What is the rms current draw of a microwave operating at 1180 W and 120 Vrms?

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Homework Help Overview

The discussion revolves around calculating the rms current draw of a microwave operating at a specified power and voltage. The original poster presents a scenario involving a microwave with a peak power consumption of 1180 W and an operating voltage of 120 Vrms.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore relationships between power, voltage, and current, questioning the correct application of rms and peak values. There are attempts to derive current using various equations, including P=VI and P=I^2R, while some express confusion over the conversion between rms and peak values.

Discussion Status

Several participants have provided insights into the relationships between different power definitions and the importance of consistency in using rms or peak values. There is ongoing exploration of how to combine equations to find the correct current draw, with some participants expressing uncertainty about their calculations.

Contextual Notes

Participants are navigating potential confusion regarding the definitions of average and peak power, as well as the implications of using different forms of voltage and current in their calculations. There is mention of a possible misunderstanding related to factors of two in the equations used.

saritche
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Homework Statement


peak power consumed by microwave is 1180 W when operated at 120 Vrms. What rms current does that microwave draw?


Homework Equations


I'm not getting the correct answer, I think I am getting I and V confused with Irms and Vrms and Imax.

The Attempt at a Solution


I=P/V
Peak=I^2R
Vrms=.707V

 
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so waht answer do you get & how?
 
I found a value for R with P=V^2/R and tried plugging that into P=I^2R, then I took that I multiplied by .707
 
best to work all in rms or all in peak, never combine the 2

note Vrms = Vp/sqrt(2) where Vp is peak V
and Irms = Ip/sqrt(2)

If P = VI, what is the relation between peak & avg power

If we say

Pavg = Vrms.Irms
Pp = Vp.Ip
 
Last edited:
thats the part I don't understand, how to combine the two equations to get the relationship between the two, are they equal because power is conserved?
 
P=VI is always true & its fine to use either (rms or peak), so you know you are working in peak or rms exclusively

if you combine the definitions below they give
Pp = 2.Pavg
this is probably where the confusion comes in

so convert to avg power then use P = VI
 
the answer is just not coming out:
I plugged in Ppeak=2(Vrms*Irms)
which is 1180=2(120)*2Irms
1180/240=2Irms
Irms=2.458

am I not getting something? or is it a number thing?
 
saritche said:
the answer is just not coming out:
I plugged in Ppeak=2(Vrms*Irms)
which is 1180=2(120)*2Irms

? you have added an extra factor of two here, any reason?

saritche said:
1180/240=2Irms
Irms=2.458

am I not getting something? or is it a number thing?
 
I assumed I had to multiply 2 times each factor 2(VI). I got it now... thank you so much for your help!
 

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