JEE Main Physics Class 12 Alternating Currents Part 1 Questions

The circuit diagram can be drawn as, Average power of an L-C-R circuit is given by
Pav=Vrmsirms‌cos‌ϕ
For Pav to be maximum, cos‌ϕ=1 ...(i)
We know that,
cos‌ϕ=

R

Z

,
where, R = resistance
and Z = impedance of L-C-R circuit.
Now,
Z=√(XL−XC)2+R2
As, cos‌ϕ=1=

R

Z

[using Eq, (i)]
⇒ R = Z
⇒R=√(XL−XC)2+R2
Squaring both sides, we get
R2=(XL−XC)2+R2
⇒(XL−XC)2=0
⇒XL=XC
Since, XC=

1

ωC

=

1

2πfC

and XL=250πΩ (given)
⇒250π=

1

2π×50×C

⇒C=

1

2π2×50×250

⇒C=4×10−6F=4μF