What Actually Happens When You Connect a Capacitor to a Battery?

“The capacitor charges” is true, but it skips the interesting part. At the instant you connect an uncharged capacitor to a battery, the capacitor initially behaves almost like a short circuit. Current jumps to its largest value, then fades as charge accumulates. In a practical circuit that brief event is controlled by resistance hiding in the battery, wires, contacts, and the capacitor itself.

This explainer follows those first instants using a 5 V source and a 100 µF capacitor. The plots come from reproducible LTspice 26 transient simulations, not hand-drawn exponential curves. The source rises from 0 to 5 V in 1 ns so the simulator sees a very fast but finite transition.

Figure 1. The practical connection model. “Series resistance” means the total effective resistance of the source, wiring, contacts, and capacitor ESR.

The first instant

Before connection, the capacitor voltage is zero. Voltage across a capacitor cannot change instantaneously unless an infinite current is available, so immediately after the switch closes it is still essentially zero. Nearly the battery’s full 5 V therefore appears across the series resistance.

That gives the initial current:

I(0+) = V/R

With 10 Ω total resistance, the initial current is 5 V / 10 Ω = 0.5 A. This is not a current that continues indefinitely. As current moves charge onto one capacitor plate and off the other, the capacitor voltage rises. Less voltage remains across the resistance, so the current falls.

For a voltage step through a resistance, the two curves are:

VC(t) = V(1 - e-t/RC),    I(t) = (V/R)e-t/RC 

You do not need to derive these equations to read them. Both contain the time constant

τ = RC

which sets the clock for the event. After one time constant, the capacitor has covered 63.2% of the distance from its initial voltage to 5 V. After three time constants it has covered about 95%, and after five it has covered 99.3%. “Fully charged” is therefore convenient shorthand: the mathematical curve approaches 5 V forever without exactly reaching it.

What is physically moving? Electrons do not cross the insulating dielectric between the capacitor plates. The source pushes electrons onto one plate and pulls them from the other, creating equal and opposite plate charge and an electric field in the dielectric. Conventional current is drawn in the opposite direction from electron motion, which is the direction marked in Figure 1. Energy is stored in that electric field, not “inside” a supply of charge particles.

At the end of the transient, the capacitor voltage nearly equals the battery voltage. The series resistance then has almost no voltage across it, so almost no current flows. In the ideal model the charged capacitor behaves like an open circuit under steady DC. A real capacitor may still draw a small leakage current, but leakage is a much slower effect and is deliberately omitted here. If the battery is disconnected after charging, the ideal capacitor retains its voltage; a practical one slowly self-discharges through leakage and whatever measurement circuit is attached.

Figure 2. The 10 Ω baseline. At 1τ = 1 ms, the voltage is 3.16 V; by 5τ = 5 ms, it is 4.966 V.

Resistance changes how violent—and how long—the event is

For C = 100 µF, changing the total series resistance produces:

Low resistance means a taller, narrower current pulse. High resistance spreads the same charging process over more time. If time is divided by each circuit’s own (\tau), however, all three capacitor-voltage curves have the same shape. Figure 3 uses normalized logarithmic time to show that common shape, and a logarithmic current scale so a 50 mA trace does not disappear beside a 5 A trace. Solid, dashed, and dotted lines make the cases distinguishable without color.

Figure 3. Resistance stretches the time axis and rescales current. The underlying exponential response does not change.

The final amount of charge is independent of that resistance:

Q = CV = (100 µF)(5 V) = 500 µC 

Current is charge per unit time, so the area under every current pulse is the same 500 µC. A smaller resistance does not require less charge; it moves that charge faster.

The central surprise: half the energy goes missing

Once charged to 5 V, the capacitor stores

EC = ½CV2 = ½(100 µF)(5 V)2 = 1.25 mJ 

It is tempting to assume the battery supplied 1.25 mJ. It actually supplied twice that amount:

Esource = VQ = V(CV) = CV2 = 2.5 mJ 

The other 1.25 mJ became heat in the total series resistance. The simulation confirms this by numerically integrating the source power, vsourcei, and resistor power, i2R, from the exported LTspice waveforms. Capacitor energy is calculated at every sample from ½CVC2. In the 10 Ω run, source energy approaches 2.5 mJ, while resistor and capacitor energy each approach 1.25 mJ.

