How Does Inductance Affect Energy Storage in AC Circuits?

[slayhy]

A 20.0-mH inductor is connected to a standard electrical
outlet (ΔVrms= 120 V; f = 60.0 Hz). Determine the
energy stored in the inductor at t = (1/180) s, assuming
that this energy is zero at t = 0.

Now when i tried to solve it, this is what i did:
ω = 2∏ f = 2∏(60.0/s) = 377 rad/s
XL = ω L = (377)(0.02)= 7.54 Ω
Irms = ΔVrms/XL = 120 V/7.54 Ω =15.9 A
Imax= √2Irms= √2(15.9 A) = 22.5 A

instantaneous current in the inductor:
i(t)=ΔVmax sin(ωt - ∏/2) / ωL
i(t)=(120√2) sin(377x(1/180) - ∏/2)
i(t)=0.206 A

U= L [i(t)]^2 [sin ωt]^2 / 2
U= 0.02 (0.206)^2 [sin (377x(1/180))]^2 / 2
U= 6.76 x 10^-3 J



But in the book's solution manual it was solved using the instantaneous current in the resistor, which has the formula: i=Imax sinωt

I don't understand why the book used current in the resistor rather than the one in the inductor.

Please help me! I would like an explanation to that.

 

[the_emi_guy]

You dropped the ωL in your 6th equation. It should be:

i(t)=(120√2) sin(377x(1/180) - ∏/2)/ωL

This is the same as:

i(t)=Imax sin(377x(1/180) - ∏/2)

You didn't need to reevaluate Imax a second time.

I'm not sure what resistor you are referring to.

 

[slayhy]

yeah you're right :)

But the thing is that i(t)= Imax sin(ωt - ∏/2) [the formula for the current in inductors] is
not equal to i(t)= Imax sinωt [the formula for the current in resistors which the book
used :S]

The book should have used i(t)= Imax cosωt for it to equal i(t)= Imax sin(ωt - ∏/2)

So I'm still not getting the same answer as the one in the book.

I don't see any resistor in the problem so why is he using the formula for resistors?
Do you get my point?

 

[Greg-ulate]

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
[itex]120 * √2 * Cos(2\pi*60Hz*t)[/itex]
Well I'm here to tell you that its really
[itex]120 * √2 * [Cos(2\pi*60Hz*t) + i Sin(2\pi*60Hz*t)] =120 * √2 * e^{i 2\pi*60Hz*t}[/itex]

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

[itex]V = IZ → I = V/Z = (2.9336 -22.1188 i) e^{i 2\pi*60Hz*t} [/itex] (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}
[itex] 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} [/itex]
ok now I know my Imax is 22.3. I take the real part of the above
[itex] i(t) = ℝ[-22.3 i e^{i 2\pi*60Hz*t}] = 22.3Sin(2\pi*60Hz*t) [/itex]
[itex] i(t = 1/180) = 19.3232Amps [/itex]

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.

 

[Nickie]

It is important to note that in an AC circuit, the current and voltage are out of phase due to the presence of inductors and capacitors. In this case, the voltage across the inductor is at its maximum value when the current is at its minimum value and vice versa. This is why the book used the current in the resistor rather than the inductor, as it represents the maximum current in the circuit.

Using the formula for instantaneous current in a resistor, i=Imax sinωt, we can find the maximum current in the circuit at t = (1/180) s:

i(t) = Imax sinωt
i(t) = Imax sin(377 x (1/180))
i(t) = Imax sin(2.09)
i(t) = Imax x 0.906
i(t) = 0.906 Imax

Since we know that the maximum current in the circuit is 22.5 A, we can solve for Imax:

0.906 Imax = 22.5 A
Imax = 24.8 A

Now we can use this value of Imax to calculate the energy stored in the inductor at t = (1/180) s:

U = L [i(t)]^2 [sin ωt]^2 / 2
U = 0.02 (24.8)^2 [sin (377 x (1/180))]^2 / 2
U = 6.76 x 10^-3 J

As you can see, the value of energy stored in the inductor is the same as the one you calculated using the instantaneous current in the inductor. The book's solution is just using a different approach to solve the problem. Both methods are correct and give the same result.