# Graphing current vs. time in an RL circut

1. Aug 7, 2010

### Sheve

1. The problem statement, all variables and given/known data

Graph I(1) and I(2) vs. time. See attachment for illustration.

2. Relevant equations

V = IR
Back EMF = -L * dI/dt

3. The attempt at a solution

I realize that an inductor is essentially the opposite of a capacitor when it comes to current and voltage: that is, the current starts small and grows with an exponential curve until it plateaus while voltage starts large and drops to nearly nothing. However, I'm not sure how to treat this in parallel with a resistor. Assuming an ideal inductor with no resistance, it should carry a lot of current at the start since the R(L) of the equation of the voltage drop over the inductor V(L) = I(L) * R(L) will be close to zero...but I know this can't be the case!

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2. Aug 9, 2010

### ehild

The current was zero before connecting the voltage source to the coil. You know Lenz's law: The back EMF induced in the coil opposes the change caused by switching on the the battery. The back EMF prevents any abrupt change, the current will grow gradually, not necessarily in an exponential way.

The voltage across the inductor is negative of the back emf:

U = L dI/dt

U is given and constant if it is an ideal voltage source. It does not matter whether you connect a resistor parallel with the coil or not. You can determine the rate of change of I through the coil. This current was zero before the voltage source was switched on. How does the current change with time?

ehild