Questions: Capacitive Elements: Behavior and Properties

Question 1

A 10 μF capacitor is in a circuit where the voltage across it is changing at a rate of 1000 V/s. What current is flowing through the capacitor?

• A. 0 A — capacitors block current flow regardless of what is happening to the voltage
• B. 0.01 A — because i = C(dv/dt) = (10 × 10⁻⁶)(1000) = 0.01 A
• C. 100 A — the high rate of voltage change causes the capacitor to act as a short circuit
• D. The current depends on the voltage level across the capacitor, not its rate of change

Answer: (B)

Current through a capacitor is i = C(dv/dt), not i = CV. The current is proportional to the *rate of change* of voltage, not the voltage itself. With C = 10 μF and dv/dt = 1000 V/s: i = (10 × 10⁻⁶)(1000) = 0.01 A. Option D is the key misconception — if current depended on voltage (not its rate of change), a capacitor would behave like a resistor. Option A is wrong because current absolutely flows when voltage is changing.

Question 2

A circuit is designed to switch the voltage across a capacitor from 0 V to 5 V instantaneously. Why is this impossible in a real circuit?

• A. Capacitors can only charge to exactly half the supply voltage due to energy conservation
• B. The formula Q = CV shows that the capacitor can't store more than CV coulombs at once
• C. Instantaneous voltage change would require infinite current (i = C·dv/dt with dt → 0), which no real source can supply
• D. The capacitor's internal dielectric resistance prevents voltage from changing faster than RC

Answer: (C)

If voltage changes by ΔV in time Δt, then i = C·(ΔV/Δt). As Δt → 0 (instantaneous change), i → ∞. No physical current source can supply infinite current — there is always some series resistance limiting current, which means the voltage change must take finite time. This is the 'voltage memory' property: a capacitor's voltage cannot jump; it must change continuously. This constraint is why capacitors are used in snubber circuits to limit voltage spikes.

Question 3

A capacitor connected to a steady DC voltage source will eventually conduct zero current in steady state.

• T. True
• F. False

Answer: True

In DC steady state, the voltage across the capacitor is constant — it has charged up to the source voltage and stopped changing. Since i = C(dv/dt) and dv/dt = 0, the current is zero. The capacitor looks like an open circuit in DC steady state. This is why coupling capacitors block DC: after the initial transient, no DC current passes through.

Question 4

A capacitor connected to a DC source acts as a short circuit (very low impedance), which is why it can quickly deliver charge.

• T. True
• F. False

Answer: False

This reverses the behavior. A capacitor in DC steady state acts as an *open circuit* — it blocks DC current entirely once charged to the source voltage. It is at *high frequencies* (rapidly changing AC) that a capacitor's impedance is low and it passes current easily. The impedance of a capacitor is Z = 1/(jωC): as frequency ω → 0 (DC), Z → ∞ (open circuit); as ω → ∞, Z → 0 (short circuit).

Question 5

Using the equation i = C(dv/dt), explain why capacitors block DC signals but pass AC signals.

Model answer: The equation says current is proportional to the rate of change of voltage. A DC signal has constant voltage, so dv/dt = 0 everywhere and i = C·0 = 0 — no current flows. The capacitor is fully charged to the DC voltage and stops passing current. An AC signal has continuously changing voltage, so dv/dt is non-zero at every moment, and current flows continuously. The faster the voltage changes (higher frequency), the larger dv/dt, and the more current flows. This is why capacitors have low impedance at high frequencies and high impedance (effectively blocking) at low frequencies including DC.

This physical intuition maps directly to the impedance formula Z = 1/(jωC): at ω = 0 (DC), Z is infinite (open circuit, blocks DC); at high ω (high-frequency AC), Z approaches zero (short circuit, passes easily). The current equation i = C(dv/dt) and the impedance formula are two representations of the same underlying property — capacitors respond to *change* in voltage, not to voltage itself.