# Homework Help: AC Power (RL)

1. Sep 22, 2012

### hogrampage

1. The problem statement, all variables and given/known data
Calculate the power absorbed by each element at t = 0-, t = 0+, and t = 200ms.

2. Relevant equations
P$_{L}(t)$ = $\frac{V_{0}^{2}}{R}e^{-Rt/L}(1-e^{-Rt/L})$
P$_{R}(t)$ = $\frac{V_{0}^{2}}{R}(1-e^{-Rt/L})$
P$_{V}(t)$ = -(P$_{L}(t)$ + P$_{R}(t)$)

3. The attempt at a solution
-10u(-t)V = -10V for t <= 0; 0V for t > 0

t = 0-: V = -10V, i = -10A
PR(0-) = (-10)2(1) = 100W
PL(0-) = 0W
PV(0-) = -100W

t = 0+: V = -10V, i = -10A
PR(0+) = (-10)2(1) = 100W
PL(0+) = 0W
PV(0+) = -100W

t = 200ms:
P$_{L}(0.2)$ = $\frac{(-10)^{2}}{R}e^{-4(0.2)}(1-e^{-4(0.2)})$ = 25W
P$_{R}(0.2)$ = $\frac{(-10)^{2}}{R}(1-e^{-4(0.2)})$ = 30W
P$_{V}(0.2)$ = -(25 + 30) = -55W

I'm not completely sure I did this problem correctly (particularly worried about the powers for 0- and 0+). If I did not, I would greatly appreciate any feedback (mistakes you see, hints, etc.). I don't want to go further in my homework until I understand this basic problem first.

Thanks!

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2. Sep 22, 2012

### Staff: Mentor

For any 2-port component the power consumed is defined to be P = I*V when the current I flows INTO the device at its + terminal.

By labeling the diagram with individual component voltage polarities and and current direction at the time of interest, and applying P = I*V for each component, you should obtain the power associated with each.

You might consider sketching v vs t for each component, too. In particular, what happens to the voltage across the inductor around time t = 0?

3. Sep 22, 2012

### hogrampage

Well, I know how power and the equations work, but this is a little different since it's AC, not DC. The main thing that is confusing me is the step-unit function of -10u(-t). I don't know how the graph would look for V vs. t for the components.

To me, at t=0, the inductor would discharge the energy stored (like a capacitor), since that is when the switch is turned off.

I was able to solve a similar problem correctly, but the time step was something like 40 + 60u(t), so it was fairly simple (u(t) instead of u(-t) is easier for me to understand).

What I really need help understanding is how u(-t) works in regards to the voltage (especially across the inductor) and current, specifically at t = 0+ and t = 0-.

4. Sep 22, 2012

### Staff: Mentor

When t is negative, the argument of u(-t) will be positive, which means the function will yield a +1 value. When t is positive, the argument will be negative and the function will yield a 0 value. In other words, for time <0 u(t) will be +1, and for time >0 it will be 0. So u(-t) represents a reflection of u(t) about the y-axis. Multiply the result by the parameter -10.
The circuit begins being driven by a source voltage of -10V for time less that 0. At time 0+ the source becomes 0V and continues at that value for all future time.

5. Sep 23, 2012

### hogrampage

So, for t < 0, the current would be:

I(t) = $\frac{V}{R}(1 - e^{-Rt/L})$ A

For t = 0$^{-}$, I could find the voltage across the inductor and the resistor, then solve for P for each component.

For t > 0, the current would be 0A (voltage source is at 0V), so how would I get the power stored in each component if there is no current flowing? Would it just be 0W for everything? If so, why would the question ask for power stored at t = 200ms?

Thanks

6. Sep 23, 2012

### Staff: Mentor

For t < 0 you should assume that the circuit has long since reached a steady state (no transients left); There will be a constant current driven by the -10V supply.

When the power supply switches to 0V at t = 0, THEN there will be changes. The current will eventually head towards zero, but there will be a transition period (transient) while the energy stored in the inductor dissipates via the resistance. Presumably t = 200ms is within the transition period, which by an engineering rule of thumb is taken to be a period of about 5 time-constants.

7. Sep 23, 2012

### hogrampage

Oh, okay. So, I should actually use the I(t) equation from my last reply for t > 0, and use V=IR for t < 0?

This is beginning to make sense (I think lol).

8. Sep 23, 2012

### Staff: Mentor

Yeah, that's more like it. Most of these sorts of problems involve first determining what the steady-state situations are pre-switching and post-switching, then working out the transient period that will bridge the two. So before the switch occurs there is a constant current flowing. A long time after the switch occurs there will be no current flowing. The "magic" between states occurs over about 5 time constants, and will involve function of an exponential form as you've been showing.

9. Sep 24, 2012

### hogrampage

Okay, I re-calculated all of the powers and I got the same answers as my first post. Are those the correct powers?

10. Sep 24, 2012

### Staff: Mentor

There are some issues.

