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

Question

An L-C circuit consists of an inductor with an inductance of 60.0 mH and a capacitor with a capacitance of 300 F . The initial charge on the capacitor is 6.50 C , and the initial current in the inductor is zero

a)What is the maximum voltage across the capacitor? v

b)What is the maximum current in the inductor? A

C)What is the maximum energy stored in the inductor? J

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

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

Explanation / Answer

the maximum voltage across the capacitor:

Vmax = Q/C = 6.5/300 = 2.17x10-2 V

the maximum current in the inductor:

Imax = Q/sqrt[LC] = 1.53x10-3 A = 1.53 mA

the maximum energy stored in the inductor:

UL = 0.5*L*Imax2 = 70.2X10-9 J = 70.2 nJ

if I' =0.5Imax, then the ne energy is,

U'L = 0.25UL

Hence, the energy stored in the capacitor is,

UC = (3/4)UL

Hence, the charge is,

Q = sqrt[(3/2)ULC] = 6.52X10-6 C = 6.52 uC

if I' =0.5Imax, the energy stored in the inductor:

U'L = 70.2 nJ/4 = 17.55 nJ

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