9 / 32 · Concept
Capacitors connect voltage change to current
Use capacitance, charge, energy, and initial conditions consistently.
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Voltage cannot jump without an impulse
An ideal capacitor stores charge according to and has current under the passive sign convention. It resists an instantaneous voltage change because that would require an impulse of current. It does not resist voltage itself: a charged capacitor can remain at a nonzero voltage while carrying zero DC current.
At a settled DC operating point, an ideal capacitor is open circuit. During a transition, it is a dynamic element and cannot simply be removed. Its initial voltage matters. A real capacitor also has leakage, equivalent series resistance, dielectric effects, and a voltage rating. The ideal model intentionally omits those properties until they are relevant to the calculation.
Worked ramp and energy
Drive a 1 μF capacitor with a constant 1 mA current. Then V/s, or 1 V/ms. Starting at 0 V, the capacitor reaches 2 V after 2 ms, provided the current source can maintain that current at the required voltage. Charge is 2 μC. Stored energy is μJ.
This energy result is different from charge: charge scales linearly with voltage, while energy scales with voltage squared. Doubling the voltage to 4 V doubles charge but quadruples stored energy to 8 μJ. Units make the distinction clear: coulombs are not joules.
Initial conditions are part of the problem
A 1 μF capacitor already at 1 V reaches 2 V after only 1 ms under the same 1 mA current. Connecting two charged capacitors abruptly is not explained by calling both “open at DC”; the transition and wiring resistance must be modeled. Likewise, shorting a charged capacitor may produce a current pulse beyond the source's ordinary steady current.
For course calculations, state capacitance, initial voltage, imposed current or surrounding resistance, and time interval. Before a real low-voltage measurement, verify polarity for polarized parts and discharge stored charge through an appropriate resistor. For changing signals, ask about current and time before applying a steady-state shortcut.
Try it yourself
What current is required to change 100 nF by 3 V in 30 μs with a linear voltage ramp?
Read the explanation
i = C ΔV/Δt = 100 nF × 3 V / 30 μs = 10 mA. This is a ramp-current prediction, excluding parasitic resistance and source limits.