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anonymous
 5 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(1e^(t/RC)
anonymous
 5 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(1e^(t/RC)

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Shayaan_Mustafa
 5 years ago
Best ResponseYou've already chosen the best response.0So you need a whole derivation? Regards, Electronics engineer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as i write the potential difference in the circuit i come to a differential equation, but my anwer is not like the book's

Shayaan_Mustafa
 5 years ago
Best ResponseYou've already chosen the best response.0do you know how to integrate or differentiate any complex thing?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0Start with Kirchoff's law for a closed circuit\[0=\epsilonI(t)RV_c(t)=\epsilon\frac{dq}{dt}R\frac q C\]where V_c is the voltage across the capacitor. A little rearrangement gives\[\frac{dq}{dt}=\frac{1}{R}(\epsilon\frac{q}{C})\to\frac{dq}{(\epsilon\frac{q}{C})}=\frac{1}{R}dt\to\frac{dq}{qC\epsilon}=\frac{1}{RC}dt\]integrate:\[\int_{0}^{Q}\frac{dq}{(qC\epsilon)}=\frac{1}{RC}\int_{0}^{t}dt\]\[\ln(\frac{QC\epsilon}{C\epsilon})=\frac{t}{RC}\to(\frac{C\epsilonQ}{C\epsilon})=1\frac{Q}{C\epsilon}=e^{\frac{t}{RC}}\]\[Q(t)=C\epsilon(1e^{\frac{t}{RC}})\] \[\text{QED}\] :D

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0and you forgot that the exponent was negative Mojtaba

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.0Why 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 ;)
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