One cycle of the power dissipated by a resistor (R = 700 Ω) is given by P(t) 55 W, 0 < t < 18.0 s P(t) = 35 W, 18.0 < t < 30 s This periodic signal repeats in both directions of time. What is the average power dissipated by the 700-Ω resistor? Pau to within three significant digits)

[tex]\frac{1}{t} \int\limits^t_0 {P(t)} \, dx \\=\frac{1}{30} (\int\limits^18_0 {55} \, dx + \int\limits^30_18 {35} \, dx )\\\\=\frac{1}{30} (55*(18-0) + 35(30-18))\\\\=\frac{1410}{30}\\[/tex]

=47W

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nafeesahmed nafeesahmed · Answer 2 · 2020-03-06T18:27:24+01:00

Answer:

[tex]P=47[/tex] [tex]W[/tex]

Explanation:

The average power dissipated in a resistor is given by

[tex]P=\frac{1}{T} \int\limits^T_0 {} \, p(t)dt[/tex]

Where T is the time taken by the sine wave to complete one cycle.

We have instantaneous power P(t) = 55 W for 0 < t < 18s and P(t) = 35 W for 18 < t < 30s

So the time period is T = 30 seconds

Now we will integrate both of the instantaneous powers

[tex]P=\frac{1}{30} (\int\limits^b_a {} \, 55dt + \int\limits^b_a {35} \, dt )[/tex]

[tex]P=\frac{1}{30} (55t +{35} t )[/tex]

[tex]P=\frac{1}{30} (55(18-0) +{35(35-18)} )[/tex]

[tex]P=\frac{1}{30} (55(18) +{35(12}))[/tex]

[tex]P=\frac{1}{30} (990 +420)[/tex]

[tex]P=\frac{1410}{30}[/tex]

[tex]P=47[/tex] [tex]W[/tex]