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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 take ALL current (even with a branch in parallel) until voltage builds. Can anyone help clear up how current behaves in this situation? Thank you.

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  • \$\begingroup\$ As Peter says BUT (imagine cap + A B C branches) - Consider rather at the start that all branches take their share of the current in inverse proportiomn to their resistance AND the cap resistance is zero so 1/zero = infinite so its share is infinite out of infinite +1/A + 1/B + 1/C. ie the others are infinitely smaller. As it charges this changes. Looking at it this way means you do not need a special case for the start point. \$\endgroup\$ – Russell McMahon Sep 11 '16 at 10:59
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Initially, the uncharged capacitor appears as a short circuit (zero Ohms) so is the lowest resistance path in the circuit, and takes all the available current.

As the capacitor charges, it effectively becomes an increasing resistance, so current is shared between the capacitor and any other paths that may be in parallel with it.

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    \$\begingroup\$ Thank you very much for answering this Peter. Appreciated. \$\endgroup\$ – Archaeus Sep 11 '16 at 2:00

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