Current decay in closed circuit with only 1 resistor

[GarethB84]

Hi this is my first post so please be nice :)

1. Homework Statement
The problem has been set up that a circuit containing a resistor and switch is being driven by a power source of unkown potential. The switch is flick to short the power source leaving only the resistor in a closed loop.
Calculate the current decay.
Initial current = 100A
Resistance = 1nLaTeX Code: \\Omega


2. Homework Equations

V=IR
Power = I2R



3. The Attempt at a Solution

I'm not too sure. The power generated, and hence the heat loss can be found to be 1e-5W so that's 1e-5 Joules per second of heat loss.

I'm not sure if we have enough information. Current decay involving inductance or capacitance is well known but just a resistor in a circuit of 'ideal' conductors?

I'm assuming it is an exponential decay of form I=I0e-t/LaTeX Code: \\tau but finding the decay constant is proving tricky. Any advice?

Thanks in advance.

 

[Defennder]

Ok I'm a little confused here. If we're dealing with ideal components then it's clear with 0 capacitance and inductance the potential across the resistor drops immediately to 0 when the switch is closed since the current stops flowing in the absence of a source. So you can't make use of power dissipated in resistor because that will give you 0. Did you omit any information? Seems like you're not given enough.

 

[GarethB84]

Hi, thanks for the reply.

This is a sub question on a rather larger topic of superconducting magnets. I understand that these magnets can be run in persistent mode, whereby the power source is removed, and due to superconductivity, the current will remain flowing indefinetly. As soon as resistance is introduced into the circuit, the current will decay- the decay time varying with the size of the resistance.

What I'm unclear on is how to find this constant, and the theory behind persistent mode. Is it true the reason for persistent flow is that the energy stored in the magnet (inductance I suppose) pushes the current round the circuit. This energy doesn't decay as no ohmic heating occurs. Am I on the right line here??

If so, then will I have to assume a sensible value of inductance and work from there??


Thanks in advance,

Gareth