This is a question on LC Circuits. The correct answer is given for the sake of r

Question

This is a question on LC Circuits. The correct answer is given for the sake of reference. To find Q1(t1), the formula is Q(t) = Qmax(cos(t + )). However, Qmax is equal to CV, so how is Qmax supposed to be found without a voltage?

A circuit is constructed with two capacitors and an inductor as shown. The values for the capacitors are: C 568 HF and C 182 HF The inductance is L 398 mH. At time t -0, the a current through the inductor has its maximum value lL(0)- 245 mA and it has the direction shown. 1) What is wo, the resonant frequency of this circuit? 135.0141333 radians/s Submit 2) What is Q1 (t1), the charge on the capacitor C1 at time t- t1 27.1 ms? The sign of Q1 is defined to be the sameas the sign of the potential difference V Va Vb at time t- t1 abo -8.9738e-4 C Submit

Explanation / Answer

Initially the current through the inductor is IL(0) = 245 mA. Now after some time the capacitors will be fully charged and the current through the inductor or the circuit would be zero.

The equivalent capacitance is C = [(568X182)/(568+182)]X10-6F = 137.83X10-6F

Charges on both the capacitors would be same as Q.

So energy of the equivalent capacitance C when it has maximum charge is,

EC = (1/2)CV2 = (1/2)Q2/C

This is equal to the initial energy stored in the inductor; which is,

EL = (1/2)LIL(0)2, when the current becomes zero, this energy of inductor is trnasferred to the equivalent capacitance of the circuit. From conservation of energy principle;

EC = EL

or,  (1/2)Q2/C = (1/2)LIL(0)2

or, Q2 = CL(IL(0))2 = (137.83X10-6F)(398X10-3H)(245X10-3A)2

so, Q = Qmax = 1.814X10-3C

Now, using the formual given by you;

Q(t) = Qmaxcos(t + )

at t = 0, Q(t) = 0, So we have;

0 = Qmaxcos(0 + )

or, cos() = 0, which makes,

= /2

So the equation becomes;

Q(t) = Qmaxcos(t + /2)

or, Q(t) = -Qmaxsint

so Q(27.1ms) = -(1.814X10-3C)sin[(135.014 rad/s)(27.1X10-3s)]

or Q(t1) = 8.97X10-4 C, will be the magnitude of charge, you can find the sign using the convention stated in the problem.....

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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