Energy stored in a RLC circuit

  • Thread starter danbone87
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  • #1
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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..
 

Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi danbone87,

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.

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.
 
  • #3
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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