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.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?
This would reverse bias the LED so it would not illuminate but the capacitor would discharge through the LED leakage and the resistor.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.
It won't, and the capacitor is backwards.How would a capacitor discharge in a circuit like the one attached?
by Jake Hertz
by Jake Hertz
by Jake Hertz
by Aaron Carman