Joined Feb 10, 2016
68
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

#### RBR1317

Joined Nov 13, 2010
573
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.

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#### thumb2

Joined Oct 4, 2015
122
My professor does not actually teach anything
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 ?

Joined Feb 10, 2016
68
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

#### thumb2

Joined Oct 4, 2015
122
i would be able to apply the formula v=L(di/dt) if i know the l
The part b of the exercise tell you the value of $$i_{L}(t)$$.

Joined Feb 10, 2016
68
so v=L (Di/dt)
by differentiation
V=L(20 x -1000 e^-1000t)

#### WBahn

Joined Mar 31, 2012
26,305
My professor does not actually teach anything and he consider that we know every thing
I'm getting a bit tired of hearing this excuse trotted out.

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
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?

Joined Feb 10, 2016
68
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

#### WBahn

Joined Mar 31, 2012
26,305
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
Again, you do NOT need to know the value of L to answer this question.

Joined Feb 10, 2016
68 #### WBahn

Joined Mar 31, 2012
26,305
What is that V(t) on the left hand side represent?

Joined Feb 10, 2016
68
voltage drops around the LR series circuit.

#### RBR1317

Joined Nov 13, 2010
573
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 $$\tau = \frac{L}{R}$$ ?

• Joined Feb 10, 2016
68
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

#### KL7AJ

Joined Nov 4, 2008
2,226
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
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

• #### oussama123443

Joined Nov 2, 2014
27
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

Joined Feb 10, 2016
68
@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.

#### WBahn

Joined Mar 31, 2012
26,305
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
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?

Joined Feb 10, 2016
68
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?
In that case if the other terminal of resister is grounded then 0-V(L)=3volts V(L) = - 3 volts

#### WBahn

Joined Mar 31, 2012
26,305
In that case if the other terminal of resister is grounded then 0-V(L)=3volts V(L) = - 3 volts
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?

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