Question

One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤t<5.0 s P ( t ) = 25 W , 5.0 ≤ t < 10 s P(t)=25 W, 5.0≤t<10 s This periodic signal repeats in both directions of time. What is the average power dissipated by the 800- Ω Ω resistor?

Answer 1

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$