Rearrange formula and work out phase angle

[leejohnson222]

Homework Statement

work out what ip, f and phase angle is, then rearrange the formula to make t the subject and find out at what point in time does 6.78 amp occur
i = Ipsin (2pi f t - phase angle)

so far i think Ip = 10 f = 1000hz
struggling with the phase angle

Relevant Equations

i = Ipsin (2pi f t - phase angle)

so far i think Ip = 10 f = 1000hz
struggling with the phase angle

 

[leejohnson222]

 

[haruspex]

I confirm your amplitude and frequency values.
To get the phase angle, plug in a value for t, read off the y value from the graph and solve the equation. You will need to do that for at least two points because of the ambiguity in evaluating $\arcsin$.

 

[kuruman]

Shouldn't the amplitude have units?

To get the phase angle, plug in a value for t, read off the y value from the graph and solve the equation. You will need to do that for at least two points because of the ambiguity in evaluating $\arcsin$.

Can the phase angle not be obtained by noting that at t = 0 the sinusoidal is at a minimum?

 

[haruspex]

Shouldn't the amplitude have units?

Can the phase angle not be obtained by noting that at t = 0 the sinusoidal is at a minimum?

I should have written "you may have to". I was trying to be general.

 

[leejohnson222]

i could be very wrong here, but i worked out the angle to be 90 degrees ?
i full cycle of the wave is 0.001 s and the delay from 0 is 0.00025 s
360 x 0.00025 / 0.001 = 90

 

[BvU]

That's not what @haruspex suggested ... but I agree.

Easy:
$I = I_p\sin (2\pi f t - \phi)\quad$ pick $t=0$ and solve $\sin(-\phi)=-1\$

Personally, I prefer to imagine the graph for $\phi = 0$ and conclude that a shift of $-{\pi\over 4}$ to the right yields the given graph. Therefore $\ -\phi=-{\pi\over 4}$

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