across a 200 V,50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is ϕ.

Assume, π35.

The value of C is -

A
100.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
100.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
100
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

V2C+V2R=Erms

V2C+1002=2002

VC=1003 V

We know that,

i=PVR=500100=5 A

Now,

VC=iXC

C=iVC(2πf)=51003×2π×50

C=1×104 F=100×106 F=100 μF

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon