Why does capacitor block dc signal at steady state?

Question

Examoogle · Accepted Answer

due to zero frequency of DC signal

Explanation: A capacitor blocks DC at steady state because its capacitive reactance is inversely proportional to frequency. Since the frequency of a DC signal is $f = 0$ Hz, the reactance approaches infinity, effectively acting as an open circuit.

Working Principle: The capacitive reactance is defined as $X_C = \frac{1}{2\pi f C}$. When a DC voltage is applied, the capacitor charges until the voltage across its plates equals the source voltage. At this point (steady state), no further charge flows, resulting in zero current and effectively infinite resistance to DC flow.

Key Formulas: $X_C = \frac{1}{2\pi f C}$ — capacitive reactance formula; $I = C \frac{dv}{dt}$ — capacitor current equation

Key Points: Steady state for a capacitor implies that charging current has ceased.; A capacitor acts as an open circuit to DC signals.; Reactance $X_C$ is measured in Ohms ($\Omega$).; Increasing frequency decreases capacitive reactance.

Advantages: Useful for blocking DC in AC amplifier coupling circuits; Effective for filtering out DC noise from signals

Disadvantages: Cannot be used for transmitting DC power; Finite breakdown voltage limits usage in high DC voltage systems

Applications: Coupling capacitors in amplifiers; DC blocking in radio frequency circuits; Smoothing filters in rectifiers

Additional Info: For AC, $f > 0$, so $X_C$ is finite and current flows through the capacitor.. Option B and D were identical in the provided list, both correctly identifying zero frequency as the cause.

Answer: B is correct — The capacitive reactance $X_C = \frac{1}{2\pi f C}$ becomes infinite when the frequency $f$ is zero, causing the capacitor to block DC current at steady state.

Exam Tip: In exams, remember that Inductors act as short circuits to DC at steady state ($X_L = 2\pi f L = 0$), while Capacitors act as open circuits.

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