I have a couple of questions on capacitor?

Question

1) Why does the resistance of a light bulb, connected to a capacitor in series, decrease as a capacitor is charged in a circuit?

2) What is the physical meaning of scale factor, pertaining to capacitor?

For Q1, I have a graph of voltage across resistor vs current across it, and I got a straight line linear graph. When I changed the resistor to light bulb, the slope of the graph became gentler after some time. Please explain this? The bulb is connected in series to a capcitor and a 3V battery.

Answer 1

The light bulb has a positive thermal resistance coefficient. As the filament heats up, the resistance increases dramatically.

If you check the resistance on a 100W light bulb at room temperature, it is about 4 ohms. According to ohms law, it should draw over 30 amps when connected to 120 volts! However, in actual operation, it draws approximately 1 amp. The resistance of the filament increases to almost 120 ohms as it heats up.

As your capacitor in series charges, the current to the bulb decreases and the bulb cools down, decreasing its resistance. DC current will not flow in a series capacitor circuit.

Although the light bulb is purely a resistive circuit, its resistance is inversely proportional to the current flow - which is directly proportional to the temperature of the filament. Simply stated, the temperature goes up, the resistance goes up - and the current goes down. Remember, E=IR. If E stays the same, and the current goes down, the resistance must increase to maintain the equality.

A simple resister, for the most part, stays constant.

Great stuff, this electronics, eh?

Answer 2

1 is wrong, the resistance of the light bulb does not change, but as the current drops from increased capacitor resistance it gets dimmer. and the voltage difference across it is lower. this is relative to the inreased resistance in the capacitor.

can't help on scale factor