Predict how a capacitor charges through resistance

The capacitor changes the voltage available to the resistor

The charging diagram starts with an uncharged capacitor connected through a resistor to a constant 6 V source. Initially the capacitor voltage is zero, so the full source voltage appears across the resistor. Current flows through the external circuit and changes the charge on the plates. It does not conduct through the ideal insulating gap.

As capacitor voltage rises, less voltage remains across the resistor. Current falls, so each next interval adds less charge. The loop relation is 6 V = I R + V capacitor. This explains the shape of the curve before an exponential formula is introduced.

Read the time constant with units

The time constant is τ = RC, where R is resistance in ohms, C is capacitance in farads and τ is time in seconds. Here C = 1000 µF = 0.001 F. At R = 2 kΩ = 2000 Ω, τ = 2000 × 0.001 = 2 s. A kiloohm is 1000 ohms; a microfarad is 10⁻⁶ farad.

For charging from zero with a constant source voltage V source, V capacitor(t) = V source [1 − exp(−t/τ)] and I(t) = (V source/R) exp(−t/τ). The dimensionless ratio t/τ measures elapsed time in units of the time constant. The exponential exp(−t/τ) is the fraction of the initial voltage gap still remaining. OpenStax: RC circuits develops these charging and discharge relations.

Predict a few points before moving the time control

With R = 2 kΩ:

Time	Number of time constants	Charging voltage	Charging current
0 s	0	0 V	3.0000 mA
2 s	1	3.7927 V	1.1036 mA
10 s	5	5.9596 V	0.0202 mA

At one time constant, 63.2% of the final voltage has been reached and about 36.8% of the initial current remains. The two percentages refer to different quantities. The current is largest just after connection, not when the capacitor is most charged.

At five time constants, voltage is approximately 99.326% of the 6 V limit. It is close, but not exactly 6 V. A display rounded to fewer digits could conceal the remaining difference. The browser does not award a physical full-charge event at an arbitrary percentage.

The time slider chooses a point on an analytic history; it does not run a clock. Changing resistance recalculates a new history with the same initial condition. It does not simulate replacing a resistor halfway through charging.

Compare equal times and equal time-constant counts

Double R to 4 kΩ. At the same elapsed 2 s, only half a time constant has passed, so capacitor voltage is approximately 2.3608 V, lower than before. At 4 s, one full time constant has passed, so voltage is again 3.7927 V. However, current is 0.5518 mA, half the current of the 2 kΩ circuit at its own one-time-constant point.

Capture those settings and explain why equal voltage fraction does not imply equal current. The same capacitance and source set the same final charge, but resistance changes both the time scale and initial current.

Transfer to a new component value

A 5 V source charges an initially uncharged 500 µF capacitor through 4 kΩ. Calculate the time constant, initial current, and capacitor voltage after 2 s. The browser retains its own 6 V and 1000 µF constants.

Check the charging transfer

τ = 4000 × 0.0005 = 2 s. Initial current is 5/4000 = 1.25 mA. At 2 s, voltage is 5(1 − e⁻¹) ≈ 3.1606 V; current is 1.25e⁻¹ ≈ 0.4598 mA. The exponential voltage fraction is the same as in the browser at one time constant, even though the absolute voltage and current differ.