Published by Patrick Mutisya · 14 days ago
Analyse the graphs of the variation with time of the potential difference (\$V\$), charge (\$Q\$) and current (\$I\$) for a capacitor discharging through a resistor.
The discharging circuit consists of a charged capacitor \$C\$ connected across a resistor \$R\$. No external emf is present.
Applying Kirchhoff’s loop rule to the discharging loop gives:
\$\$
VR + VC = 0 \quad\Rightarrow\quad IR + \frac{Q}{C}=0 .
\$\$
Since \$I = -\dfrac{dQ}{dt}\$ (the current is the rate of loss of charge), we obtain the first‑order differential equation:
\$\$
R\left(-\frac{dQ}{dt}\right)+\frac{Q}{C}=0
\;\;\Longrightarrow\;\;
\frac{dQ}{dt} = -\frac{Q}{RC}.
\$\$
Integrating the differential equation yields the exponential decay law:
\$\$
Q(t)=Q_0\,e^{-t/\tau},\qquad \tau = RC .
\$\$
Because \$V = Q/C\$, the potential difference across the capacitor also decays exponentially:
\$\$
V(t)=V0\,e^{-t/\tau},\qquad V0=\frac{Q_0}{C}.
\$\$
The current is the time derivative of charge (with a minus sign):
\$\$
I(t)= -\frac{dQ}{dt}= \frac{Q0}{RC}\,e^{-t/\tau}= I0\,e^{-t/\tau},
\qquad I0=\frac{V0}{R}.
\$\$
The quantity \$\tau = RC\$ is called the time constant. It has the following significance:
| Quantity | Expression | Value at \$t=\tau\$ |
|---|---|---|
| Charge \$Q\$ | \$Q_0 e^{-t/\tau}\$ | \$0.37\,Q_0\$ |
| Voltage \$V\$ | \$V_0 e^{-t/\tau}\$ | \$0.37\,V_0\$ |
| Current \$I\$ | \$I_0 e^{-t/\tau}\$ | \$0.37\,I_0\$ |
All three quantities share the same exponential shape, differing only in their initial values. Typical features to note on the graphs:
| Feature | Charge \$Q(t)\$ | Voltage \$V(t)\$ | Current \$I(t)\$ |
|---|---|---|---|
| Initial value (t=0) | \$Q_0\$ | \$V_0\$ | \$I0 = V0/R\$ |
| Mathematical form | \$Q_0 e^{-t/\tau}\$ | \$V_0 e^{-t/\tau}\$ | \$I_0 e^{-t/\tau}\$ |
| Physical meaning | Amount of stored charge | Potential difference across the plates | Rate of charge flow through the resistor |
| Behaviour at large \$t\$ | \$\rightarrow 0\$ | \$\rightarrow 0\$ | \$\rightarrow 0\$ |
| Units | Coulombs (C) | Volts (V) | Amperes (A) |
Given: \$C = 10\,\mu\text{F}\$, \$R = 5\,\text{k}\Omega\$, initially \$V_0 = 12\,\$V.
Solution:
\$I(2\tau) = I_0 e^{-2} = 2.4\times10^{-3}\,e^{-2} \approx 3.25\times10^{-4}\,\$A.
The discharge of a capacitor through a resistor is described by simple exponential laws for charge, voltage and current. The time constant \$\tau = RC\$ governs the rate of decay. By recognising the characteristic shape of the \$Q\$–\$t\$, \$V\$–\$t\$ and \$I\$–\$t\$ graphs, students can extract quantitative information such as \$\tau\$, initial conditions, and the state of the circuit at any given time.