# Circuit: Total Energy into a component over interval

In summary, the conversation discusses finding the total energy absorbed by an electric component from time t1 to t2. The equation for power is used to find the expression for the total energy, and a hint is given for evaluating the expression for a specific value of t2.

## Homework Statement

The voltage at terminal a relative to terminal b of an electric component is given by $v(t) = 20\cos(120\pi t)$ Volts and the current into a is $i(t) = -4\sin(120\pi t)$ amps. Find 1) the total energy that flows into component from time t1 to t2 and 2) in particular find the energy absorbed when t2 = t1 + 1/15.

## Homework Equations

Power = i(t)*v(t)

## The Attempt at a Solution

So I set Power = dE/dt = i(t)*v(t) and integrated to yield the final expression

$$\DeltaE_{t_1\rightarrow t_2} = \frac{1}{6\pi}\left[\cos(240\pi t)\right]_{t_1}^{t_2}\qquad(1)$$

I believe that this expression takes care of part 1). However, for part 2), I am unclear on how to evaluate the expression from t1 to t1 + 1/15. This becomes:

$$\DeltaE_{t_1\rightarrow t_2} = \frac{1}{6\pi}\left[\cos(240\pi t_1+1/15) - \cos(240\pi t_1) \right]$$

Is there some sort of trig trick I a can use to evaluate this? Or somehow use the fact the a cosine function is periodic?

Just need a hint here

$$\DeltaE_{t_1\rightarrow t_2} = \frac{1}{6\pi}\left[\cos({\color{red}240\pi t_1+1/15}) - \cos(240\pi t_1) \right]$$
Maybe it's just a typo within LaTeX, but you didn't multiply through the 240π quite correctly.
Is there some sort of trig trick I a can use to evaluate this? Or somehow use the fact the a cosine function is periodic?

Just need a hint here
Here's a hint: Note that

$$\cos (\omega t + \theta) = \frac{e^{j(\omega t + \theta)} + e^{-j( \omega t + \theta)}}{2}$$

Try to express your equation exponentially. You'll be able to pull out a complex exponential (that's not a function of t1), and functions as a complex constant. Recalling that [itex] e^{j \theta} = \cos \theta + j \sin \theta [/tex], the whole thing can be reduced in this particular problem.