Show that during the charging of the capacitor in an RC circuit, the charge of the capacitor, as a function of time, is: q(t) = C(1 – e-t/RC)
the character that doesn't show up is the emf (E)
The current is
i = dq/dt = C*E*(1/RC)*e^(-t/RC)
= (E/R)*e^(-t/RC)
iR + q/C = E*e^(-t/RC)+ E*(1 – e^-t/RC)
= E
Therefore the sum of the voltage drop across the capacitor and the resistor equals E at all times, as required.
thank you very much drwls, this is a question from a sample exam, getting ready for my finals tomorrow!
Good luck with finals! I took a lazy approach to your question, showing that what you wrote is the solution, rather than solving the general differential equation.
thanks. It's all good, I understand your approach and I really appreciate it!
To show that during the charging of a capacitor in an RC circuit, the charge of the capacitor as a function of time is given by the equation q(t) = Cε(1 - e^(-t/RC)), we will need to use the basics of RC circuits and calculus.
First, let's consider an RC circuit where a resistor R and a capacitor C are connected in series, and a voltage source ε is connected across them. When the circuit is connected to the voltage source, the capacitor starts to charge.
The behavior of charging a capacitor in an RC circuit can be described using the equation Q(t) = q(t) + q(0) * e^(-t/RC), where Q(t) represents the charge supplied by the external source, q(t) is the charge accumulated on the capacitor at time t, and q(0) is the initial charge on the capacitor.
Since we are interested in the charging process, assuming the initial charge q(0) is zero, the equation simplifies to Q(t) = q(t).
Now let's use the basic relationship between charge, voltage, and capacitance:
Q(t) = C * V(t),
where V(t) is the voltage across the capacitor at time t. The voltage across the capacitor is related to the voltage source and the resistor by Ohm's Law:
V(t) = ε * (1 - e^(-t/RC)).
This equation represents the voltage across the capacitor as a function of time during the charging process. It starts from 0 and exponentially approaches ε as time goes to infinity.
Substituting this back into the equation for charge Q(t) = C * V(t), we get:
q(t) = C * V(t),
q(t) = C * ε * (1 - e^(-t/RC)).
This equation represents the charge on the capacitor as a function of time during the charging process. It starts from 0 and exponentially approaches Cε as time goes to infinity.
Therefore, we have shown that during the charging of a capacitor in an RC circuit, the charge of the capacitor, as a function of time, is q(t) = Cε(1 - e^(-t/RC)).