Question

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

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

b) At what frequency would the peak current be 2.3 A?
Hz

Answer 1

For a I got 1.242 A using i= (V/XL)*sin(2(pi)(f)(t)-pi/2)

Answer 2

\[i=\frac{ V }{ Xl}=\frac{ V }{ wl }=\frac{ V }{ 2*\pi*f*l }\]

Answer 3

Sorry, but using that aren't you only finding the maximum value because you are using the impedance?

Answer 4

Someone help pleaseeeee :(

Answer 5

\[2 \pi fL = \frac{ V }{ I }\]

Answer 6

\[I = \frac{ V }{ 2\pi fL }\]

Answer 7

for 2nd point.. \[f = \frac{ V }{ 2\pi L I }\]

Answer 8

that is wrong though I have tried it. My online assignment is not telling me it is right. Plus I believe I need to find the instanteneous current at 5 ms not just the peak current no?

Answer 9

I got this formula from this relation.... \[X _{L} = \frac{ V }{ I } and \rightarrow X _{L} = 2\pi fL\]

Answer 10

What's right answer do you know?

Answer 11

i don't think you have to find instantaneous current... its Root mean suqare current we can find just..

Answer 12

If the frequency is 90 Hz, what is the current at 5 ms? it says at 5 millisecond and no I don't know the answer :( wouldn't t=5ms means instantaneous?

Answer 13

WHen we have just one value of time.. it means we have total time... if 2 or 3 values of time are given then the first time will be use for instantaneous current...

Answer 14

anyway the answer the equation you gave me is 1.306 A and it is wrong apparently :/

Answer 15

Okay try this one also.....
\[I = \frac{ V }{ X _{L} } ( 1 - e ^{^{\frac{ -Rt }{ L }}} )\]

Answer 16

Just want to point out the equation above cannot be applied as we don't have a resistor across the circuit, unless you're talking about inductance reactance

Answer 17

Inductive Reactance is present when an inductor is used.. the is X = 2 pi f L

Answer 18

This problem is not clear. First for the 5ms value, we have to assume that it means 5 ms from initial start which we must further assume the waveform started at zero. 90 Hz is a period of 1/90 or 11 ms. With these assumptions we would assume the angle is 5/11 of its cycle (not quite at 1/2 cycle where the value would be at 0 volts. But.....with all of these assumptions I don't want to fool with it, the approaches taken by the earlier posters may be what it is about, just take voltage and divide by the inductive reactance.
But why did they bring in the 5 ms??