In an AC circuit, does a capacitor current lead or lag voltage?

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Multiple Choice

In an AC circuit, does a capacitor current lead or lag voltage?

Explanation:
A capacitor’s current leads its voltage in an AC circuit because the current is driven by how fast the voltage is changing, not by the voltage value itself. For an ideal capacitor, i = C dv/dt. If the voltage is a sinusoid v(t) = Vp cos(ωt), then dv/dt = -ωVp sin(ωt) = ωVp cos(ωt + 90°). So the current is i(t) = ωCVp cos(ωt + 90°), which means it reaches its peak a quarter cycle earlier than the voltage. In phasor form, I = jωC V, showing a +90° phase shift relative to V. The lead by 90 degrees holds for an ideal capacitor regardless of frequency (though the current magnitude changes with frequency due to the 1/(ωC) impedance).

A capacitor’s current leads its voltage in an AC circuit because the current is driven by how fast the voltage is changing, not by the voltage value itself. For an ideal capacitor, i = C dv/dt. If the voltage is a sinusoid v(t) = Vp cos(ωt), then dv/dt = -ωVp sin(ωt) = ωVp cos(ωt + 90°). So the current is i(t) = ωCVp cos(ωt + 90°), which means it reaches its peak a quarter cycle earlier than the voltage. In phasor form, I = jωC V, showing a +90° phase shift relative to V. The lead by 90 degrees holds for an ideal capacitor regardless of frequency (though the current magnitude changes with frequency due to the 1/(ωC) impedance).

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