Charging of capacitors in series

[Nimbus2000]

Hello everyone, I have a doubt about charging of capacitors in series. Suppose I connected two capacitors of same value, say,1 mF in series and put a bulb in series with them and applied a voltage 10V across the series. In steady state, the voltage across each capacitor will be 10/2=5V. Right?
The bulb will glow for a short time while charging the capacitors. Now, if I changed the capacitance to 1nF, still there will be 5V across each capacitor in steady state. Right? The bulb will glow for shorter time now(may not even be visible). So, as we go on reducing the capacitance, the glowing period of the bulb goes on decreasing but in steady state, voltage across each capacitance remains same 5V. Now, instead of capacitors, I just put two open switches in the circuit, will the voltage across each switch be 5V in steady state? I know the capacitance of the switch will be very small, but not 0. If no, how does the capacitance value affect this? Because, the voltage will be 5V, same for 1mF capacitor and 1nF capacitor. Instead if I just put two open switches, will it be same? Thanks in advance.

 

[tech99]

Hello everyone, I have a doubt about charging of capacitors in series. Suppose I connected two capacitors of same value, say,1 mF in series and put a bulb in series with them and applied a voltage 10V across the series. In steady state, the voltage across each capacitor will be 10/2=5V. Right?
The bulb will glow for a short time while charging the capacitors. Now, if I changed the capacitance to 1nF, still there will be 5V across each capacitor in steady state. Right? The bulb will glow for shorter time now(may not even be visible). So, as we go on reducing the capacitance, the glowing period of the bulb goes on decreasing but in steady state, voltage across each capacitance remains same 5V. Now, instead of capacitors, I just put two open switches in the circuit, will the voltage across each switch be 5V in steady state? I know the capacitance of the switch will be very small, but not 0. If no, how does the capacitance value affect this? Because, the voltage will be 5V, same for 1mF capacitor and 1nF capacitor. Instead if I just put two open switches, will it be same? Thanks in advance.

For the voltage sharing to happen, you must assume a value of capacitance, even for the case of the open circuit switch. And with these very small values of capacitance, you will need a current sensor capable of measuring the very short pulse and a voltmeter having extremely high resistance.

 

[Svein]

In steady state, the voltage across each capacitor will be 10/2=5V. Right?

Maybe. Since no steady state current is flowing, there is no way to predict the voltage across each capacitor. For real capacitors the leakage current will determine the voltage across them.

Now, instead of capacitors, I just put two open switches in the circuit, will the voltage across each switch be 5V in steady state?

Again, there is no way to tell (and if you measure the voltage, you are interfering with the circuit).