Deriving the Exponential Capacitor Decay equation

Ben Duffield, , , , , ,

If you’ve studied capacitors, you’ll have (very probably) come across the decay equation:
Q = Q_0 e^\frac{-t}{RC} where R is resistance, C is capacitance, Q is the charge on the capacitor at time t and Q_0 is the initial charge.

Where does this come from?

Firstly from the basics of circuits, we know that V = IR (yes, I’m using the british convention with regards the symbol for voltage, I understand elsewhere U is used instead).
V = IR \Rightarrow I = \frac{V}{R}

Now when we discharge a capacitor through a circuit, the e.m.f.s are supplied by the capacitor.
Q = CV \Rightarrow V = \frac{Q}{C}

Now substitutiting for V in the above equation gives
I = \frac{V}{R} \Rightarrow I = \frac{\frac{Q}{C}}{R} = \frac{Q}{RC}

So we now have an equation for the current in the circuit, but wait, where does this current come from? It’s the capacitor discharging – this current is equal to the rate of discharge of the capacitor, so:
-I = \frac{{\mathrm d}Q}{{\mathrm d}t}
So substituting the above expression for I
-\frac{Q}{RC} = \frac{{\mathrm d}Q}{{\mathrm d}t}

Ooh look, a differential equation.
Seperate the variables giving:
\int -\frac{1}{RC} \, {\mathrm d}t = \int \frac{1}{Q} \, {\mathrm d}Q
\Rightarrow \frac{-t}{RC} = ln(Q) + ln(k) where k is some constant.
\Rightarrow e^\frac{-t}{RC} = kQ
\Rightarrow Q = \frac{1}{k}e^\frac{-t}{RC}

If we define the initial charge to be Q_0 – the charge when t = 0
\mathrm{ let } t = 0 \Rightarrow Q_0 = \frac{1}{k}
so we now can replace \frac{1}{k} giving:

Q = Q_0 e^\frac{-t}{RC}

MY WORK HERE IS DONE

:D

Deriving the Exponential Capacitor Decay equation

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