Hello,

I am a new hobbyist and I have a question which has conflicting answers online. I'm currently learning about rc timing circuits- more specifically- using a resistor/capacitor network to switch on a transistor after a certain time period. Charging and placing the capacitor in parallel with the base of the transistor is clear, but I am having trouble understanding the branch current calculations. When the switch is closed, I know caps act as a short immediately until charge and voltage build. What I don't understand is why I am taught by some that current takes every path available based on ohms law, while a reputable lecturer says that the initial 'short' from a capacitor will divert ALL current (even from a branch in parallel) until voltage builds. Can anyone help clear up how current behaves in this situation? Thank you.

 

"When the switch is closed, I know caps act as a short immediately until charge and voltage build. What I don't understand is why I am taught by some that current takes every path available based on ohms law, while a reputable lecturer says that the initial 'short' from a capacitor will divert ALL current (even from a branch in parallel) until voltage builds. Can anyone help clear up how current behaves in this situation? Thank you."


They are both correct. The voltage in a capacitor can not change instantly, so if it is at zero volts initially then immediately after switching in a power source it will still be zero volts. That means a parallel resistor, for example, will see zero volts across it immediately after switch-on and the current through the resistor will be determined by ohms law: I = V/R or I = 0 volts /R ohms = 0 amps. So, for an instant, all the current is going through the capacitor, I = C dV/dt.

 

Thank you very much for this reply. Appreciated.

 

Thank you very much for this reply. Appreciated.

Your welcome

 

@Archaeus
Imagine the charges as shoppers in your local supermarket's long and slow moving checkout lines (checkers are resistors, the slower they check shoppers out, the higher the resistance). Suddenly a new lane opens, and shoppers (charges) flow in to fill the lane until its just as backed up as the rest. Shoppers are still moving through all lanes. But if an empty lane opens, you better believe everybody is going to try to take advantage of it. Not a complete metaphor, but it might help.
Also, check out the audio part (Capacitor Testing) of this article: http://www.allaboutcircuits.com/technical-articles/inspecting-electronic-circuits-with-an-audible-circuit-probe/ It might help to hear the potential difference across the capacitor change (it is discharging in the example sound files).
Mark