What is the relationship between current and voltage in an inductive load?

[MarcL]

Homework Statement

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?

Homework Equations

i_l= I_l(ωt-ø) and I_l = V_l / X_l

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.

 

[NascentOxygen]

The problem seems inadequately specified.

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?

What is the "correct" answer? How did you work out your answer?

 

[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. :/

 

[NascentOxygen]

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

 

[MarcL]

First I used this formula:

i_l = I_l sin (ωt-ø) which always gave me 1.306 when replacing I_l by V/X_l


However, the right answer was obtained by using


i_l = I_l 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.

 

[NascentOxygen]

First I used this formula:

i_l = I_l sin (ωt-ø) which always gave me 1.306 when replacing

I_l by V/X_l

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.

However, the right answer was obtained by using

i_l = I_l 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.

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

 

[MarcL]

.403576 A was the answer.

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. :)

 

[NascentOxygen]

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!

 

[MarcL]

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

 

[NascentOxygen]

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

That's what is needed to get their answer.