6.1.3 Charging and discharging capacitors

Discharging capacitors
Capacitors discharge through a constant-ratio pattern known as exponential decay.

Before a capacitor is allowed to discharge through a resistor (e.g if a switch is open at t=0)...

• The p.d. (V) across the capacitor or resistor is V₀
• For the capacitor Q = V C so charge stored is the capacitor is Q = V₀C
• For the resistor V=IR so the current in the resistor is I = V₀/R

When a switch is closed the capacitor discharges through the resistor. This means that, as time goes by, the charge stored by the capacitor decreases (so the p.d. across it also decreases). Since the p.d. decreases, the current in the resistor decreases accordingly and eventually p.d., the charge stored in the capacitor, and the current in the resistor are all 0. They all show exponential decay over time meaning we can use similar equations for each quantity...

V = V₀e^-t/CR

I = I₀e^-t/CR

Q = Q₀e^-t/CR 

Where I₀, V₀, and Q₀ is the maximum current/p.d./charge at t=0.

The time constant of a capacitor-resistor circuit is CR (in seconds). It is a measure of how long the exponential decay will take in a particular capacitor-resistor circuit. It's symbol is τ (tau) and it is measured in seconds (s).When t = CR, the p.d. across the resistor/capacitor is given by...

V = V₀e^-t/CR = V₀e^-CR/CR = V₀e^-1 = 0.37 V₀

Okay so we need to be able to model exponential decay. Provided the capacitor and resistor in a circuit are in parallel then they will have the same p.d. We known that Q = V C (charged stored by a capacitor) and V = I R (current stored in a circuit). Since V is the same for both components we can form the following equation for current:

I = V/R = Q/(CR)

Usually, I = ΔQ/Δt. However, for a capacitor, I = -ΔQ/Δt (this shows that the charge on the capacitor decreases with time). We can now write the above equation as:

-ΔQ/Δt = Q/(CR)

Meaning that...

ΔQ/Δt = -Q/(CR)

Q = Q₀e^-t/CR is simply a solution to this equation.

This equation (ΔQ/Δt = -Q/(CR)) can be used to model the decay of Q (charge on a capacitor)...

• Start with a known value Q₀ of for the time constant CR
• choose a time interval Δt which is small compared to CR
• Calculate the charge (ΔQ) leaving the capacitor in the time interval (Δt)...
• ΔQ = (Δt/(CR)) x Q
• Calculate Q left on capacitor by subtracting ΔQ from the previous charge
• Repeat many times

Straight line graphs

We know that the p.d. (V) across a discharging capacitor is V = V₀e^-t/CR. If we take logs  (to the base e) of both sides we get...

lnV = ln(V₀e^-t/CR) = lnV₀ + ln e^-t/CR = lnV₀ - (t/(CR))

We can plot ln V against t and the gradient will be -1/(CR) and the y-intercept will be lnV₀.

Charging capacitors

Okay so all that stuff above is all well and good if we want to talk about discharging capacitors, but what about actually charging them in the first place? Well, when we close a switch in a circuit the capacitor will start to charge. The p.d. across the capacitor (V_C)will increase from 0. The p.d. across the capacitor (V_C) and resistor (V_R) must always add up to V₀ (K2) so it follows that V_R must decrease as V_C increases.  After a while V_R falls to 0 as the capacitor is fully charged (V₀ = V_C).

We know that in a circuit current decreases exponentially as the capacitor discharges through the resistor...I = I₀e^-t/CR. Since V = IR and R is constant we know that...

V_R = V₀e^-t/CR

At any time (t) V₀ = V_R + V_C  so...

V_C = V₀ - V₀e^-t/CR

Therefore...

V_C = V₀ (1 - e^-t/CR)

We can also use this equation with charge Q on a capacitor (as well as p.d. V across a capacitor).