This 50% loss is not a special property of 10 Ω. For an initially uncharged capacitor connected to an ideal fixed-voltage step through any positive resistance, the same energy split results. Reducing resistance raises peak current and peak heating power while shortening the heating event. Those effects cancel in the time integral, leaving the same total resistive loss.

That result can feel paradoxical because a tiny resistor seems unable to consume much energy. But its current is enormous: instantaneous resistor power is (i^2R). As (R) shrinks, the initial current grows as (1/R), the initial power grows as (1/R), and the duration shrinks in proportion to (R). The product of power and time remains finite.

There is also a simple graphical way to see why the source supplies twice the stored energy. During charging, each increment of charge (dq) leaves a fixed 5 V source, so the source pays (5,dq). The capacitor’s own voltage, however, rises from 0 to 5 V while that charge arrives. Its average voltage during the process is therefore 2.5 V, and its accumulated field energy is (2.5,Q=\tfrac12CV^2). The difference between the fixed source voltage and the rising capacitor voltage appears across the resistance at every instant. Integrating that voltage difference times current gives the other half.

Figure 4. Left: cumulative energy in the 10 Ω run. Right: the finite 10 mΩ limiting case is extremely tall and narrow, but its current-time area remains (CV).

The “missing half” is specific to direct charging from a fixed-voltage step through resistance. It is not a universal tax on every way of charging a capacitor. Carefully controlled current, an inductor-based converter, or energy-recovery circuitry can change the energy story. Those are different circuits with different source waveforms and additional energy-storage elements.

What happens if resistance is exactly zero?

The separate near-ideal simulation uses 10 mΩ, not zero. Its time constant is only

τ = (10 mΩ)(100 µF) = 1 µs 

and its current begins near 500 A. This is a limiting-case illustration, not a claim that an ordinary 5 V battery can supply 500 A. The 5 ns maximum timestep resolves the narrow transient, while the source’s documented 1 ns rise time keeps the simulated input finite.

An exactly zero-resistance connection driven by a mathematically instantaneous voltage step is different in kind. The formulas predict an infinitely tall, zero-width current impulse whose area remains

∫ i(t) dt = CV = 500 µC 

Ordinary finite-current waveforms no longer describe that singular idealization. An “exact 0 Ω” LTspice trace would invite a physical interpretation the model does not deserve, which is why none is shown here.

Real circuits always supply mechanisms that tame the singularity. Batteries have internal impedance and current limits. Capacitors have equivalent series resistance. Wires and contacts contribute resistance, and contact bounce can complicate the first microseconds. Wiring and component inductance also prevents current from changing infinitely fast; with capacitance, that inductance can produce overshoot or ringing. At high currents, these supposedly small parasitics become the circuit.

Reproduce the LTspice results

The downloadable inputs are the resistance-sweep schematic and the 10 mΩ schematic. Equivalent text netlists are included for inspection. Both use a 0-to-5 V PULSE source with 1 ns rise and fall times, a 100 µF capacitor with IC=0, waveform compression disabled, and saved traces V(in), V(cap), and I(V1).

The sweep directive is:

.step param Rseries list 1 10 100
.tran 0 60m 0 2u
.option plotwinsize=0
.save V(in) V(cap) I(V1)

The near-ideal run replaces the sweep with R1=10m and uses:

.tran 0 8u 0 5n

From PowerShell, run LTspice in batch ASCII mode, then invoke the dependency-free Node renderer:

& "$env:LOCALAPPDATA\Programs\ADI\LTspice\LTspice.exe" -b -ascii ltspice\capacitor_charge_sweep.asc
& "$env:LOCALAPPDATA\Programs\ADI\LTspice\LTspice.exe" -b -ascii ltspice\capacitor_charge_near_ideal.asc
node assets\render.js

The script parses stepped raw data, derives the energy integrals, writes the measurement table, and regenerates all four SVG figures. The generated .raw and .log files are retained beside the schematics so the numerical source behind every plotted line can be inspected.