Questions: Capacitors and Capacitance

Question 1

A capacitor has been connected to a 9V DC source long enough to fully charge. How much current is flowing through the capacitor?

• A. 9 × C amperes, where C is the capacitance in farads
• B. Zero, because dv/dt = 0 when voltage is constant
• C. A small but nonzero leakage current equal to V/R
• D. A current that has decayed exponentially to a negligible value

Answer: (B)

The voltage-current relationship is i = C(dv/dt). At DC steady state, voltage is constant, so dv/dt = 0 and therefore i = 0. The capacitor acts as an open circuit — no current flows through it. This is the fundamental reason capacitors block DC.

Question 2

A capacitor with capacitance C is charged to voltage V, storing energy U = ½CV². The voltage is then doubled to 2V. By what factor does the stored energy change?

• A. 2× — stored energy is proportional to voltage
• B. 4× — stored energy scales with voltage squared
• C. √2× — stored energy scales with the square root of voltage
• D. The stored energy is unchanged; only the charge distribution shifts

Answer: (B)

U = ½CV². Doubling V gives U = ½C(2V)² = ½C·4V² = 4·(½CV²). Energy quadruples. This nonlinear relationship matters practically: a capacitor at twice the voltage stores four times as much energy — and releases it just as suddenly if discharged.

Question 3

A capacitor with higher capacitance usually stores more energy than one with lower capacitance.

• T. True
• F. False

Answer: False

Stored energy is U = ½CV² — it depends on both capacitance and voltage. A small capacitor at high voltage can easily outstore a large capacitor at low voltage. For example, C = 1 μF at 1000V stores ½J; C = 1 F at 1V stores only 0.5J as well, but at 0.1V stores only 5 mJ. Capacitance alone does not determine stored energy.

Question 4

In an AC circuit, a capacitor's impedance decreases as frequency increases, so high-frequency signals pass through more easily than low-frequency ones.

• T. True
• F. False

Answer: True

Impedance Z_C = 1/(jωC). As frequency ω increases, Z_C decreases — the capacitor opposes high-frequency signals less. At DC (ω = 0), impedance is infinite (open circuit). At very high frequency, impedance approaches zero (short circuit). This frequency-dependent behavior is why capacitors are used to block DC bias while passing AC signal components.

Question 5

Why does a capacitor block DC but allow AC signals to pass? Explain in terms of the voltage-current relationship i = C(dv/dt).

Model answer: DC means constant voltage — dv/dt = 0 — so the current i = C·0 = 0. No current flows; the capacitor is an open circuit. AC means continuously changing voltage, so dv/dt ≠ 0 and current flows. The higher the frequency, the faster the voltage changes, and the larger the current for the same voltage amplitude — which is why high-frequency signals encounter lower impedance.

The key is that current through a capacitor reflects the rate of voltage change, not voltage level. A steady voltage, no matter how large, produces no current. A rapidly oscillating voltage, even if small in amplitude, produces large current. This is the physical basis of capacitive coupling and high-pass filtering.