Forums » General Electronics Discussion

Hi everyone! I'm new to CircuitLab.

I know how to do it only if it's parallel with a ressistor

E.g. : If a current source of 12mA is in parallel with a ressistor of 330Ohms, Resulting equivalent Voltage source will be 3.96V (0.012 x 330) in series with the 330Ohms ressistor.

But for the same circuit above, what happens if a new capacitor (1uF) is fixed in series with the 330Ohms ressistor

[ (Capacitor in series with ressistor) // current source]

Please try to give a link of a diagram if possible.

Thanks!

Please draw your circuit in CL.

Then there's no confusion over what you are asking about.

You can also then run simulations on it to see how it behaves.

In the meanwhile here's a demonstration of the equivalence of Thevenin and Norton voltage sources.

If you put a capacitor in series with the resistor in a Norton circuit then at DC you have open circuited the current source and so, in theory, the voltage at the output will eventually fly off to infinity as the current source charges up the capacitor.

In practice, because the capacitor will have some parallel leakage resistance (Rpar) and the voltage measurement device will have some finite resistance (Rload), the output voltage, V(out), will rise to:

V(out) = I(source)*(Rpar//Rload)

where Rload is the total parallel resistance of whatever load you apply (which may just be the air) and the input resistance of the device you use to measure the output voltage across the load.

If you put a capacitor in series with the resistor in a Thevenin circuit then at DC you have open circuited the series circuit between the ideal voltage source and the output and so the voltage at the output will be zero.

In theory the voltage at the output will be the voltage source plus whatever the initial voltage across the capacitor and , given ideal components (no leakage across the capacitor and infinite load resistance at the output) it will stay that way forever.

However, in practice, the capacitor will have some leakage resistance (Rpar) and the load will be some finite resistance (Rload) so the actual output voltage will fall to:

V(out) = V(source)*Rload/(Rpar+Rload)

where Rload is the total parallel resistance of whatever load you apply (which may just be the air) and the input resistance of the device you use to measure the output voltage across the load.

See this link about voltmeters in CL:

https://www.circuitlab.com/forums/analog-design/topic/b95fc6f8/can-a-voltmeter-be-given-infinite-resistance-for-education/

Thank you very much! It helped a lot!