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 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

B)

Maximum current in the inductor is

I = Q / sqrt ( LC )

= (6.50*10^ -6 C ) / sqrt ( (60*10^ -3 H )(200*10^ -6 F ))

=0.00187 A

= 1.87 mA

c)

energy stored in the inductor is

E = ( 1/ 2) L I ^2

= (0. 5 ) (60*10^ -3 H ) ( 0.00187 A ) ^2

= 1.0567*10^ -7 J

d)

Angular frequency

w = 1/ sqrt ( LC ) = 1/ sqrt ( 200*10^ -6 F ) ( 60*10^-3 H) = 288.67

when the current in the inductor is half of maximum value, charge on the capacitor is also decreases to half

Q = 6.50 / 2 = 3.25 micro C.

e)

Energy stored in the inductor is

E = ( 1/ 2) L I ^2

= ( 1/ 2) L ( I /2 ) ^2

= (60*10^ -3 H ) (0.00187 A) ^2 / 8

= 2.622*10^ -8 J

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