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.

\[  I=40.0 e^{-\frac{0.5}{0.25}}=5.41 \ \mathrm{A}  \]