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Inductor Current

Could someone check to see if I have worked this out correctly

The current (๐‘–๐ฟ) through a 250 mH inductor (L) has a relationship with time (t) as follows:
๐‘–๐ฟ = 1/L โˆซ sin(100t) dt
Determine the inductor current when the time is 1.5 seconds.

iL= 1/0.25 cos(100*1.5)
= 2.797A

Thanks
Reply 1
Original post by thomas0611
Could someone check to see if I have worked this out correctly

The current (๐‘–๐ฟ) through a 250 mH inductor (L) has a relationship with time (t) as follows:
๐‘–๐ฟ = 1/L โˆซ sin(100t) dt
Determine the inductor current when the time is 1.5 seconds.

iL= 1/0.25 cos(100*1.5)
= 2.797A

Thanks

When you integrate sin, you need to get the sign correct and account for the 100 multiplier (frequency).
Reply 2
Original post by mqb2766
When you integrate sin, you need to get the sign correct and account for the 100 multiplier (frequency).

So it should be -cos?
(1/25010^-3) * (-cos150/100) = -0.0279A

Does that look correct?
Reply 3
Original post by thomas0611
So it should be -cos?
(1/25010^-3) * (-cos150/100) = -0.0279A

Does that look correct?

Integral of sin is -cos, however not sure what youve done with the 100 and 10^(-3). If you have
sin(wt)
what is its integral? You can validate by differentiating what you end up with using the chain rule. Obv w=100 for your example.
(edited 7 months ago)
Reply 4
Original post by mqb2766
Integral of sin is -cos, however not sure what youve done with the 100 and 10^(-3). If you have
sin(wt)
what is its integral? You can validate by differentiating what you end up with using the chain rule. Obv w=100 for your example.

So it should be -1/2cos(wt) Is that the correct integral
Reply 5
Original post by thomas0611
So it should be -1/2cos(wt) Is that the correct integral

If you differentiate that using the chain rule, what do you get? No idea where the 1/2 comes from.
Reply 6
Original post by mqb2766
If you differentiate that using the chain rule, what do you get? No idea where the 1/2 comes from.

Sorry should be -1/wcos(wt)

=-1/100cos(150)

I'm unsure on the exact step to differentiate this. Would I do the brackets first?
Reply 7
Original post by thomas0611
Sorry should be -1/wcos(wt)

=-1/100cos(150)

I'm unsure on the exact step to differentiate this. Would I do the brackets first?

I think this is correct now:
1/0.25-(1/100)cos(100t) = 0.03464A
(edited 7 months ago)
Reply 8
Original post by thomas0611
Sorry should be -1/wcos(wt)

=-1/100cos(150)

I'm unsure on the exact step to differentiate this. Would I do the brackets first?

-1/wcos(wt) is correct as the integral for sin(wt), but you should be able to verify this using the chain rule as its
dy/du du/dt
where here u=wt so dy/du = sin(wt) and du/dt = w so you get back to where you started. So just use that in the original expression with the 1/L multiplier and you should be good. Im getting a different value compared to #7, so show each part if necessary.
(edited 7 months ago)
Reply 9
Original post by mqb2766
-1/wcos(wt) is correct as the integral for sin(wt), but you should be able to verify this using the chain rule as its
dy/du du/dt
where here u=wt so dy/du = sin(wt) and du/dt = w so you get back to where you started. So just use that in the original expression with the 1/L multiplier and you should be good. Im getting a different value compared to #7, so show each part if necessary.

1/l -1/100cos(100t)+c
c=0
-4*1/100cos(100*1.5)
-4/100cos(150)
=0.0346A

That is my working. Should the calculator be set in radians?

This would give answer as -o.02797A
Reply 10
Original post by thomas0611
1/l -1/100cos(100t)+c
c=0
-4*1/100cos(100*1.5)
-4/100cos(150)
=0.0346A

That is my working. Should the calculator be set in radians?

This would give answer as -o.02797A

Yes for radians. When you differentiate or integrate trig functions like that youre assuming theyre in radians. Agree with the value.
Reply 11
Original post by mqb2766
Yes for radians. When you differentiate or integrate trig functions like that youre assuming theyre in radians. Agree with the value.

Thanks for your help!
Reply 12
Original post by mqb2766
Yes for radians. When you differentiate or integrate trig functions like that youre assuming theyre in radians. Agree with the value.

I have similar calculations but am struggling with 0.04cos. where does this come from?
Reply 13
Original post by thomas0611
So it should be -cos?
(1/25010^-3) * (-cos150/100) = -0.0279A
Does that look correct?

Agreed. the (-) sign indicates a lagging current w.r.t. Voltage although not mentioned
Original post by JTRLC
I have similar calculations but am struggling with 0.04cos. where does this come from?

My understanding is that sin(100t)dt integrated becomes -1/100cos(100t) + C

So you have 1/L * 1/100 = 1/0.25 * 1/100 = -0.04 which then you can sub back in

I have the same answer as the poster above (27.97mA) but what I don't understand in why we can just disregard the constant C. Why are we to assume C=0 or just remove from our workings? Is it because is a definitive integral as we have t=1.5?

Could anyone help?
Sorry for double post but just in case anyone comes back to this post in future. I submitted my assignment based on the answer with my calculator in radians mode but I had feedback from my tutor that this was incorrect. I've gotten away with it as my methodology is correct I still got the mark I was looking for, but the final answer was wrong due to calculator mode. One to watch.
Original post by AdamMc90
Sorry for double post but just in case anyone comes back to this post in future. I submitted my assignment based on the answer with my calculator in radians mode but I had feedback from my tutor that this was incorrect. I've gotten away with it as my methodology is correct I still got the mark I was looking for, but the final answer was wrong due to calculator mode. One to watch.

Hi AdamMc90,

Did your tutor say what the answer should have been or what the mode of the calculator should have been, i was under the impression that it also should have been in radians mode and have also got the same answer as above.

Thanks.

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