# Am I allowed to use KLV and KCL with a capacitor in the circuit?

Gold Member
I'd like to explain how the original circuit I posted works. I'll repost its image if I may:

http://img853.imageshack.us/img853/9302/diodescabal.jpg [Broken]

Once the switch is closed (and the push button remains unpressed) all the current flows towards the capacitor until it fully charges, then afterwards current flows through the resistor and LED's to the ground. When it passes the 5 mA for each diode they light up (in different times since the resistors don't have the same value). P is there to cause a shortcircuit thereby discharging the capacitor.

How did I do?

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vk6kro
Current starts to flow in the LEDs when the voltage across the capacitor rises above their turn-on voltage.

For white LEDs this is about 3.5 volts but it is different (lower) for other colours.

When you actually see the LEDs lighting up depends on the efficiency of the LED and the background lighting.
In a dark room, you may see the light at 1 or 2 milliamps.

Is that capacitor actually 500 MILLI Farads? ie half a Farad?

Gold Member
Current starts to flow in the LEDs when the voltage across the capacitor rises above their turn-on voltage.

For white LEDs this is about 3.5 volts but it is different (lower) for other colours.
Let me see if I have it straight. At the first moment the switch is closed, all the current flows towards the capacitor. The more time goes by, the more current starts flowing through the diodes (and resistors) until all the current flows through them. So the capacitor in fact just steals a bit from the current influx in the beginning until the diodes get it?

Is that capacitor actually 500 MILLI Farads? ie half a Farad?
Yes

vk6kro
The capacitor will charge up to almost the supply voltage because the diodes draw very little current compared to the 3.75 amps that the capacitor initially draws.

The final voltage across the capacitor will be caused by the supply voltage (15 volts) minus the voltage drop due to the very small current drawn by the diodes. This will be something less than 32 mA so the drop will only be about 128 mV.

If you were to really build this, you would need to limit the current in the switch across the capacitor because this could be very large (depending on the internal resistance of the capacitor) and may damage the switch.

A good way to do this would be to put a small resistor between the capacitor and the 4 ohm resistor, then shorting that junction of the two resistors to ground. About 1 ohm should be adequate protection for the switch.

This way, the capacitor could fully discharge in a few seconds.

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Gold Member
OK, thanks for this, but I'm trying to gauge this process based on time. What happens in the first moment it's closed, till the steady state. I was hoping my explanation above is correct?

vk6kro
In the first moment, there is not enough voltage present to turn on the LEDs. So the only current will be about 3.75 amps flowing into the capacitor and charging it up.

As the voltage across the capacitor gets greater, this 3.75 amps decreases to (15 - Vcap) / 4.

Eventually, one or both of the LEDs will start drawing a little bit of current and some time after that, they will be visibly lit up.

Gold Member
In the first moment, there is not enough voltage present to turn on the LEDs. So the only current will be about 3.75 amps flowing into the capacitor and charging it up.

As the voltage across the capacitor gets greater, this 3.75 amps decreases to (15 - Vcap) / 4.

Eventually, one or both of the LEDs will start drawing a little bit of current and some time after that, they will be visibly lit up.
How can you say that there is not enough voltage at the first moment if voltage is constant at 15 V?

NascentOxygen
Staff Emeritus