Calculation of RMS Voltage and Current

Click For Summary

Discussion Overview

The discussion revolves around the calculation of RMS voltage and current in a theoretical circuit resembling an on-off switch with a 50% duty cycle. Participants explore the implications of these calculations in the context of power delivery and the characteristics of the waveforms involved.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that the Vrms for a constant 5V input is 5V, while the Irms for a square wave with a 50% duty cycle is proposed to be I/√2 = 1/√2.
  • Another participant argues that Prms = Vrms * Irms is only valid for purely resistive loads and that the voltage across the resistor is not constant, thus Vrms cannot be simply 5V.
  • A later reply suggests that to find the average power, one must integrate V(t)*I(t) over one cycle, indicating that Vrms would not be 5V in this scenario.
  • One participant seeks clarification on whether there is a term "Prms" or if it is simply average power, noting that integrating V*I over time yields energy.
  • Another participant points out that the expression I/√2 is valid only for sine waves, challenging its application to the square wave scenario presented.

Areas of Agreement / Disagreement

Participants express differing views on the validity of using Vrms and Irms calculations for the given circuit. There is no consensus on the correct approach to calculating power or the definitions of terms like Prms.

Contextual Notes

Participants highlight limitations in their calculations, particularly regarding the assumptions about waveform shapes and the conditions under which certain formulas apply. The discussion remains focused on theoretical considerations without resolving the discrepancies in power calculations.

vinayakbhat82
Messages
6
Reaction score
0
I've a theoretical circuit that resembles an on-off switch with a duty cycle of 50%. The voltage input to the circuit is a constant 5v. The current drawn from the source is 1A when switch is ON and 0A when OFF. What is the Vrms and Irms?

From what I know:
Vrms for constant 5v is 5v.
Irms for above current (square wave with 50% duty cycle) would be I/√2 = 1/√2.
This imples P=Vrms*Irms = 5/√2

However, if we separately do some calculations for power, it would be like 5*1*0.5 (V*I*duty chle) = 5/2.

Can this discrepancy be explained please? Please let me know what am I missing.
 
Physics news on Phys.org
vinayakbhat82 said:
I've a theoretical circuit that resembles an on-off switch with a duty cycle of 50%. The voltage input to the circuit is a constant 5v. The current drawn from the source is 1A when switch is ON and 0A when OFF. What is the Vrms and Irms?

From what I know:
Vrms for constant 5v is 5v.

Prms = Vrms * Irms is only valid for a purely resistive load.

If you have a battery, a switch and a resistor in series, you can consider V and I across the resistor, and V won't be constant, so Vrms isn't 5V.

You can also consider V and I across the resistor and switch, and then Prms = Vrms * Irms
isn't valid and you have to do an integral of V(t)*I(t) over one cycle.
 
Hey willem2,

Thanks for the reply.

Yes, I did have that idea earlier, but I was trying to answer the following: If I've a blackbox that has the circuit you mentioned i.e. switch and a resistor with no access to internal nodes and I need to have a DC equivalent of the original source that delivers the same power, how would I arrive at the numbers if I was provided with the voltage and current waveforms at the boundary. By reverse engineering, we may say that Vrms needs to be 5/√2 in this case, however, I wanted to understand the rationale behind doing so.

I completely agree to your point that the rms voltage across the resistor would not be 5v, but would rather be 5/√2.

Is there a term called Prms or is it just P/ If so, please help me understand. Also I believe, integrating V*I over time will give you Energy.
 
vinayakbhat82 said:
Is there a term called Prms or is it just P/ If so, please help me understand. Also I believe, integrating V*I over time will give you Energy.

There's no Prms, just average power.

The integral indeed gives the energy used in one period, so you'll have to divide by the length of the period to get the average power.
 
Irms for above current (square wave with 50% duty cycle) would be I/√2 = 1/√2.

This is valid only for a sine wave. ie the I/√2
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 16 ·
Replies
16
Views
14K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K