An L - C circuit consists of an inductor with an inductance of 60.0mH and a capa

Question

An L-C circuit consists of an inductor with an inductance of 60.0mH and a capacitor with a capacitance of 200uF . The initial charge on the capacitor is 6.50uC , and the initial current in the inductor is zero.

Part A

What is the maximum voltage across the capacitor?

Part B

What is the maximum current in the inductor?

Part C

What is the maximum energy stored in the inductor?

Part D

When the current in the inductor has half its maximum value, what is the charge on the capacitor?

Part E

When the current in the inductor has half its maximum value, what is the energy stored in the inductor?

Explanation / Answer

Solution: Given L = 60 mH, C = 200 F and Q = 6.5 C.

Part A: Maximum voltage across the capacitor V = q/C = 0.032 V.

Part B: maximum currant in the circuit imax = q? = q/(LC)1/2 = 1.876 A.

Part C: maximum energy stored in the inductor = Limax2 = 0.105 J.

Part D: When the current in the inductor has half its maximum value, the energy stored in inductor is E = 0.026 J. Rest of the energy (0.079 J) will be stored in the capacitor (q2/2C).

So, q = 5.62 C.

Part E: When the current in the inductor has half its maximum value, the energy stored in inductor is E = 0.026 J.

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