TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Given, applied emf is
E=30‌sin‌200‌t
In given circuit,
L=0.05H,C=500µF
R=10Ω,ω=200rads−1
Now, inductive reactance, XL=Lω=0.05×200
=10Ω
Also, capacitive reactance,
XC=‌

1

Cω

=‌

1

500×10−6×200

=10Ω
As XL=XC,
circuit is in resonance, so amplitude of current is
Imax=‌

Emax

R

=‌

30

10

=3A