## anonymous 4 years ago how can i prove that the charge in a charging capacitor in a DC circuit which has a battery, a resistor and a capacitor equals to: Q=Cε(1-e^(t/RC)

1. anonymous

So you need a whole derivation? Regards, Electronics engineer.

2. anonymous

as i write the potential difference in the circuit i come to a differential equation, but my anwer is not like the book's

3. anonymous

do you know how to integrate or differentiate any complex thing?

4. TuringTest

Start with Kirchoff's law for a closed circuit$0=\epsilon-I(t)R-V_c(t)=\epsilon-\frac{dq}{dt}R-\frac q C$where V_c is the voltage across the capacitor. A little rearrangement gives$\frac{dq}{dt}=\frac{1}{R}(\epsilon-\frac{q}{C})\to\frac{dq}{(\epsilon-\frac{q}{C})}=\frac{1}{R}dt\to\frac{dq}{q-C\epsilon}=-\frac{1}{RC}dt$integrate:$\int_{0}^{Q}\frac{dq}{(q-C\epsilon)}=-\frac{1}{RC}\int_{0}^{t}dt$$\ln(\frac{Q-C\epsilon}{-C\epsilon})=-\frac{t}{RC}\to(\frac{C\epsilon-Q}{C\epsilon})=1-\frac{Q}{C\epsilon}=e^{-\frac{t}{RC}}$$Q(t)=C\epsilon(1-e^{-\frac{t}{RC}})$ $\text{QED}$ :D

5. TuringTest

that was fun !

6. TuringTest

and you forgot that the exponent was negative Mojtaba

7. TuringTest

Why have I still not received a medal for this? I usually don't complain about that kind of thing, but this was a cool derivation! Hook it up people ;)