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In a half wave rectifier only a single diode is present. One end of the secondary wire of the transformer is connected to the p side of diode while the other to the load resistor. The n side is connected to the load resistor. When the diode is reverse biased no current passes through it. But current does pass through the other wire of the secondary which eventually meets the load resistor . So wouldn't the output in this case be an ac current?

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  • $\begingroup$ Would Electrical Engineering be a better home for this question? $\endgroup$ – Qmechanic Nov 15 '17 at 6:31
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    $\begingroup$ "But current does pass through the other wire of the secondary which eventually meets the load resistor" - why do you say that? If I understand your setup correctly, there is (essentially) no current through the series resistor, diode, and secondary winding when the diode is 'off' (reverse biased). $\endgroup$ – Alfred Centauri Nov 15 '17 at 11:56
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Yes, it would be no DC in a strict sense. If the resistor is ohmic and the diode ideal, the current would look like a sine wave, where one half is set to zero. Though it would not be an AC, if AC is interpreted as strictly sinusoidal (with full wave). Sometimes this kind of waveform is called "pulsed DC", especially if a smoothing electrolyte capacitor is put in parallel to the resistor.

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  • $\begingroup$ It's DC with a massive ripple. It would not be "AC" because the flat part of the waveform is at zero, the half-sine is above zero. If you were to feed the output through a capacitor, you'd be back to AC. because the capacitor blocks the DC. $\endgroup$ – hdhondt Nov 15 '17 at 5:56
  • $\begingroup$ If a capacitor were in series with the resistor, it would be loaded up to the maximum voltage of the rippled DC. Afterwards no current would flow anymore, since the diode is always in current-free mode. $\endgroup$ – xeeka Nov 15 '17 at 6:20
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A current does not flow simply because you make a connection; you have to make two connections to form a circuit. When the diode is reverse biased it is like a switch being open, current cannot flow. When the diode is forward biased it turns on to complete your circuit.

The output through the load looks like this:

Half-wave rectified current

The diode blocks one polarity of the AC cycle but allows the other polarity through.

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