Capacitor Fundamentals & V-I Relationship

The component that fights voltage change and stores energy in an electric field.

Circuit AnalysisElectrical Engineering Year 1

⏱️ About 16 min

Flip a switch and a capacitor's voltage barely budges — it stubbornly resists sudden change. That single property shapes every transient you will ever analyze.

💡

The big idea: A capacitor stores energy in an electric field, and its current depends not on the voltage across it but on how fast that voltage is changing.

🎯 By the end, you'll be able to

Define capacitance in terms of charge, voltage, and physical structure
Apply the i=C dv/dt relationship to time-varying voltages
Explain why capacitor voltage cannot change instantaneously
Use the integral form to recover voltage from current history
Identify capacitor behavior in DC steady state as an open circuit

📎 Helpful to know first

Practical Op-Amp Limitations

What Is a Capacitor?

A capacitor is a two-terminal device that stores energy in an electric field. Its physical structure is simple: two conductive plates separated by a thin insulating layer called a dielectric. When a voltage is applied across the plates, charge accumulates — positive on one plate, negative on the other — and the electric field between the plates stores energy.

The amount of charge Q stored per volt V applied defines the capacitance C, measured in farads (F):

\[ Q = C\,V \]

Charge stored is proportional to applied voltage; the constant of proportionality is capacitance C.

The Defining V-I Relationship

Since current is the rate of flow of charge (i = dQ/dt) and Q = CV, differentiating gives the capacitor's defining equation:

\[ i(t) = C\,\frac{dv(t)}{dt} \]

Capacitor current is proportional to the rate of change of voltage, not the voltage itself.

🔑 Current depends on dv/dt, not v

This is the single most important difference between a capacitor and a resistor. A resistor's current depends on the voltage across it right now. A capacitor's current depends on how fast the voltage is changing. A capacitor sitting at 1000 V with no change draws zero current; a capacitor at 0 V whose voltage is rising rapidly can draw a large current.

Voltage Cannot Change Instantaneously

If the capacitor voltage were to jump instantaneously, dv/dt would be infinite, requiring infinite current — a physical impossibility. Therefore:

✨ Voltage continuity rule

A capacitor's voltage cannot change instantaneously. If v(0−) = 5 V just before a switching event, then v(0+) = 5 V immediately after. This continuity condition is the foundation of all transient analysis with capacitors.

The Integral Form

Sometimes you know the current history and need the voltage. Rearranging i = C dv/dt and integrating gives:

\[ v(t) = \frac{1}{C}\int_{0}^{t} i(\tau)\,d\tau + v(0) \]

Voltage at time t equals the initial voltage plus the accumulated charge (scaled by 1/C) from the current that has flowed since.

DC Steady State: The Open-Circuit Analogy

In DC steady state, all voltages and currents are constant. If v is constant, then dv/dt = 0, and therefore i = C dv/dt = 0. A capacitor with zero current flowing through it behaves like an open circuit. This is how you analyze capacitors in DC steady-state circuits: replace them with an open and solve the remaining resistive network.

⚠️ Open circuit only at true steady state

A capacitor acts as an open circuit only when all transients have died out and every voltage is truly constant. During a transient — while dv/dt is nonzero — current flows and the capacitor is very much active.

📝 Worked example: A 10 μF capacitor has the voltage v(t) = 5 sin(100t) V applied across it. Find the current i(t).

Start with the defining relationship: i = C dv/dt.
Differentiate v(t) = 5 sin(100t): dv/dt = 5×100 cos(100t) = 500 cos(100t) V/s.
Multiply by C = 10×10⁻⁶ F: i = 10×10⁻⁶ × 500 cos(100t) = 0.005 cos(100t) A.
Express in milliamperes: i(t) = 5 cos(100t) mA.

✓ i(t) = 5 cos(100t) mA

✏️ Practice: A 20 μF capacitor has a voltage that increases linearly at 500 V/s. What is the current through the capacitor (in amperes)?

Solution

Since voltage changes linearly, dv/dt = 500 V/s (constant).
Apply i = C dv/dt = 20×10⁻⁶ × 500 = 0.01 A.

Check your understanding

1. A capacitor has 12 V across it and the voltage is not changing. What is the current through the capacitor?

Since dv/dt = 0 (voltage is not changing), i = C dv/dt = 0 regardless of the voltage magnitude or capacitance value.

2. Why can't a capacitor's voltage change instantaneously?

An instantaneous voltage change means dv/dt approaches infinity, which by i = C dv/dt requires infinite current — physically impossible.

✅ Key takeaways

A capacitor stores energy in an electric field between two conductive plates separated by a dielectric.
Capacitance C (in farads) relates charge to voltage: Q = CV.
The defining V-I relationship is i = C dv/dt — current depends on the rate of voltage change, not voltage itself.
Capacitor voltage cannot change instantaneously because that would require infinite current.
In DC steady state (dv/dt = 0), a capacitor acts as an open circuit.
The integral form v(t) = (1/C)×∫i dt + v(0) recovers voltage from current history.

➡️ Now meet the inductor — the dual of the capacitor, where the roles of voltage and current are swapped.

Want to test yourself on this? Try the Electrical Aptitude test →