A 1.0 E -6 F capacitor has a charge of 10.0 micro C. It is connected to a 1.0 H inductor at t = 0. Compare the potential energy of the capacitor to the potential energy of the inductor at t = 10s. Which energy is greater?

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire

Recognitions:
Homework Help
 Quote by Partap03 A 1.0 E -6 F capacitor has a charge of 10.0 micro C. It is connected to a 1.0 H inductor at t = 0. Compare the potential energy of the capacitor to the potential energy of the inductor at t = 10s. Which energy is greater?
Welcome to PF.

Connected how?

And what are your thoughts on it?

 First of all thanks. My thoughts I know that potential energy is basically voltage across each component. Also, C = Q / (change in V) ; C is the capacitance, Q is the charge, and V is volts. solving for change in V and we get V = Q / C Am I going in the right direction?

Recognitions:
Homework Help

 Quote by Partap03 First of all thanks. My thoughts I know that potential energy is basically voltage across each component. Also, C = Q / (change in V) ; C is the capacitance, Q is the charge, and V is volts. solving for change in V and we get V = Q / C Am I going in the right direction?
First of all I don't know how it's connected. Is the inductor in || to the capacitor? What is the circuit?

 Quote by LowlyPion First of all I don't know how it's connected. Is the inductor in || to the capacitor? What is the circuit?
That's all the problem says. Lets assume that it is in series then how would I approach the problem?

Recognitions:
Homework Help
 Quote by Partap03 That's all the problem says. Lets assume that it is in series then how would I approach the problem?
If it is in series then I think it matters what the rest of the circuit would be.

If it is in || then one is a short to the other isn't it?

Isn't the decay constant pretty swift?

 Tags capacitor, circuit help, electricitymagnetism, homework help, physics help