# Inductors in an AC circuit

by slayhy
Tags: circuit, inductors
 P: 72 Inductors in an AC circuit An inductor driven by an AC source conducts current that is +90 degrees out of phase with the voltage. Since the problem states that the stored energy is zero, implying that the current is zero at time t = 0, then the equation used to find the current looks the same as a resistor with a value of 7.54 Ω. Once you know the current you will be able to solve for the stored energy. your expression i(t)=(120\sqrt{2}) sin(377x(1/180) - ∏/2) is not right since it disagrees with the statement that the energy is zero at time t = 0. The part of learning electronics that is most frustrating is not knowing or being confident enough to recognize the shortcuts that are given in the problem. In this problem, they attempt to make it easy on you by asking for the value at t = 1/180, since that is 1/3 of a period of the 60Hz source. The condition that I = 0 at time t = 0 tells you the phase of the current must be 0 or ∏ at time t = 0, so the phase 1/3 of a period after that is 2∏/3 or 5∏/3. After that all you need to know is Imax, which you already found. Check your formula for the energy stored in an inductor. Despite the simplifications that they use to introduce students to the topic, I think it just confuses you down the line. This problem is worded in a way to avoid using complex numbers, but complex numbers make the math so much easier... or at least consistent. As soon as you add a resistor to this problem, you can no longer solve it easily using previous method. You already know that the voltage waveform from a standard electrical outlet is $120 * √2 * Cos(2\pi*60Hz*t)$ Well I'm here to tell you that its really $120 * √2 * [Cos(2\pi*60Hz*t) + i Sin(2\pi*60Hz*t)] =120 * √2 * e^{i 2\pi*60Hz*t}$ With that knowledge you can find the current even in a more complicated problem such as with added resistance, for example, what if there was some non-zero resistance in the wires? Let R be the resistance of the wires, 1Ω, in series with the inductor. Z = Zl + R = i ω l + 1 where Zl is the complex impedance of the inductor l $V = IZ → I = V/Z = (2.9336 -22.1188 i) e^{i 2\pi*60Hz*t}$ (thank you mathematica) uh oh, now I have some phase angle that is not as simple as +90, its actually 82.4 now. Sorry if I've lost you but I'm going to finish this now that I've started. I've got to adjust my expression for current so that i(t = 0) = 0 I rotate my new expression for the current by multiplying by e-i θ $i(t) =e^{-i \frac{2\pi}{360}7.6}(2.9336 -22.1188 i) e^{i 2\pi*60Hz*t} = -22.3 i e^{i 2\pi*60Hz*t}$ ok now I know my Imax is 22.3. I take the real part of the above $i(t) = ℝ[-22.3 i e^{i 2\pi*60Hz*t}] = 22.3Sin(2\pi*60Hz*t)$ $i(t = 1/180) = 19.3232Amps$ Ok well now I know what the current is so I can say all sorts of things like what is the energy stored in the inductor and what is the power dissipated in the wires and so on (wow 249 Watts). I hope I didn't confuse you more.