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## Homework Statement

A voltage of [tex] V = V_0 sin(\omega t) [/tex] is applied to an inductor. Find an expression for the current and hence the power.

## Homework Equations

I have used the equation: [tex] V = -L \frac{di}{dt} [/tex]

and for power: [tex] P = IV [/tex]

## The Attempt at a Solution

I tried to find I(t) by integrating both sides of the first equation:

[tex] V_0 \int_{0}^{t}\sin({\omega t}) dt = -L \int_{0}^{I} dI [/tex]

[tex] \frac{V_0}{\omega} (1-\cos({\omega t})) = -LI [/tex]

Giving the expression for the current as:

[tex] I = -\frac{V_0}{L\omega}(1-\cos{(\omega t})) [/tex]

Then for the power:

[tex] P = IV = V_0 sin(\omega t) \times -\frac{V_0}{L\omega}(1-\cos{(\omega t})) [/tex]

[tex] P = -\frac{V_0^2}{L\omega}(\sin(\omega t))(1-\cos{(\omega t})) [/tex]

However, the answer only provides the power, which is given by: [tex] P = -\frac{V_0^2}{L\omega}(\sin(2\omega t)) [/tex]

I believe that the answer is missing a factor of a 1/2 in there, but I am still not sure how to arrive there. I believe that I must have gone wrong in the current calculation, but upon trying other methods (eg. differentiating enegry for inductor wrt t, it all leads to the same integral)

Any help is greatly appreciated.