- #1
Sheve
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Homework Statement
Graph I(1) and I(2) vs. time. See attachment for illustration.
Homework Equations
V = IR
Back EMF = -L * dI/dt
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!