\[
U = \tfrac12 C V^{2}.
\]
\[
\boxed{\tau = RC}\quad\text{(seconds)}.
\]
It is the time required for the charge or the voltage on the capacitor to fall to \(\frac1e\) (≈ 37 %) of its initial value.
Consider a capacitor initially charged to \(V{0}\) (\(Q{0}=CV_{0}\)) that is suddenly connected across a resistor \(R\).
\[
V{C}+V{R}=0.
\]
\[
\frac{Q}{C}+IR=0.
\]
\[
I=-\frac{dQ}{dt}.
\]
\[
\frac{Q}{C}-R\frac{dQ}{dt}=0\;\Longrightarrow\;
\frac{dQ}{dt}=-\frac{Q}{RC}.
\]
\[
\int{Q{0}}^{Q}\frac{dQ'}{Q'}=-\int_{0}^{t}\frac{dt'}{RC}
\;\Longrightarrow\;
\ln\!\left(\frac{Q}{Q_{0}}\right)=-\frac{t}{RC}.
\]
\[
\boxed{Q(t)=Q_{0}\,e^{-t/RC}}.
\]
\[
\boxed{V(t)=V_{0}\,e^{-t/RC}}.
\]
\[
\boxed{I(t)=\frac{V_{0}}{R}\,e^{-t/RC}}.
\]
\endol>
\[
U(t)=\tfrac12 C[V(t)]^{2}
=\tfrac12 C V_{0}^{2}\,e^{-2t/RC}
=U_{0}\,e^{-2t/\tau},
\]
where \(U{0}= \tfrac12 C V{0}^{2}\) is the initial energy.
Because of the factor 2, the energy falls twice as fast as the voltage (or charge).
| Time | Voltage (or Charge, or Current) | Fraction of Initial Value |
|---|---|---|
| \(t = \tau\) | \(V = V_{0}e^{-1}\) | 0.368 ≈ 37 % |
| \(t = 2\tau\) | \(V = V_{0}e^{-2}\) | 0.135 ≈ 14 % |
| \(t = 5\tau\) | \(V = V_{0}e^{-5}\) | 0.007 ≈ 0.7 % |
The same percentages apply to the charge \(Q(t)\) and the current \(I(t)\) because they share the identical exponential factor.
\[
\log{10} V = \log{10} V{0} -\frac{t}{\tau}\log{10}e,
\]
whose gradient \(m = -\frac{1}{\tau}\log_{10}e\) (≈ \(-0.434/\tau\)).
\[
\tau = -\frac{1}{m}\,\log_{10}e \approx -\frac{0.434}{m}.
\]
\[
\frac{V}{V_{0}} = e^{-t/\tau}
\;\Longrightarrow\;
t = -\tau\ln\!\left(\frac{V}{V_{0}}\right).
\]
Example: for \(V = 0.20V_{0}\), \(t = -\tau\ln(0.20) \approx 1.61\tau\).
Problem: A semi‑log plot of voltage versus time for a discharging capacitor shows a straight line with a gradient of \(-25\;\text{s}^{-1}\) when the vertical axis is natural‑log (ln) scale. Find τ and the time at which the voltage has fallen to 10 % of its initial value.
\[
\tau = \frac{1}{25\;\text{s}^{-1}} = 0.040\;\text{s}.
\]
\[
t = -\tau\ln(0.10)=0.040\times2.303 = 0.092\;\text{s}\approx2.3\tau.
\]
Problem: A \(10\;\mu\text{F}\) capacitor is discharged through a \(2\;\text{k}\Omega\) resistor. The initial voltage is \(12\;\text{V}\). Find τ and the voltage after \(3\;\text{s}\).
\[
V(3\;\text{s}) = 12\,e^{-3/0.020}=12\,e^{-150}\approx 0\;\text{V}.
\]
After 3 s (≈150τ) the capacitor is essentially fully discharged.
Problem: A \(4.7\;\mu\text{F}\) capacitor charged to \(5\;\text{V}\) discharges through a \(1\;\text{M}\Omega\) resistor. Find τ, then the voltage, current and remaining energy after \(2\tau\).
\[
U = \tfrac12 C V^{2}= \tfrac12 (4.7\times10^{-6})(0.68)^{2}\approx1.1\times10^{-6}\;\text{J}.
\]
This is about 1.9 % of the initial energy \(U_{0}=5.9\times10^{-5}\;\text{J}\).
| Quantity | Symbol | Formula / Definition | Typical Example |
|---|---|---|---|
| Capacitance | \(C\) | Given or measured | 10 µF |
| Resistance | \(R\) | Given or measured | 2 kΩ |
| Time constant | \(\tau\) | \(\tau = RC\) | 0.020 s |
| Voltage (time t) | \(V(t)\) | \(V_{0}e^{-t/\tau}\) | ≈0 V after 3 s |
| Charge (time t) | \(Q(t)\) | \(Q_{0}e^{-t/\tau}\) | 0.68 µC after 2τ (example) |
| Current (time t) | \(I(t)\) | \(\dfrac{V_{0}}{R}e^{-t/\tau}\) | 6.8 × 10⁻⁷ A after 2τ |
| Energy (time t) | \(U(t)\) | \(\tfrac12 C V_{0}^{2}e^{-2t/\tau}\) | ≈1.1 µJ after 2τ |
A labelled schematic showing a charged capacitor \(C\) connected across a resistor \(R\) with the direction of discharge current indicated, together with a plot of \(V\) versus \(t\) that marks the points \(t=\tau\), \(2\tau\) and \(5\tau\). (Insert as a figure in your notes.)
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