Let me establish some facts I have understood by myself: The polarity of resistor does not change, as the current from A.C source flowing through does not change direction. The D.C potential difference value fluctuates from zero to maximum. To reduce this fluctuation a smoothing capacitor is used.

What I don't understand is:

  1. Is the equation \$t=RC\$ applicable to both charging and discharging of capacitor?
  2. Is the time constant the same in both cases?
  3. If so then why does the capacitor charge fully in time \$t\$ as the potential difference rises to maximum initially but discharge slowly as the potential difference falls? Its time constant for charging and discharging is different apparently! It's greater for discharging and less for charging. Why is this?
  4. Initially, as the capacitor charges won't it cause some of the current to be diverted towards itself and less current would flow through resistor, as they are arranged in parallel?
  • 2
    \$\begingroup\$ What is "p.d" ? \$\endgroup\$ – Phil Frost Aug 12 '14 at 11:43
  • \$\begingroup\$ p.d. = potential difference. \$\endgroup\$ – Andy aka Aug 12 '14 at 11:53
  • \$\begingroup\$ The t = RC formula establishes a time constant. When the voltage on the RC network suddenly changes (is subject to a "step function"), the time constant tells us how much time is required to reach 63% of the way between the old voltage and the new voltage. This applies in other kinds of physical systems. \$\endgroup\$ – Kaz Aug 12 '14 at 14:38
  • \$\begingroup\$ If the resistor is understood to be the load of the capacitor, we do not consider it to be in parallel. It is in fact in series with the capacitor: we regard the capacitor and the load resistor to be a circuit. It just happens that the rectifier is connected to this circuit such that it is parallel to the R and to the C. \$\endgroup\$ – Kaz Aug 12 '14 at 14:52
  • \$\begingroup\$ The time constants for charging and discharging are not the same, because the capacitor charges from the rectifier, but discharges through the load. The valve-like action of the rectifier diodes ensure the one-way flow of current. \$\endgroup\$ – Kaz Aug 12 '14 at 14:54

The charge time of the capacitor is hardly affected by load resistance at all (except in extremes). The charging of the capacitor is determined by the resistance of the forward conducting diodes in the bridge rectifier - this resistance is likely to be around 1 ohm or less.

The discharge time of the capacitor is determined only by the load resistor which is probably many ohms hence charge and discharge times will be different.

A more exact answer would consider the leakage inductance of the transformer feeding the bridge and the cable resistance feeding the transformer but, in general, the equivalent forward resistance of the diodes in the bridge ensure the cap charges quickly (compared to the discharge due to load resistance.

Here's a picture that shows an applied AC voltage and a single diode charging up a capacitor: -

enter image description here

After a few cycles of AC the capacitor is starting to receive full charge. Remember the picture is a half-wave rectifier so the negative excursions of the AC aren't contributing. With a significant load resistance there will be a droop/decay on the capacitor voltage between conduction periods of the diode but, under normal circumstances the voltage will have a trend that is asymptotic with the peak AC voltage minus 1x diode volt drop of about 0.7 volts.

Here's a picture that shows the droop in the capacitor voltage between diode conduction periods. This is a full-wave picture and assumes the diode is perfect: -

enter image description here

  • \$\begingroup\$ that really sorts out everything but could you also please help me out in the last question about the current flowing through the resistor \$\endgroup\$ – Sara Aug 12 '14 at 12:02
  • \$\begingroup\$ If the voltage supply is near perfect and the diodes were very low impedance when conducting, the current supplied to both charge the capacitor and power the load resistor would be high enough so that both are adequately serviced. As impedances rise (diode and supply) there will come a point when the peak voltage that the cap can be charged to is limited by the load current and therefore the output voltage will not reach maximum potential. \$\endgroup\$ – Andy aka Aug 12 '14 at 12:17
  • \$\begingroup\$ just wanted to clear it up that if the capacitor was not included in the circuit more current pass through the resistor compared to when the capacitor is there \$\endgroup\$ – Sara Aug 12 '14 at 12:27
  • \$\begingroup\$ With the capacitor in-circuit, the load resistor will be sustained at a much higher average voltage than circumstances when the cap is not there. \$\endgroup\$ – Andy aka Aug 12 '14 at 12:30

Yes, t = RC applies, but the diode(s) in the rectifier are nonlinear elements. Their effective resistance changes with the polarity of the voltage across them.

In both cases, the resistance value you use in the equation is the parallel combination of the load resistor and the diodes. However, when charging, the diodes are conducting — have a low resistance — and so the RC time constant has a low value that's completely dominated by the diodes. When discharging, the diodes are not conducting (high resistance), and the RC time constant is dominated by the load resistance.


The time constant is a property of the circuit. The time constant is (not) the time needed for charging the capacitor (nor) the time needed to discharge the capacitor.

enter image description here

From the figure above, there are three time intervals that you should take care about, each is different than the other. The time interval represented by the green line is the (charging time). The time interval represented by the red line is the (discharging time). The time interval represented by the purple line is the (time constant). In the charging process, it represents the time needed to charge the capacitor from zero to 63% of its capacity. In the discharging process, it represent the time needed to discharge the capacitor from full to 37% of its capacity. The time constant is the same in both processes.


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