When the voltage source is at 0V, how can it be producing (or absorbing) power? It'll behave like a piece of wire.

At time 0+, the current in the circuit cannot yet have changed its magnitude nor its direction. Since the voltage source is at 0V, what voltage will be driving that current? Can you label the components with their potentials for time 0+?

Can you sketch the potential across the inductor for time t > 0? How about the current?

11. Sep 24, 2012

### hogrampage

I must be misunderstanding some of the concepts. Here is what I came up with:

t = 0+:
V = 0V
I = 0A
VL = -10e$^{-4t}$
VR = 0V

t = 0-:
V = -10V
I = -10A
VL = 0V
VR = -10V

t > 0:
VL = -10e-4t V
I = $\frac{V}{R}(1 - e^{-4t})$

t < 0:
VL = 0V
I = -10A

I think I understand how to find stuff for t < 0 and t > 0, but t = 0 is not making sense to me. I don't understand how to find the voltage drops and current(s) at t = 0+ and t = 0-.

Last edited: Sep 24, 2012
12. Sep 24, 2012

### Staff: Mentor

The current through an inductor cannot change instantaneously. If its current was 10A at time 0-, then it must still be 10A at time 0+.

13. Sep 24, 2012

### hogrampage

Well, by I, I meant the current through all of the components. In my book, there's an example with a series RC circuit and the current is different at 0- and 0+.

I guess I still don't get how to find the power absorbed by each component at 0- and 0+. Wouldn't the power absorbed by each of them be 0W, since there's no current flowing? I'm completely lost :| (the book - Engineering Circuit Analysis 8ed Hayt - is terrible at explaining these things).

Are my calculated powers correct for t = 200ms, or are those also wrong?

Last edited: Sep 24, 2012
14. Sep 24, 2012

### Staff: Mentor

There MUST be current flowing at t = 0+ if there was current flowing at t = 0-. The current passing through an inductor cannot change instantaneously. The inductor will develop ANY voltage required in order to prevent instantaneous current change, borrowing energy from its magnetic field to do so. Hence the formula, E = L di/dt.

Inductors behave akin to mass when a force is applied. When you apply a force to a mass it changes velocity gradually, not instantaneously. The larger the mass, the slower the change for a given applied force. The same happens with current passing through an inductor. The larger the inductor, the more gradual the current change.

So, what voltage must the inductor produce in order to maintain the same current at t = 0+?

15. Sep 24, 2012

### hogrampage

Would it be -10V?

EDIT: If I use V = L$\frac{di}{dt}$ and plug in 0 for t, I get -12.5V.

Last edited: Sep 24, 2012
16. Sep 24, 2012

### Staff: Mentor

Without a point of reference I can't answer directly. I can say that the magnitude of the potential will indeed be 10V. You've got to orient it on the inductor so that it maintains the current's direction over the t=0 transition.

After the transition the inductor voltage will decay and the current will decay as the stored energy of the inductor wanes. The "usual" decay formula applies to both.

17. Sep 24, 2012

### hogrampage

So, in other words, the power absorbed by the inductor at t = 0- and t = 0+ is actually -100W (see my reason for negative below), not 0W? And, the power absorbed by the resistor is then 0W (no voltage across it)?

The current, according to the problem, flows in a clockwise direction, but it is negative. Does that mean the inductor is generating power, while the source is absorbing power?

Last edited: Sep 24, 2012
18. Sep 24, 2012

### Staff: Mentor

A resistor absorbs power so long as current flows though it. The resistor is what will eventually dissipate the energy of the circuit that was stored in the inductor.

The inductor will produce power while it supplies the energy to keep current flowing (it stored up power in its magnetic field during the time t < 0). Eventually the energy all dissipated and the potential and current will taper off to zero.

If a component has a potential across it and a current flows out of the (relatively) positive lead, then it is producing power. If the current flows into the positive lead then it is absorbing power Think of the way current and potential drop work for a resistor; current flowing through the resistor produces a potential drop in the direction of the current. Thus the current is flowing into the relatively positive lead and so power is being absorbed by the resistor.

The problem was set up to get you to think about what things like -10u(-t) mean in terms of the potential they produce --- what potential will actually drive the circuit and what the "true" direction of the current will be. You can always simplify things by drawing equivalent circuits that correspond to the various time periods and which have clear source values and current directions indicated.

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19. Sep 24, 2012

### hogrampage

Okay, those images make things a lot clearer. I will definitely start drawing more diagrams from now on :).

I am, however, having issues figuring out which equation(s) to use for the voltage across the inductor for t < 0. I know it will eventually be 0V, but what would be the equation? Or, would it just be 0V for all t < 0?

20. Sep 24, 2012

### Staff: Mentor

For time t < 0 it is assumed that the circuit has reached steady state; it's been that way for some indefinite but relatively long time (theoretically perhaps an infinite amount of time). So there are no transients left, just steady-state conditions. Current is not changing so the voltage across the inductor must be zero (E = L di/dt, and di/dt is zero).