# Video: Identifying Circuits That Produce Half-Wave Rectification

Diagram (a) shows a circuit that can be used to rectify an alternating current. If the input voltage is that shown in diagram (b), which of the following graphs shows the output voltage as measured by the voltmeter in the circuit diagram?

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### Video Transcript

Diagram (a) shows a circuit that can be used to rectify an alternating current. If the input voltage is that shown in diagram (b), which of the following graphs shows the output voltage as measured by the voltmeter in the circuit diagram? [A] Diagram 1 [B] Diagram 2 [C] Diagram 3 [D] Diagram 4

Okay, so in this question, we’ve got diagram a, which shows the circuit that we’re considering, and diagram b, which is the input voltage that’s produced by the AC source in this circuit. And as well as this, we’ve been told that the circuit in diagram a can be used to rectify an alternating current. Now, based on this information, we need to work out which one of these four graphs here shows the output voltage as measured by the voltmeter in the circuit.

Okay, so to answer this question, let’s first consider what we have in the circuit diagram in diagram a. What we’ve got is an AC source, a diode, a resistor, and a voltmeter. Now, the AC source, the diode, and the resistor are in series. And the voltmeter is measuring the potential difference across the resistor. Now, the AC source is producing a sinusoidal potential difference across the circuit. And if we had a circuit where there was no diode, but instead we just had an AC source, the resistor, and the voltmeter, then the sinusoidally varying potential difference produced by the AC source would be the same as the potential difference across the resistor. In that situation, we would have a sinusoidally varying current through the resistor as well.

However, coming back to the circuit in diagram a, that’s not what we’re going to see because remember we have a diode. Now, a diode is a circuit component that only allows current flow through it in one direction. Now, we can choose to say that anytime the potential difference produced by the AC source is positive, that is going to generate a current flow in this direction that’s clockwise around the circuit. And then, we can see that that current flow is actually allowed to be sustained because current can flow in this direction through the diode. We can see which direction a diode allows a current through it by looking at the arrow in that circuit diagram.

And so, anytime the potential difference is positive produced by the AC source, there is a current in the clockwise direction in the circuit. Therefore, there is a current through the resistor in this direction in the circuit. And then for the resistor, we can recall that the potential difference across the resistor which is the value measured by the voltmeter is equal to the current through that resistor multiplied by the resistance of the resistor itself. So the point is that anytime the potential difference from the source is positive, a current is flowing clockwise through the circuit which is allowed to flow because of the diode. And so, there is a nonzero current for the resistor. And if there’s a nonzero current through the resistor, then the potential difference also is nonzero, where that potential difference is the voltage measured by the voltmeter. And this is because we’re looking at Ohm’s law just for the resistor and the resistance of the resistor is a constant.

Therefore, we can say that the voltmeter will measure the same potential difference across the resistor as is produced by the source whenever the source is actually producing a positive voltage. However, as soon as we look at the negative portion of the curve, then it’s a whole different story because when the voltage from the AC source becomes negative, that negative voltage tries to generate a current in the circuit in the opposite direction to before. In other words, this AC source is now trying to set up a current in the counterclockwise direction, which would be fine if it weren’t for the diode because the diode does not allow a current through it in the opposite direction. And hence, there is no current in this circuit because there aren’t any other branches for the current to flow through that don’t end up at the diode at some point.

And so, whenever the potential difference from the source is negative, there is no current in the circuit. So the current of the resistor is zero. And then coming back to Ohm’s law once again, if the current is zero, then the potential difference across the circuit, which is remember being measured by the voltmeter, is also going to be zero. So in other words, for the entirety of the time that the potential difference source is producing a negative voltage, the voltage measured by the voltmeter itself is going to be zero because there is zero potential difference across the resistor.

Then, we go back once again to the positive part of the cycle. And in that situation, once again, we will have a positive potential difference measured by the voltmeter. And that potential difference will follow the voltage produced by the AC source. And once again, we return to the negative part which is going to result in a zero voltage measured by the voltmeter. And so, all in all, what we expect to see is something like this when it comes to the potential difference measured by the voltmeter. And at this point, we can identify this as a half-wave rectified potential difference.

But more importantly, out of the four diagrams that we have to choose from, we can see that the first of these diagrams correctly matches what we’re expecting to see in terms of the voltage measured by the voltmeter in diagram a.