Discussion in 'Homework Help' started by Asad ahmed1, Feb 10, 2016.

Feb 10, 2016
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My professor does not actually teach anything and he consider that we know every thing I took a look at the previous paper and i found this in it the inductor value is not given so how to find voltage across it the Formula v=Ldi/dt still be applicable to this

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2. ### RBR1317 Active Member

Nov 13, 2010
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What would be the best way to find the voltage across the inductor if you know the current through the resistor?

Can you find the value of L from the circuit time constant, τ? What is the energy stored in the inductor at t=0? Over time, all that energy will be dissipated in the resistor.

Last edited: Feb 10, 2016
3. ### thumb2 Member

Oct 4, 2015
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This is what many students say..

I suppose in part a (which is not shown in attachment), the RL circuit was connected to a power supply. Isn't it ?

Feb 10, 2016
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no it isnt but i would be able to apply the formula v=L(di/dt) if i know the l but how can i solve this

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5. ### thumb2 Member

Oct 4, 2015
89
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The part b of the exercise tell you the value of $i_{L}(t)$.

Feb 10, 2016
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so v=L (Di/dt)
by differentiation
V=L(20 x -1000 e^-1000t)

7. ### WBahn Moderator

Mar 31, 2012
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4,917
I'm getting a bit tired of hearing this excuse trotted out.

Why do you need to know the value of the inductor?

What is the relationship between the voltage across the resistor and the voltage across the inductor?

What do you need that isn't given in order to find the voltage across the resistor?

What do you need that isn't given in order to find the power dissipated in the resistor?

Feb 10, 2016
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I need to know the value of inductor so i would put the formula v=Ldi/dt and find the value of v and for the power i can use p=i^2R
voltage across resistor is v=Ir and v across inductor is v=Ldi/dt

9. ### WBahn Moderator

Mar 31, 2012
18,080
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Again, you do NOT need to know the value of L to answer this question.

Feb 10, 2016
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11. ### WBahn Moderator

Mar 31, 2012
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What is that V(t) on the left hand side represent?

Feb 10, 2016
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voltage drops around the LR series circuit.

13. ### RBR1317 Active Member

Nov 13, 2010
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Why do you refer to the voltage across the resistor as $I \times R$ implying that $I$ is a constant when the current in the circuit is given as $i(t)=20 e^{-1000t}$ or alternately as $i(t)=20 e^{-t/\tau}$ where $\tau = \frac{L}{R}$ ?

Feb 10, 2016
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i can get the tau from this and i also have r so i can get the value of L
L=0.001x10 = 0.01H

15. ### KL7AJ AAC Fanatic!

Nov 4, 2008
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I'm sorry to say I've had a few teachers like that...I hope I never fall into that category. But yes, di/dt applies here. The voltage will actually be an exponential curve.
I also highly recommend BUILDING the circuit! You really get a lot more insight out of this when you have the actual hardware.
Keep up the good work...sooner or later you'll end up with a great teacher. I've had a few of them too!
eric

16. ### oussama123443 New Member

Nov 2, 2014
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i assume that the circuit was connected to a source prior to having this circuit. you need to plug the time at which you want to find the voltage , you should that exponential decay equation to get the voltage at any time

Feb 10, 2016
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@KL7AJ In the beginning of semester i asked him why we use the arrow into the node can we make all currents leave from node so i end up teaching him the KCl :/ now Iam taking lectures from the net and sites like this and reading books .thanks for the reply it made my day.

18. ### WBahn Moderator

Mar 31, 2012
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You do NOT need to find the value of L. Forget it. It is not needed to answer these questions. Stop focusing on finding something that you do not need!

If I tell you that the voltage across the resistor is 3V, what is the voltage across the inductor?

Feb 10, 2016
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In that case if the other terminal of resister is grounded then 0-V(L)=3volts V(L) = - 3 volts

20. ### WBahn Moderator

Mar 31, 2012
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Good, except that doesn't matter whether the "other terminal" is grounded or not. Notice that which terminal is being referred to in the first place is not specified, nor is the direction of the current. You get to pick.

So, for the given voltage, what is the voltage across the resistor as a function of time?