# Calculate the average power dissipation in Joules

I'm self-studying this problem, and it seems that have a problem solving this.

The only thing I didn't understand is this solution is how to calculate the average power dissipation. It's never mentioned in class and I can't find any formula even to proceed.

I know that:

$$p_{avg} = \dfrac{V_m \cdot I_m}{2}$$

And I don't see how I can apply this formula to the problem. And I don't think it is helpful anyway since it gives the average power in Watts.

So, how can I calculate the average power dissipation?

• Hint: JOULES measures energy. Power is energy/time. (1Joule/sec = 1 WATT) – JIm Dearden Sep 14 '14 at 18:30
• Remember Cal 2? Yeah, you'll need that. – Ignacio Vazquez-Abrams Sep 14 '14 at 18:30

The J is probably a typo, unit of power is watt and not joule. To calculate signal power: $P=\frac{1}{T}\int\limits_0^T{u(t)i(t)\,dt}=\frac{1}{T}\int\limits_0^T{\frac{u^2(t)}{R}dt}=\frac{4}{TR}\int\limits_0^{T/4}{(10t)^2dt}=\frac{4\cdot 100}{4\cdot 100}\int\limits_0^{1}{{t^2}\,dt}=\left[\frac{t^3}{3}\right]_0^1=\frac{1}{3}-0=\frac{1}{3}\,\mathrm{W}$
In part (b) you solved for the instantaneous power $p(t)$.
$$\dfrac{1}{T}\int_t^{t+T}p(t)\mathrm{d}t$$