CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

When an L–R combination is connected in series with 12V – 50 Hz supply, a current of 0.5A flows through the combination. The current differs in phase from applied voltage by π3 rad. Then resistance is

A
24Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12Ω
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 12Ω
Z=VI=120.5=24Ω
Cos 60=RZ
R=Z2=242=12Ω

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Alternating Current Applied to an Inductor
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon