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# Physics/Peak current and rms and frequency

Question
Current through a 50ohm resistor is I=0.8sin(240t). What is the rms current? At what frequency does it vary? What is the power dissipated in the resistor?

I don't know how to approach this question since it's an equation. I do know the 0.8 refers to the current, but I'm not sure how to find the others.

Hello Sona,

The peaks of the current, if you watched on an oscilloscope, would be +0.8 A and -0.8 A. When the waveform is a pure sinewave (and that would be true here), RMS is the peak value multiplied by the square root of 1/2. Therefore
Irms = 0.8*0.707 Arms (or just A if it is understood that we are talking about an rms measurement.)

The 240t part gives the information to calculate frequency. As time advances, the argument for the sine function increases. So if you plot points, I versus t, you get a sine wave. When t advances from zero until 240t is equal to 2*pi, it has gone through one cycle of the oscillation. So
if 240*t = 2*pi, then t = 2*pi/240 = 0.026 sec
That is the period, T, of the oscillation, so the frequency is given by
f = 1/T

One advantage of an RMS measurement, is for power calculations. If this were a DC current, power is just P=I*R^2. If it is an AC current it can be more complicated because the amplitude keeps changing. But if you have an RMS measurement for I, then you can use
P=I*R^2

I hope this helps,
Steve

Physics

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