In a Full-wave rectification plus smoothing

[Mark Harder]

The first question has been answered. Draw a full-wave rectified circuit and you will see that the (positive) current can't flow backward. Kirchoff's voltage law supplies the answer to your second question. The law says that the voltage across a circuit is the sum of the voltages of the series components of the circuit. In the power supply circuit, there are no series connections between the + and - poles of the output of the supply, which applies to the resistor and the capacitor both.. Therefore, the voltage across them will be the same at all times. You said somewhere here that the currents through the capacitor and the resistor will add, and that's absolutely true.
Please note that the load to which the supply is delivering current is also a resistance in parallel to the capacitor and the supply's resistor. So the current out of the capacitor is distributed through both resistances as if they were one. You need to apply the parallel resistance rule to determine the current delivered by the supply to its own resistor and that of the load. That is why resistors added to power supply outputs have high values, in the megohm range commonly, so that most current is directed to where it's needed to do work.
A less formal way to think about it is this: Suppose the voltage across the capacitor is greater than that across the output resistance. Since the current can't flow backward, all of it will flow through the resistance. Then you must consider that Ohm's law says that it will create a voltage drop across the resistor (of the same polarity as the capacitor.) The current flowing out of the capacitor is initially high and the capacitor will lose voltage rapidly. On the other hand, the voltage added to the output resistance by that rush of currrent will be rapidly increasing Thus, the 2 voltages will approach each other rapidly, at which point they will be equal. The circuit will always act to equalize the capacitor and resistance voltages.

 

[BvU]

The first question has been answered. Draw a full-wave rectified circuit and you will see that the (positive) current can't flow backward. Kirchoff's voltage law supplies the answer to your second question. The law says that the voltage across a circuit is the sum of the voltages of the series components of the circuit. In the power supply circuit, there are no series connections between the + and - poles of the output of the supply, which applies to the resistor and the capacitor both.. Therefore, the voltage across them will be the same at all times. You said somewhere here that the currents through the capacitor and the resistor will add, and that's absolutely true.

So far, so good. Yes, one can add currents in one's mind. Kirchoff current laaw is clear in that current from AC source + current ex capacitor = current into resistor load. So if current from AC source = 0 THEN current ex capacitor = current into resistor load

Please note that the load to which the supply is delivering current is also a resistance in parallel to the capacitor and the supply's resistor. So the current out of the capacitor is distributed through both resistances as if they were one.

No. the diodes block current ex capacitor from going into the supply's resistor. I dislike this complication, but yes, real AC sources do have internal resistance.

You need to apply the parallel resistance rule to determine the current delivered by the supply to its own resistor and that of the load. That is why resistors added to power supply outputs have high values, in the megohm range commonly, so that most current is directed to where it's needed to do work.

This is nonsense and it is going to confuse potato no end.

A less formal way to think about it is this: Suppose the voltage across the capacitor is greater than that across the output resistance. Since the current can't flow backward, all of it will flow through the resistance. Then you must consider that Ohm's law says that it will create a voltage drop across the resistor (of the same polarity as the capacitor.) The current flowing out of the capacitor is initially high and the capacitor will lose voltage rapidly. On the other hand, the voltage added to the output resistance by that rush of current will be rapidly increasing Thus, the 2 voltages will approach each other rapidly, at which point they will be equal. The circuit will always act to equalize the capacitor and resistance voltages.

More formally: Q=CV AND V = IR . The wiring between C and R is supposed to have no resistance, so there is no voltage difference over those wires. therefore V_capacitor = V_{load resistor} at all times.

Potato, where are you now ? All clear or is the mist getting worse ?