The Student Room Group

inductor current (HNC maths)

The current (𝑖𝐿) through a 100 mH inductor (L) has a relationship with time (t) as follows:
𝑖𝐿 = 1/L ∫ cos(300𝑑) 𝑑t
Determine the inductor current when the time is 0.75 seconds.

ive managed to integrate this equation fine to
i_l = (1/L) (1/300) sin(300t) + C
however i know now i can input the values for L and t however im unsure how to calculate the constant C
Original post by Jeffers999
The current (𝑖𝐿) through a 100 mH inductor (L) has a relationship with time (t) as follows:
𝑖𝐿 = 1/L ∫ cos(300𝑑) 𝑑t
Determine the inductor current when the time is 0.75 seconds.

ive managed to integrate this equation fine to
i_l = (1/L) (1/300) sin(300t) + C
however i know now i can input the values for L and t however im unsure how to calculate the constant C

The cos() in the original question must represent the instantaneous voltage across the resistor and Im presuming its "switched on" at t=0 when the current is zero.
Reply 2
Original post by mqb2766
The cos() in the original question must represent the instantaneous voltage across the resistor and Im presuming its "switched on" at t=0 when the current is zero.

Yea ive gotten to that aswell ,
Im pretty sure my integration is correct im just confused and what to do with the C constant at the end as i dont know how to work out value for it in this case
Original post by Jeffers999
Yea ive gotten to that aswell ,
Im pretty sure my integration is correct im just confused and what to do with the C constant at the end as i dont know how to work out value for it in this case


The current is 0 at time 0, which gives c = ...
did you ever get anywhere with this?
Reply 5
Original post by jack hare
did you ever get anywhere with this?

Similar to the other thread, what have you done .... ?
did you get anywhere with this mqb2766?
iL= 1/L ∫ sin(100t) dt time=1.5 seconds 250mH=0.25

1/0.25 cos(100*1.5)


I am completely lost tbh with you

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