In LC circuit the inductance L = 40 mH and capacitance C = 100 μ F. If a voltage V(t) = 10sin(314 t) is applied to the circuit, the current in the circuit is given as:

Question

In LC circuit the inductance L = 40 mH and capacitance C = 100 μ F. If a voltage V(t) = 10sin(314 t) is applied to the circuit, the current in the circuit is given as: (A) 10 cos 314 t (B) 0.52 cos 314 t (C) 5.2 cos 314 t (D) 0.52 sin 314 t

Tardigrade Admin · Accepted Answer

XL = ω L = 314 Ã 40 Ã 10-3 = 12.56 Ω XC = (1/ω C) = (1/314× 100× 10-6) = (104/314) = 31.84 Ω Phasor Vm = Im(XC - XL) 10 = Im(31.84 - 12.56) Im = (10/19.28) = 0.52A I = 0.52 sin (314t+(π/2))

Q. In $L C$ circuit the inductance $L = 40 m H$ and capacitance $C = 100 μ F$ . If a voltage $V (t) = 10 s in (314 t)$ is applied to the circuit, the current in the circuit is given as:

$10 cos 314 t$

$0.52 cos 314 t$

$5.2 cos 314 t$

$0.52 s in 314 t$

Q. 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:

10cos314t

0.52cos314t

5.2cos314t

0.52sin314t

XL​=ωL=314×40×10−3=12.56Ω XC​=ωC1​=314×100×10−61​ =314104​=31.84Ω Phasor Vm​=Im​(XC​−XL​) 10=Im​(31.84−12.56) Im​=19.2810​=0.52A I=0.52sin(314t+2π​)

Q. In $L C$ circuit the inductance $L = 40 m H$ and capacitance $C = 100 μ F$ . If a voltage $V (t) = 10 s in (314 t)$ is applied to the circuit, the current in the circuit is given as:

$10 cos 314 t$

$0.52 cos 314 t$

$5.2 cos 314 t$

$0.52 s in 314 t$