Hi,

I'm struggling with this task for my preperation for a exam next week.
Can sombody please help me?

An AC voltage of amplitude 150 volts is impressed across a pure resistance of 100 ohms. Calculate the current and the power dissipation when the frequency of the voltage is 60 cycles/sec.

 

I already gave you some tips.

Why are you posting the same question again?

 

Since the load is purely resistive the frequency doesn´t matter at all.
Amplitude means peak voltage, so you need to calulate the average voltage, then the equation is P=V*V/R and I=V/R
The current you calculated is completely wrong, VR/R is still V, not I. (unless VR means Vavg as in average voltage, then it would be true)

 

Is it possible to say that VR= Io R sin (wt) = 106 sin (2 pi *60) = 31 V

Also, this is completely wrong too. V0*sin (wt) is the instantenous voltage at a certain point in time t, so where did the t go in the second part? It should be Io*R sin(wt) = Io R sin (2*pi*60*t) so if you wanted to know the voltage at t=260ms this equation would be useful.

 

I think when he writes VR he is referring to V across resistor R, i.e. V with subscript R.

Sloppy notation.

 

Thanks for reply. So you can just say. I = 1.5A, and that the power disspation is
P = IVcos 1 =225 W

Is that correct?

 

No, the average current is 150*0.707/100=1.06A and the power dissipation is average voltage times average current = 112W. And no cosinus nor sinus in there.

 

Aha, thanks a lot for the help!

 

Power dissipation = 112W is correct.

But do not say average voltage times average current.

The correct answer is rms voltage times rms current.

Average voltage and rms voltage are not the same.

 

Thanks for all the replies