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In L C circuit with the inductance L = 40 mH and capacitance C = 100 μF. If a voltage V(t) = 10 sin (314 t) is applied to the circuit, the current in the circuit is given as -
- 10 cos 314 t
- 0.52 sin 314 t
- 5.2 cos 314 t
- 0.52 cos 314 t
Correct answer: 0.52 cos 314 t
Solution
X_L = wL = 314*0.04 = 12.6 ohm; X_C = 1/(wC) = 1/(314*100e-6) = 31.8 ohm. Net X = 12.6 - 31.8 = -19.3 ohm (capacitive), |Z| = 19.3, so I_max = 10/19.3 = 0.52 A. Capacitive circuit -> current leads V by 90deg, so i = 0.52 cos(314 t).
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