JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, inductance, L=100mH
=100×10−3H
Capacitance, C=100µF
=100×10−6F
Resistance, R=120Ω
Maximum voltage, V0=30V
Angular frequency, ω=100rads−1
Impedance, Z=√(XL−XC)2+R2
Here, XL=ωL=100×100×10−3=10Ω
∴‌‌XC‌‌=‌

1

ωC

=‌

1

100×100×10−6

=100Ω
⇒‌‌Z‌‌=√XC−XL)2+R2
‌‌=√(100−10)2+(120)2=√(902+1202=153.3Ω
As we know that, peak current,
I0=‌

V0

Z

=‌

30

153.3

=0.195A∼eq0.2A
Resonance frequency,
∴‌‌f=‌

1

2π√LC

=‌

1

2π√100×10−3×100×10−6

⇒‌‌f=‌

1

2π√10−5

=‌

1

2π×10−2√10−1

=‌

100π

2π

=50Hz