# Urgent first order RL circuit problem

Discussion in 'Homework Help' started by Favour, Jul 16, 2008.

1. ### Favour Thread Starter Member

Jul 15, 2008
10
0
Hello
Kindly assist me with a solution to the attached first order RL problem. Looking forward to your response.

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May 16, 2005
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3. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
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My apologies for having prematurely closed this thread. The second file was similar enough to the first I thought it was the same one. Comparing them side by side, they are obviously different. I have re-opened this thread.

4. ### Ratch New Member

Mar 20, 2007
1,068
4
Favour,

Possible error. A resistor connected to the 2A source is valued at 2K ohms. Is that correct, given that all the other resistors are much, much smaller? Ratch

5. ### theamber Active Member

Jun 13, 2008
318
0
To analyze this is a complex calculation with differential equations. First you need to simplify the circuit with one resistance and one impedance with the Thevenim equivalent.
Also note that after the switch is closed at t>infinity the inductance is equivalent to a short circuit. And the current through the inductance prior to closing the circuit was 0.
The natural response of t=0 is exactly equal to the steady state response with the switch closed, but should be oposite in sign.
The time constant with any resistor in series with an inductor is t=L/R or in general t=1/|s|.

6. ### Ratch New Member

Mar 20, 2007
1,068
4
Favour,

Well, lets see. Since I did not get any feedback from you on whether that 2k ohm resistor is correct, I am going to assume that the resistance is really 2 ohms. The inductor shorts everything east of its position. This means that it forms a current divider with 4 ohms and 2 ohms. Therefore at t=0 the steady state current through the inductor is 2/3 A, and Vo is zero. At t=infinity, the inductor will again short out Vo to zero.

Now that we know the initial conditions at t=0, we can calculate the transient starting from t=0. The best way to do this is a loop equation using Laplace transforms. It is easy to learn to apply them compared to learning to solve differential equations. Ratch

7. ### Favour Thread Starter Member

Jul 15, 2008
10
0
Ratch,

Sorry, I was offline. You are correct, it is 2 ohms and not kilo ohms as indicated.
How did you arrive at current through the inductor to be 2/3A? Made several attempt without success.
Guys,

8. ### Ratch New Member

Mar 20, 2007
1,068
4
Favour,

By using the current division formula. See http://www.tpub.com/content/doe/h1011v1/css/h1011v1_134.htm

While the switch is open, we have two branches across the current source. One branch is 4 ohms and the other which includes the inductor is 2 ohms. Therefore the current through the inductor branch is 2*(4/(4+2)) = 4/3 . As you can see, I made an arithmetic error in my previous post. Ratch

9. ### Favour Thread Starter Member

Jul 15, 2008
10
0
Looking at the problem, at what time does Vo reduce to 50% of its value at t=0? Please help!

Favour

10. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Show us your attempts at solving the problem.