Why RC Curves Are Exponential

Overview

This note explains why RC charging and discharging curves are exponential.

It is an enrichment note. For the main topic, you usually only need to use the exponential equations, not derive them.

The key idea is simple:

• the capacitor p.d. depends on charge, $V_C=Q/C$
• the resistor current depends on the remaining driving p.d.
• as charge changes, the current changes
• the rate of change is therefore proportional to how far the capacitor is from its final state

That proportionality produces an exponential curve.

Core Ideas

The derivation uses three relations:

• capacitor p.d.: $V_C=Q/C$
• resistor p.d.: $V_R=IR$
• current as rate of charge flow: $I=dQ/dt$

The mathematical trigger for an exponential curve is a differential equation of the form:

$$\frac{dx}{dt}=-kx$$

where the rate of change of a quantity is proportional to the quantity still present.

Definition

In this note, an RC exponential is a time dependence containing:

$$e^{-t/RC}$$

where $RC$ is the time constant of the circuit.

Why It Matters

The main topic asks you to use RC exponential equations. This enrichment note explains why those equations have that form, so the curves are not treated as arbitrary facts.

Exam Relevance

For normal H2 exam work, the important skill is applying the RC equations correctly.

This derivation is useful when you want to understand:

• why the curves are exponential rather than straight lines
• why the same time constant $RC$ appears in both charging and discharging
• why current is largest at the start of charging and then decreases
• why a capacitor approaches its final p.d. gradually

Do not spend exam time deriving these equations unless the question explicitly asks for a derivation.

Key Representations

The two most important representations are:

• discharge: the remaining charge decreases exponentially
• charge: the missing charge decreases exponentially

Discharging a Capacitor

Suppose a capacitor initially stores charge $Q_0$ and discharges through a resistor $R$.

At any time:

$$V_C=\frac{Q}{C}$$

The resistor p.d. is equal in magnitude to the capacitor p.d., so the current magnitude is:

$$I=\frac{V_C}{R}$$

Therefore:

$$I=\frac{Q}{RC}$$

During discharge, the charge on the capacitor is decreasing. If positive current is chosen as the current leaving the positive plate, then:

$$I=-\frac{dQ}{dt}$$

So:

$$-\frac{dQ}{dt}=\frac{Q}{RC}$$

or:

$$\frac{dQ}{dt}=-\frac{Q}{RC}$$

This means the rate at which charge decreases is proportional to the charge remaining.

Separating variables:

$$\frac{1}{Q}dQ=-\frac{1}{RC}dt$$

Integrating:

$$\ln Q=-\frac{t}{RC}+\text{constant}$$

Using $Q=Q_0$ when $t=0$ gives:

$$Q=Q_0{e}^{-t/RC}$$

Since $V_C=Q/C$, the capacitor p.d. has the same exponential factor:

$$V_C=V_0{e}^{-t/RC}$$

The current magnitude also decays with the same exponential factor:

$$I=I_0{e}^{-t/RC}$$

Charging a Capacitor

Now suppose an initially uncharged capacitor charges through a resistor from a DC supply of p.d. $V_0$.

At any time, Kirchhoff’s loop rule gives:

$$V_0=IR+V_C$$

Since:

$$V_C=\frac{Q}{C}$$

and:

$$I=\frac{dQ}{dt}$$

we get:

$$V_0=R\frac{dQ}{dt}+\frac{Q}{C}$$

Rearrange:

$$R\frac{dQ}{dt}=V_0-\frac{Q}{C}$$

or:

$$\frac{dQ}{dt}=\frac{C{V}_0-Q}{RC}$$

The final charge is:

$$Q_{\infty }=C{V}_0$$

So:

$$\frac{dQ}{dt}=\frac{Q_{\infty }-Q}{RC}$$

This says the charging rate is proportional to the charge still missing from the final value.

Let:

$$u=Q_{\infty }-Q$$

Then:

$$\frac{du}{dt}=-\frac{dQ}{dt}$$

So the equation becomes:

$$\frac{du}{dt}=-\frac{u}{RC}$$

This has the same form as exponential decay:

$$u=u_0{e}^{-t/RC}$$

For an initially uncharged capacitor, $Q=0$ at $t=0$, so $u_0=Q_{\infty }$.

Therefore:

$$Q_{\infty }-Q=Q_{\infty }{e}^{-t/RC}$$

Hence:

$$Q=Q_{\infty }\left(1-e^{-t/RC}\right)$$

Since $V_C=Q/C$:

$$V_C=V_0\left(1-e^{-t/RC}\right)$$

The current is largest at the start and then decays:

$$I=I_0{e}^{-t/RC}$$

where, for a simple series RC charging circuit:

$$I_0=\frac{V_0}{R}$$

Why the Time Constant Is $RC$

In both charging and discharging, the exponential factor is:

$$e^{-t/RC}$$

So the natural time scale is:

$$\tau =RC$$

After one time constant:

$$e^{-1}\approx 0.368$$

Therefore:

• a discharging quantity has about $36.8\%$ remaining
• a charging quantity has reached $1-e^{-1}\approx 63.2\%$ of its final value

Physical Interpretation

The exponential shape is not arbitrary.

For discharge:

• a larger remaining charge gives a larger p.d.
• a larger p.d. gives a larger current
• a larger current removes charge faster
• as charge falls, the removal rate also falls

For charge:

• the supply initially gives a large current
• as $V_C$ rises, less p.d. remains across the resistor
• current decreases
• the capacitor approaches its final p.d. gradually

Key Results

Discharging:

$$Q=Q_0{e}^{-t/RC}$$ $$V_C=V_0{e}^{-t/RC}$$ $$I=I_0{e}^{-t/RC}$$

Charging:

$$Q=Q_{\infty }\left(1-e^{-t/RC}\right)$$ $$V_C=V_0\left(1-e^{-t/RC}\right)$$ $$I=I_0{e}^{-t/RC}$$

Links

• Electric Capacitance
• DC Circuits