# Current through inductor in a circuit with a sinusoidal voltage source

1. Nov 9, 2012

### AbbeAbyss

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

http://www.wifstrand.se/albert/stuff/p7.7-1.png [Broken]

2. Relevant equations

vL = L * (diL / dt).

All of the inductors can be reduced to a single 8 H inductor, so

iL = (6 / 8) * ∫ cos 100t dt = 0.0075 sin 100t

if you integrate from 0 to t.

3. The attempt at a solution

The formula states that iL = iL(t0) + ∫ vL dt if you integrate from some initial condition current t0 to t. I get the correct answer if I integrate from 0 to t, but I fail to see what iL(t0) is and how we can use 0 for the lower bound.

It seems as if we can just assume that t0 = 0 and that iL(t0) = 0, if so, why is that?

Last edited by a moderator: May 6, 2017
2. Nov 9, 2012

### rude man

You have di/dt = V/L so
i = ∫(V/L)dt + constant (indefinite integral + a constant)

To evaluate the constant, asume V = 0 for t < 0. Then, at t = 0, a sudden voltage V = 6V is applied to L. What is the current then at t = 0? What is the constant? Do you see now why your last statement is correct?

Last edited by a moderator: May 6, 2017
3. Nov 10, 2012

### AbbeAbyss

Yes, I see now that at t = 0 (or rather t = 0+ denoting the exact moment the voltage 6 * cos 0 is applied) the current is 0 since the current through an inductor can't change instantaneously, thanks.