In DC steady-state, the impedance of an ideal capacitor as ω approaches zero is what?

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Multiple Choice

In DC steady-state, the impedance of an ideal capacitor as ω approaches zero is what?

Explanation:
In DC steady-state, a capacitor blocks DC because its impedance grows without bound as frequency goes to zero. For an ideal capacitor, Z = 1/(jωC). When ω → 0 (which corresponds to DC), the denominator goes to zero, so the magnitude of Z tends to infinity. That means the capacitor behaves as an open circuit to DC—the current through it drops to zero after any transient charging. This is why, in DC conditions, an ideal capacitor presents infinite impedance. The other options imply a finite or zero value, or depend on C, which isn’t the case in the ω → 0 limit.

In DC steady-state, a capacitor blocks DC because its impedance grows without bound as frequency goes to zero. For an ideal capacitor, Z = 1/(jωC). When ω → 0 (which corresponds to DC), the denominator goes to zero, so the magnitude of Z tends to infinity. That means the capacitor behaves as an open circuit to DC—the current through it drops to zero after any transient charging. This is why, in DC conditions, an ideal capacitor presents infinite impedance. The other options imply a finite or zero value, or depend on C, which isn’t the case in the ω → 0 limit.

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