A square pulse train (i.e. square wave) with low and high voltage values of VL= 0 and VH= 3.1 volts is applied to a DC voltmeter. The voltage measured by the instrument is 0.1 volts. The same signal is applied to an 87 Ω resistor. Find the average (real) power dissipated on the resistor in mW. Round off your answer to two decimal places.

This was on old homework question and I'm studying for my finals, but this problem is killing me. My professor said to find the duty cycle to answer, but even then I'm not sure what good it does.

The answer is 3.56 Watts at least by the homework website.

Thanks

 

First ask if the answer can possibly be correct. Consider:

We apply a DC voltage of 3.1V to an 87Ω resistor. What is the power dissipated in the resistor? If you then start switching the voltage off part of the time, do you expect this power to go up or down? So would you expect to be able to get 3.56W of power into this resistor with such a waveform?

But let's forget about the supposed correct answer (are you sure it isn't supposed to be 3.56mW?). Consider the following questions:

Q1) Let's assume that the waveform is at VH for a fraction of the time, call it 'x', and at VL the rest of the time. In terms of VH, VL, and x, what voltage, Vdc, will a DC voltmeter read?

Q2) If this waveform is applied to a resistor R, how much power, P, will be dissipated, in terms of VH, VL, and x?

Q3) Can you take the two expression from the above and eliminate x from them so that you have a single equation for P in terms of VH, VL, and Vdc?

 

My bad on the Watts, can't type. But I did figure it out, thanks.
Big thing is to make sure I am in the same time units, seconds. I was in different units.

 

My bad on the Watts, can't type. But I did figure it out, thanks.
Big thing is to make sure I am in the same time units, seconds. I was in different units.

Which is why you should always, always, always track your units through every step of your work. It's easy to make mistakes, and tracking your units will let you catch the majority of them.