Solve Part (a) Simple RL Circuit Problem

In summary, the voltage across an inductor will be 1000V if the current through the inductor is zero at the beginning of the cycle. The function that connects the initial state of Vo to the final state of Vo is an exponential that starts at 0V and rises to a plateau at 1000V.
  • #1
geft
148
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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?
 
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  • #2
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
I see now. Thanks!
 
  • #4
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)
 
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  • #5
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
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
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
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
geft said:
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?

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.

attachment.php?attachmentid=41417&stc=1&d=1322754936.jpg
 

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  • #10
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?
 
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  • #11
geft said:
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?

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

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:
[tex] f(t) = f_o + \Delta e^{-t/\tau} [/tex]
[tex] g(t) = g_o + \Delta (1 - e^{-t/\tau}) [/tex]
where fo and go are the offsets and Δ the start to finish change.

attachment.php?attachmentid=41419&stc=1&d=1322760673.jpg


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
Thank you so much for taking the time to write all that. Very informative!
 
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1. What is a simple RL circuit?

A simple RL circuit is an electrical circuit that contains a resistor (R) and an inductor (L) connected in series. It is a basic circuit used to study the behavior of inductors in response to changing currents.

2. How do you solve a simple RL circuit problem?

To solve a simple RL circuit problem, you can follow these steps:

  1. Determine the values of the resistor (R) and inductor (L) in the circuit.
  2. Calculate the total resistance (RT) of the circuit by adding the resistance of the resistor and the inductor (RT = R + XL).
  3. Find the time constant (T) of the circuit by dividing the inductance (L) by the total resistance (T = L/RT).
  4. Use the time constant to find the current (I) in the circuit at a given time (t) using the formula I = I0 * (1-e-t/T), where I0 is the initial current.
  5. Repeat the previous step for different times to plot a current vs. time graph.

3. What is the significance of solving a simple RL circuit problem?

Solving a simple RL circuit problem helps us understand the behavior of inductors and their role in electrical circuits. It also allows us to calculate important values such as current, voltage, and time constant, which are essential for designing and analyzing more complex circuits.

4. What is the difference between a simple RL circuit and a complex RL circuit?

A simple RL circuit contains only one resistor and one inductor, while a complex RL circuit can have multiple resistors and/or multiple inductors connected in various configurations. Solving a simple RL circuit is relatively straightforward, but solving a complex RL circuit requires advanced techniques such as Kirchhoff's laws and network analysis.

5. Can a simple RL circuit be used in real-life applications?

Yes, simple RL circuits are commonly used in electronic devices such as power supplies, filters, and oscillators. They are also used in various industries for tasks such as voltage regulation and signal processing. Understanding simple RL circuits is crucial for designing and troubleshooting these applications.

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