# Capacitor in RC

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## Main Question or Discussion Point

Hello! I am bit confused about kirchoff's 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?

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

Dale
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Hello! I am bit confused about kirchoff's 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