# Charged capacitor doesn't allow impulses to pass

I tried this circuit and was confused about the theory of capacitors.

Here's what I'm supposed to do: on pressing S1, the capacitor is charged and the led blinks for a second. On pressing S2, the capacitor discharges.

When I try to press S1 without discharging the capacitor again, the led doesn't blink. Now capacitors are supposed to allow change in voltages across them i.e. the led should blink when the capacitor is charged, right? Why doesn't it?

Initially, the voltage on the + plate of the capacitor is 0V.

Pressing S1 connects it to a 9V source; current flows, charging the capacitor.

On releasing S1, the voltage on the + plate of the capacitor remains at 9V.

Pressing S1 again connects a 9V source to a plate charged to a potential of 9V, so no current flows. The LED does not light again.

If you attach a highish value resistor (10k-100k) between the + plate and ground, it will discharge the capacitor between presses of S1, so it will light again for each press of S1.

• How would this translate in terms of impedance? – Prabhpreet Jul 5 '13 at 9:56
• Impedance of what exactly? – pjc50 Jul 5 '13 at 10:25
• Impedance of the capacitor. – Prabhpreet Jul 5 '13 at 10:54
• Impedance of a capacitor depends on the frequency of the signal; I don't think you can compute a sensible, meaningful impedance number for sharp on/off switching of low or no particular frequency. While the signal is constant you can consider it "high impedance". – pjc50 Jul 5 '13 at 10:59
• the square edge generated by closing a switch is essentially a square wave and consists of some low frequency component and a (theoretically) infinite number of higher frequency harmonics. The capacitors present a low impedance to the high frequency harmonics, and a higher impedance to the lower frequencies, and an infinite impedance to the lowest frequency possible, DC. – Phil Frost Jul 5 '13 at 11:46