Kirchhoff's Law with Capacitors: Voltage, Resistance & Charge

[Silviu]

Hello! I am bit confused about kirchhoffs law with capacitors. So if I have a battery and resistance, I obtain V-RI=0. But from what I read, if I have a capacitor and resistance (so discharging capacitor), the equation reads
R(dq/dt) + q/c = 0, which is equivalent to RI + q/c =0. Why in this case we have "+" between terms, if the capacitor acts like a battery?

 

[BvU]

If the capacitor discharges, the current lowers the charge, so $I = - dV/dt = -{1\over C}{dq\over dt}$

 

[Dale]

Hello! I am bit confused about kirchhoffs law with capacitors. So if I have a battery and resistance, I obtain V-RI=0. But from what I read, if I have a capacitor and resistance (so discharging capacitor), the equation reads
R(dq/dt) + q/c = 0, which is equivalent to RI + q/c =0. Why in this case we have "+" between terms, if the capacitor acts like a battery?

In addition to the above, it could also come from using the passive sign convention for the capacitor and the active sign convention for the battery. The convention used doesn't affect any of the physics.

The passive sign convention means that a positive current goes from the positive side to the negative side so P=IV is the power received by the element. The active sign convention means that a positive current goes from the negative side to the positive side so P=IV is the power produced by the element.

 

[vanhees71]

To get the right sign, it's much easier to think in terms of actual current flow. E.g., a positive current towards the positively charged plate of the capacitor leads to an increasing voltage. It's much simpler to think about the physics than learn the quite complicated conventional rules by engineers, which usually leads to confuse themselves ;-).