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

A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of the source is constant at 10 V. Box P contains a capacitance of 1 μF in series with a resistance of 32 Ω. Coil Q has a self inductance of 4.9 mH and a resistance of 68 Ω in series. The frequency is adjusted so that maximum current flows in P and Q.

The impedance of Q at this frequency is

A
200 Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1350 Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
55 Ω
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
9524 Ω
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 9524 Ω
For impedence in coil Q
Z=R22+(XL)2 ....(i)
Now if the maximum current flows through P and Q, then circuit is at resonance. Also at resonance ω=1LC
ω=14.9×103×106ω=1057
So we can find XL as
XL=ωL=1057×4.9×103=70 Ω
Putting values in eq(i)
Z=682+702=9524 Ω

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