Homework Help: Simple RL circuit

1. Nov 30, 2011

geft

I'm stuck at part (a) of this question: http://i.imgur.com/9IqKu.png

It seems that the source is DC voltage, and since the inductor acts as a short circuit in DC, Vo is basically 1000V? So the sketch is basically a straight line? Is there something I'm missing here?

2. Nov 30, 2011

Staff: Mentor

The voltage across an inductor: v = L.di/dt

For DC conditions, meaning when all branches of the circuit have a steady current and voltage, then di/dt =0

But in your case, the current is changing, so you are not dealing with DC conditions. In the problem you face, we are interested in the transient conditions, before the steady state DC conditions settle in.

3. Nov 30, 2011

geft

I see now. Thanks!

4. Nov 30, 2011

geft

Is this correct?

Ldi/dt + iR = Vs

di/dt + iR/L = Vs/L

i(0) = Vs/R = 1000/100 = 10 A

i(t) = 10exp(-1000t)

VL(t) = Ldi/dt = (100m)(10)(-1000)exp(-1000t) = -1000exp(-1000t)

VR(t) = VS - VL(t) = 1000 + 1000exp(-1000t)

Last edited: Nov 30, 2011
5. Nov 30, 2011

Staff: Mentor

Not quite. The initial current through the inductor will be zero --- inductors don't like sudden changes in current!

If the initial current is zero, then surely the initial voltage drop across the resistor must also be zero.

6. Nov 30, 2011

geft

I mistakenly assumed the inductor acts like a short circuit at t=0 since there is no magnetic field.

Anyway, if i(0) = 0, then i(t) = 0, which means VL(t) = 0. So VR(t) = VS = 1000 V?

Isn't this what I brought up in my original post? How can I make VR a function of t?

7. Nov 30, 2011

Staff: Mentor

The current through the inductor is going to increase over time to a maximum value set by the resistance. So the voltage across the resistance must likewise increase over time accordingly.

8. Dec 1, 2011

geft

So my working is correct, except I(0) should be left as it is in order to get Vout in terms of t? Wouldn't that be the steady state response then? What should I do to obtain the transient response?

9. Dec 1, 2011

Staff: Mentor

Often the easiest way to obtain the transient response is to:

1. Determine the initial conditions at time t = 0+.
2. Determine the final steady state situation (t → ∞)
3. Determine the time constant for changes that occur
4. Connect 1 and 2 with the appropriate exponential functions

In your case you have an initial inductor current of zero, and since the inductor and resistor are in series, Vo is zero initially. After a long time the inductor will look like a short circuit and the current through it will be a constant value determined by the Vs and the resistor.

So the exponential function that connects the initial state of Vo to the final state of Vo will be an exponential function that starts at 0V and rises to a plateau at Vs.

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10. Dec 1, 2011

geft

So the function is actually obtained by observation and not by calculation?

Sorry, but I seem to be having trouble getting the exponential. I keep getting those which divide by 0 at t = 0. Is this a log graph?

Last edited: Dec 1, 2011
11. Dec 1, 2011

Staff: Mentor

Usually the student slogs through the differential equations a few times before enlightenment is reached or is bestowed upon them The equations describing the response of a first order circuit always have one of two forms: a "decaying" exponential $e^{-t/\tau}$ or a "rising" exponential $1 - e^{-t/\tau}$.

Sometimes the functions are offset by some amount. For example a rising exponential may start at some initial value other than zero, or a decaying one end up at some voltage other than zero. In both cases it's the Δ between the start and end that you use as the amplitude of the exponential. In general then you have:
$$f(t) = f_o + \Delta e^{-t/\tau}$$
$$g(t) = g_o + \Delta (1 - e^{-t/\tau})$$
where fo and go are the offsets and Δ the start to finish change.

In this case you have the g(t) case, the voltage starts at zero so the offset go = 0, and the total change is Δ = (1000V - 0V) = 1000V.

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12. Dec 1, 2011

geft

Thank you so much for taking the time to write all that. Very informative!

Last edited: Dec 1, 2011