RC Circuit Charging

Equivalent forms

Q(t) = C V_0 (1 - e^{-t/RC})

I(t) = \frac{V_0}{R} e^{-t/RC}

The time constant τ = RC is the unique combination of resistance and capacitance with units of seconds — a fingerprint emerging straight from dimensional analysis.

Symbol	Name	SI unit	Dimension	Range
$V$	Capacitor voltageoutput Voltage across the capacitor at time t	V	M·L²·T⁻³·I⁻¹	0 – 20
$V₀$	Source voltage EMF of the battery driving the circuit	V	M·L²·T⁻³·I⁻¹	1 – 24
$R$	Resistance Series resistance	Ω	M·L²·T⁻³·I⁻²	100 – 100000
$C$	Capacitance Capacitance of the storage element	F	M⁻¹·L⁻²·T⁴·I²	0.000001 – 0.001
$t$	Time Time elapsed since switch closed	s	T	0 – 10

Unit systems

Symbol	SI	CGS	Imperial
$V$	V	statV	V
$R$	Ω	s/cm	Ω
$C$	F	cm	F
$t$	s	s	s

Where it holds

Valid for ideal linear RC circuits with constant R, C and a step EMF. Breaks down for nonlinear loads, lossy dielectrics, or when stray inductance becomes significant at high frequencies.

Dimensional analysis

$[\tau ] = [R][C] = (\Omega )(F) = (V$ /A $)(A\cdot s/V) = s \checkmark$

Discovery

Hermann von Helmholtz / William Thomson (Kelvin) · 1853

Kelvin analyzed transient currents in the first transatlantic telegraph cables (1850s) and derived the RC exponential decay — pivotal for understanding signal smearing in long cables.

Try this

Why does a turn-signal blink at a steady rate — what sets the rhythm?

A 100 μF capacitor charges through a 10 kΩ resistor from a 12 V battery. How long does it take to reach 8 V?

Research status: stable

Real-world applications

555-timer circuits in clocks and oscillators
Camera flash recharge timing
Debounce circuits for mechanical switches
RC low-pass filters in audio and signal processing

Common misconceptions

The capacitor never truly reaches $V_{0}$ — it just gets arbitrarily close
Current does not drop to zero instantly; it decays with the same $\tau as$ voltage rises
$\tau = RC$ has units of seconds; this is not a coincidence — Ohms $\times$ Farads = seconds

Experimental verification

Trace V(t) on an oscilloscope across a known R and C: the time to reach 63.2% of

V_{0}

equals RC to within component tolerance (typically

\pm 5

%). Demonstrated in every undergraduate electronics lab.

Derivation

Kirchhoff's loop:

V_{0} = IR + Q/C

.

With

I = dQ/dt

, this is

RC dQ/dt + Q = CV_{0}

, a first-order linear ODE.

Solution

Q(t) = CV_{0}(1 - e^(-t/RC))

; dividing by C gives V(t).

Limiting cases

t \to 0

⟶

V \to 0

At the instant the switch closes, the capacitor is uncharged and acts like a wire.

t \to \infty

⟶

V \to V_{0}

Eventually the capacitor reaches the source voltage and current stops flowing.

t = \tau

⟶

V \approx 0.632\,\mathrm{V}_{0}

After one time constant the capacitor has charged to

(1 - 1/e) \approx 63.2

% of its final voltage.

What if…

What if R is doubled?

$\tau$ doubles. The capacitor charges to the same final voltage but takes twice as long to get there.

What if the battery is replaced by a step from 0 to

V_{0}

then back to 0?

You see a charging exponential followed by a discharging exponential with the same $\tau$ — symmetric in shape.

What if you place two capacitors in parallel?

Effective C doubles, $so \tau$ doubles. The circuit becomes 'sluggish' — useful for filtering out fast spikes.

1

Time to reach 8 V from a 12 V source

Given ·

V₀:: 12
R:: 10000
C:: 0.0001
V target:: 8

Find · t

Steps

$\tau = RC = 10^{4} \times 10^{-4} = 1\,\mathrm{s}$
$Set V(t) = 8$ : $8 = 12(1 - e^(-t/1))$
Rearrange: $e^(-t) = 1 - 8/12 = 1/3$
Take ln: $t = \ln 3 \approx 1.099\,\mathrm{s}$

Result ·

t \approx 1.10\,\mathrm{s}

2

Initial charging current

Given ·

V₀:: 9
R:: 4700
C:: 0.00001
t:: 0

Find · I(0)

Steps

At $t = 0$ the capacitor voltage is 0, so it acts like a wire
All of $V_{0}$ drops across R
Apply Ohm's law: $I = V_{0}/R = 9/4700 = 0.00191$ A

Result ·

I(0) = V_{0}/R = 9/4700 \approx 1.91\,\mathrm{mA}

Ohm's Law→Energy Stored in a Capacitor→LC Oscillation Frequency→