1
\$\begingroup\$

Why does the capacitor discharge (used as a battery to provide voltage to the resistor) after the peak of the alternating voltage (after the voltage reaches the highest value and goes on to decrease its value)?

Why after the peak? Why does it automatically discharge?

enter image description here

I thought there was a potential difference between the capacitor and the resistor or something?

\$\endgroup\$
5
  • 2
    \$\begingroup\$ Hint: What is the point of a power supply? What does it do? \$\endgroup\$
    – user253751
    Feb 15 at 1:15
  • \$\begingroup\$ A capacitor is like a very tiny rechargeable battery. \$\endgroup\$
    – td127
    Feb 15 at 1:47
  • 1
    \$\begingroup\$ Think of it like a bucket with a hole in it. Water (charge) flows out. Bigger hole, the steeper the slope. \$\endgroup\$
    – StainlessSteelRat
    Feb 15 at 2:35
  • \$\begingroup\$ What is your general understanding of what happens if you connect a charged capacitor to a load? Does it charge more, discharge, or nothing happens? \$\endgroup\$
    – Justme
    Feb 15 at 5:12
  • \$\begingroup\$ @Justme i believe it discharges as there are potential difference across the two plates and the charge only have the path of flowing through the resistor . \$\endgroup\$
    – Crown Raven
    Feb 16 at 10:45

1 Answer 1

Reset to default
2
\$\begingroup\$

When the voltage Vin reaches the maximum, it charges the capacitor with this maximum voltage value. When Vin starts to drop, the voltage across the capacitor becomes greater than the voltage Vin and the diode cuts-off (it is reverse biased). Then the capacitor starts discharging through the resistor, the only path for the capacitor's stored electrical charges . The capacitor discharge time depends on the values ​​of R and C (time constant = RxC). As shown in the [Vout x t] graph, the time constant is much larger than the voltage period Vin (time from one voltage peak to another).

\$\endgroup\$
2
  • \$\begingroup\$ how do we know that the voltage across capacitor is lower than vin? as i understand, the voltage across the capacitor is determined by how many charges it stored on one plate, it keeps accumulating. can u expain it with more detail please(sry im too dumb) \$\endgroup\$
    – Crown Raven
    Feb 16 at 10:43
  • \$\begingroup\$ The capacitor only accumulates charge when the source voltage (Vin) is greater than the voltage on the capacitor plus the voltage on the diode ("C charging" in the graph). When the voltage Vin starts to decrease (sinusoid decrease) the capacitor does not receive any more charges, because the diode is cut off (Vc >Vin). So it discharges on the resistor parallel to the capacitor. \$\endgroup\$
    – Luizzz
    Feb 23 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.