1. May 27, 2014

### MarcL

1. The problem statement, all variables and given/known data

An ideal inductor L = 88 mH is connected to a source whose peak potential difference is 65 V.

If the frequency is 90 Hz, what is the current at 5 ms?

2. Relevant equations

il= Il(ωt-ø) and Il = Vl / Xl

3. The attempt at a solution

So I did get the answer right, by chance. I would like to understand... Why does the definition of an inductive load specifies that its current is lagging behind the voltage of the inductor but to get the right answer, I didn't minus my angular frequency and the phase constant.

2. May 27, 2014

### Staff: Mentor

Maybe the arrangement has a switch which applies the sinusoid when it's at some particular level, OR

maybe the question asks for the instantaneous value of the current 5ms after its peak?

3. May 27, 2014

### MarcL

Sorry if I didn't explain it well. As a start, the statement is copy pasted from my online assignment so there was no changed to it.

However, how I worked it out was by chance, I forgot to minus the phase constant ( where ø= pi/2)
because the sin function of the current from an inductor lags behind the velocity by 90 degrees, therefore it is ωt-ø to get the phase of the current from the inductor no? ( i don't know if this made sense). I mean this is what I get from my book. :/

4. May 27, 2014

### Staff: Mentor

Yes, the current lags the sinusoidal voltage by 90°. What answer did you give that was considered right?

5. May 27, 2014

### MarcL

First I used this formula:

il = Il sin (ωt-ø) which always gave me 1.306 when replacing Il by V/Xl

However, the right answer was obtained by using

il = Il sin (ωt) which confuses me to no end because then, when do I use the phase constant? I thought it was always part of the formula and the concept of the ac circuit.

6. May 27, 2014

### Staff: Mentor

Yes, that gives the peak value of the current. Because the circuit is nothing more than pure inductance, you also can say that this current sinewave is 90° behind the voltage sinewave.

It sounds like an assumption is being made that time=0 will be when the current sinewave passes through zero. That would make ɸ=0.

I'm still waiting to hear what answer you gave for the current at 5ms. We can then work backwards to determine what the question should have been. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]

Last edited by a moderator: May 6, 2017
7. May 27, 2014

### MarcL

It is kinda scary to think some assumptions were made without mentioning them because my final is coming up and this is our last assignment...

But yeah, let's see where I went wrong / question went wrong. :)

8. May 27, 2014

### Staff: Mentor

Yes, they are taking the origin as the start of the current's sinewave waveform, 1.306sin(2π90t)

They are side-stepping the phase difference between voltage and current by using the current waveform as the reference. That makes things easier!

9. May 27, 2014

### MarcL

wait, how did you know that? because they skipped the waveform? how do we know that from reading the question?

10. May 27, 2014

### Staff: Mentor

That's what is needed to get their answer.