Inductor Current

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

The current (𝑖𝐿) through a 250 mH inductor (L) has a relationship with time (t) as follows:
𝑖𝐿 = 1/L ∫ 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).

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.

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.

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?

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.

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.