How to Calculate Energy Stored in an RLC Circuit?

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SUMMARY

The discussion focuses on calculating the energy stored in an RLC circuit, specifically involving a 21.0 µF capacitor charged by a 160.0 V power supply and a 0.220 mH inductor. At time t = 0 ms, the energy stored in the capacitor is calculated using the formula U = 0.5 * C * V^2, resulting in 1/2 * 21.0 µF * (160 V)^2. At t = 1.30 ms, the energy in the inductor is determined using the formula U = 0.5 * L * I^2, where I is derived from the current equation i = -ωQsin(ωt).

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danbone87
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Homework Statement



A 21.0 \mu F capacitor is charged by a 160.0-{\rm V} power supply, then disconnected from the power and connected in series with a 0.220-{\rm mH} inductor.

Calculate the energy stored in the capacitor at time t = 0{\rm{ ms}} (the moment of connection with the inductor).

Calculate the energy stored in the inductor at t = 1.30 ms.

Homework Equations



i=-omegaQsin(omega*t)

and U=.5LI^2

I need to either find charge q or a resistance value to find I with and I'm stuck..
 
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Hi danbone87,

danbone87 said:
A 21.0 \mu F capacitor is charged by a 160.0-{\rm V} power supply,

The way the capacitor is initially connected by itself to the battery will give you its initaial Q value.

danbone87 said:
I need to either find charge q or a resistance value to find I with and I'm stuck..

If I'm reading the problem correctly, there is no resistor; just a capacitor connected to an inductor.
 
A 21.0 \mu F capacitor is charged by a 160.0-{\rm V} power supply, then disconnected from the power and connected in series with a 0.220-{\rm mH} inductor.

Calculate the energy stored in the capacitor at time t = 0{\rm{ ms}} (the moment of connection with the inductor).

Calculate the energy stored in the inductor at t = 1.30 ms.ans= q=cv=22u*160m gives u charge
energy=.5*22u*160m*160m=1/2(c*v*v)
now
VL=Ldi/dt
 

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