Finding the phase angle of current through a capacitor?

[x86]

Homework Statement

Calculate the current in the capacitor shown in the figure below if the voltage input is V(t) = 29cos(377t-30) V, C = 1 micro Farad

Homework Equations

i = C dv/dt

The Attempt at a Solution

I essentially differntiate V(t) and multiply by 1 * 10^-6. I then get a negative amplitude, so I add 180 degrees to the phase angle to get a positive amplitude,

i = 1*10^-6*29*377 * (-sin(377t - 30) = 0.0109sin(377t + 150)

What is the phase angle? The correct answer is 60 degrees. I'm confused about how to get this. I'm getting 150 degrees.

 

[Delta2]

You should convert the current equation to cosine form using familira trigonometric identities.

 

[x86]

You should convert the current equation to cosine form using familira trigonometric identities.

Does it always have to be converted to cosine?

If I convert it to cosine, the phase angle changes from 150 to 240

 

[Delta2]

Current and Voltage have to be both in sine or cosine (and with the same sign of amplitude) to be able to discuss about the phase angle.

You ve made a mistake in the conversion, it is sin(x)=cos (x-90)=-cos (x+90).

 

[x86]

Current and Voltage have to be both in sine or cosine (and with the same sign of amplitude) to be able to discuss about the phase angle.

You ve made a mistake in the conversion, it is sin(x)=cos (x-90)=-cos (x+90).

I have sin(x+150) in my above equation. I want to convert it to cosine
EDIT: I seem to have some confusion.

IN my book it says cos(x) = sin(x+90) (just as you said).

This means if I want to convert to cosine from sin,

cos(x+60) = sin(x+60+90)

 

[Delta2]

You are correct now, so it will be cos (x+60)=sin(x+150), (x=377t), there you got a phase angle of 60, THOUGH when we talk about phase angle we usually mean the phase difference between voltage and current which is 90 in this case.