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.
Electrical – why the power absorbed by RL load in half wave rectifier is I^2R not V^2/R nor VI
powerrectifier
Related Solutions
Either can work correctly if designed properly. If you have a dumb rectifier supply feeding a 7805, then all the rectifier part needs to do is guarantee the minimum input voltage to the 7805 is met.
The problem is that such a power supply only charges up the input cap at the line cycle peaks, then the 7805 will drain it between the peaks. This means the cap needs to be big enough to still supply the minimum 7805 input voltage at the worst case current drain for the maximum time between the peaks.
The advantage of a full wave rectifier is that both the positive and negative peaks are used. This means the cap is charged up twice as often. Since the maximum time since the last peak is less, the cap can be less to support the same maximum current draw. The downside of a full wave rectifier is that it takes 4 diodes instead of 1, and one more diode drop of voltage is lost. Diodes are cheap and small, so most of the time a full wave rectifier makes more sense. Another way to make a full wave rectifier is with a center tapped transformer secondary. The center is connected to ground and there is one diode from each end to the raw positive supply. This full wave rectifies with only one diode drop in the path, but requires a heavier and more expensive transformer.
A advantage of a half wave rectifier is that one side of the AC input can be directly connected to the same ground as the DC output. That doesn't matter when the AC input is a transformer secondary, but it can be a issue if the AC is already ground-referenced.
Assuming sinusoide waveform:
Diode drop voltage shorten the duty cycle δ. Diode drop voltage depends on Iavg (see diode datasheet Iavg vs Vf curve)
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Best Answer
\$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.