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Calculation of RMS Voltage and Current

  1. Nov 18, 2011 #1
    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.
  2. jcsd
  3. Nov 18, 2011 #2
    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.
  4. Nov 18, 2011 #3
    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.
  5. Nov 18, 2011 #4
    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.
  6. Nov 18, 2011 #5


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