
z=R2+(XC−XL)2
R=0
Z=XC−XL
=ωC1−ωL
=314×100×10−61−314×40×10−3
=31.84−12.56
=19.28Ω

i=ZV0sin(314t+2π)
∴i=ZV0cos(314t)⇒i=19.2810cos(314t)
⇒i=0.52cos(314t)
JEE Main 2020 — Physics Electromagnetism
In LC circuit the inductance L=40mH and capacitance C=100μF. If a voltage V(t)=10sin(314t) is applied to the circuit, the current in the circuit is given as:
Held on 9 Jan 2020 · Verified 6 Jul 2026.
0.52cos(314t)
10cos(314t)
5.2cos(314t)
0.52sin(314t)
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