RC circuit: With a DC power supply

Question:
Consider an RC circuit with a DC voltage supply as shown in the figure. There are four identical resistors ($R$=1.00 mega $\mathrm{\Omega}$) connected to one capacitor ($C$= 1.00 micro F). The voltage of the power supply is 10$\times$10$^6$ V. If the capacitor is fully charged and then the power supply is removed, find the current at $t=$0.5 s.

Answer:
All the four resistors are connected in parallel. The equivalent resistance of multiple parallel resistors is given by
\[
\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_{\mathrm{1}}}+\frac{1}{R_{\mathrm{2}}}+\frac{1}{R_{\mathrm{3}}}+\cdots
\]
In this case, we obtain
\[
\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{4}{R}
\]
Thus,
\[
R_{\mathrm{eq}}=\frac{R}{4}=\frac{1.00\times 10^6}{4}=2.50\times 10^5 \mathrm{\Omega}
\]
From Kirchoff's law, each circuit element consumes the voltage, and the total of it and supplied voltage must be zero. The capacitor consumes voltage, $\frac{Q}{C}$ where $Q$ is the total charge. The resistor consumes $RI$ because of Ohm's law. Since we consider the discharging process, voltage from the power supply is zero. Then, we have
\[
-RI-\frac{Q}{C}=0
\]
We know $I=\frac{dQ}{dt}$, so
\[
\frac{dQ}{dt}+\frac{Q}{RC}=0
\]
Solve this differential equation:
\[
Q=Q_0 e^{-\frac{t}{RC}}
\]
The charges and current flow are equivalent in terms of time, so can express it as
\[
I=I_0 e^{-\frac{t}{RC}}
\]
$RC$ is known as the time constant, $\tau=2.50\times 10^5 \times 1.00 \times 10^{-6}=0.25$ s. $I_0$ is the current right before the voltage source is removed; namely, the maximum current. It can be calculated by Ohm's law.
\[
I_0=\frac{V}{R}=\frac{10\times 10^6}{2.50\times 10^5}=40.0 \ \mathrm{A}
\]
Then, we can find the current at $t=0.5$ s.
\[
I=40.0 e^{-\frac{0.5}{0.25}}=5.41 \ \mathrm{A}
\]

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