Hey everyone,

I don't think I was particularly clear in my last question, my apologies.

How would a capacitor discharge in a circuit like the one attached?

 

Hey everyone,

I don't think I was particularly clear in my last question, my apologies.

How would a capacitor discharge in a circuit like the one attached?

In this circuit, current would flow as the capacitor charged for the first time when connected to the voltage source (V1). This would cause the LED (D1) to illuminate briefly and fade away as the capacitor charged and the current flow dropped.

Once charged and in steady state, the LED would be off and no current would flow in the circuit.

EDIT: Erroneous comment regarding reverse biasing the LED removed. Thanks everyone for the correction. See Post #3 below.

Capacitors appear as open circuits to DC voltage, so current could only flow during voltage transients such as charge and discharge.

 

If the voltage source were disconnected and replaced with a jumper, the capacitor would discharge and again, the diode would illuminate briefly until the capacitor became discharged.

This would reverse bias the LED so it would not illuminate but the capacitor would discharge through the LED leakage and the resistor.

 

Your post was perfectly clear.
The cap in post #1 is connected backwards.

 

Hi mcardoso, I don't agree with your statement "If the voltage source were disconnected and replaced with a jumper, the capacitor would discharge and again, the diode would illuminate briefly until the capacitor became discharged. " When the battery is replaced by a jumper the LED would be reverse biased so it would not illuminate. Most LEDs when reverse biased with 9 volts would be destroyed. If it was a normal diode when reverse biased almost no current would flow so the capacitor would discharge very slowly.

Les.

 

How would a capacitor discharge in a circuit like the one attached?

It won't, and the capacitor is backwards.

 

To better understand, replace the led with a volt meter.