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why the power absorbed by RL load (resistor in series with inductor) in half wave rectifier is I^2R not V^2/R nor VI. V , I are with RMS value.

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    \$\begingroup\$ Who says it isn't? \$\endgroup\$ – The Photon Dec 26 '18 at 20:20
  • \$\begingroup\$ i calculated the rms value of I & V and applied in these formulas and got different results , plus there is no linear relation ship between I & V. \$\endgroup\$ – 0xDEADC0DE Dec 26 '18 at 20:46
  • \$\begingroup\$ Current in an RL circuit is affected by impedance, not just resistance, so current must be derived from voltage and impedance. Power dissipation (not reactive or apparent power) is related only to resistance, and thus can be calculated based on this current, but if the current had been calculated without the inductance, it would be incorrect. The E^2/R formula doesn't work on an RL circuit because it's bypassing impedance, which must factor in. Is there a chance this is the source of your misconception? \$\endgroup\$ – K H Dec 26 '18 at 23:07
  • \$\begingroup\$ Probably because you're not being specific in what is defined as what. The power dissipated ("absorbed" may work here, but is misleading) is dissipated in the resistor, and is equal to \$V_R^2 / R\$. But if you define \$V\$ as something other than \$V_R\$, it will, of course, not be the correct voltage to use for calculating power dissipated in the resistor. \$\endgroup\$ – TimWescott Dec 26 '18 at 23:58
  • \$\begingroup\$ i got the silly mistake i made lol. V^2/R = I^2R only if V is across the resistance. \$\endgroup\$ – 0xDEADC0DE Dec 27 '18 at 9:04
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\$V^2/R\$ is also correct if you use RMS volts.

To understand why V*I is not correct, it’s good to understand how average power of a periodic signal is calculated.

The instantaneous power is always v(t)*i(t).

To determine the average power you integrate it over some interval and divide by the time for the average. If the signal is periodic you can take that integral over a single cycle and you know that each cycle is the same in steady-state.

If the load happens to be a fixed resistance, you can substitute for one of i(t) or v(t) and get, say, the integral of \$v(t)^2\$/R. We call the square root of that integral (divided by the interval time) the RMS voltage. The resistance is fixed so it can be applied later, for any resistance.

There is no physical meaning to RMS volts * RMS amperes.

You should also note that the usefulness of RMS current or voltage is more-or-less limited to when you have resistive loads- for example it does not help much if the load is a diode.

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  • \$\begingroup\$ but if the load is RL in half wave rectifier, the current and the voltage have totally different shapes (no linear relationship nor shifting). how is it v^2/R is same as i^2R \$\endgroup\$ – 0xDEADC0DE Dec 26 '18 at 20:55
  • \$\begingroup\$ The relationship between voltage and current for an ideal resistor is always described by Ohm's law, so the shape of voltage and current waveforms (across/through a resistor) will be always be identical, just scaled by a constant (R). \$\endgroup\$ – Spehro Pefhany Dec 26 '18 at 21:01
  • \$\begingroup\$ The load is Resistor in series with inductor (RL) not just Resistor. circuit : i.ytimg.com/vi/iGY8c8bukf8/maxresdefault.jpg \$\endgroup\$ – 0xDEADC0DE Dec 26 '18 at 21:04
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    \$\begingroup\$ The RMS voltage across the resistor is just the RMS current through the combination multiplied by the resistance. The RMS current value allows you to calculate the power dissipation in the resistor. Other than that, the RMS voltage/current may not be of much use. Actually I see a point of confusion here- in your question you refer to RL which you (we now know) means resistance and series inductance, but typically a simple load resistor is often referred to as \$R_L\$. \$\endgroup\$ – Spehro Pefhany Dec 26 '18 at 21:07
  • \$\begingroup\$ Sorry for misleading. by The RMS voltage i meant the sum of inductor and resistor voltages as are they both the load,as shown in the previous picture . by textbook "Power Electronics by Daniel W Hart" states that the power is I^2 R There is no linear relationship that's why i'm confused. \$\endgroup\$ – 0xDEADC0DE Dec 26 '18 at 21:19

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