Q.A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?

In a pure capacitor the current leads the applied voltage by π/2, i.e. current is ahead (forward).

For a capacitor driven by $V=V_0\sin\omega t$, the charge is $q=CV$ and the current is

$I=\frac{dq}{dt}=\omega C V_0\cos\omega t=\omega C V_0\sin\!\left(\omega t+\tfrac{\pi}{2}\right).$

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In a pure capacitor the current leads the applied voltage by a quarter cycle, i.e. a phase difference of 90°.

For an a.c. circuit containing only a capacitor, the charging current is maximum when the voltage is zero and vice versa. The current leads the voltage by

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In a pure capacitor, $I$ leads $V$ by $\pi/2$.

In an AC circuit containing only a capacitor, the current leads the applied voltage by a phase angle of $90°$ ($\pi/2$ radians) — this follows from $I = C\dfrac{dV}{dt}$, since the current is proportional to the rate of change of voltage, which is largest when V is crossing zero and zero when V is at its peak. In the phasor diagram, the voltage phasor $V_0$ is drawn along the reference (horizontal) axis, and the current phasor $I_0$ is drawn rot …

In a purely capacitive AC circuit current leads voltage by $90^\circ$, so $\cos\phi=0$ and the average power consumed is zero — option (ii).

Concept. Average AC power is $P=V_{rms}I_{rms}\cos\phi$, where $\phi$ is the phase angle between voltage and current.

Why zero. In a circuit with only a capacitor, the current leads the voltage by exactly $\phi=90^\circ$. Then …

In a purely capacitive AC circuit, current leads voltage by a phase angle of 90°.

For a pure capacitor, if $V = V_0\sin\omega t$ then the charge $q = CV_0\sin\omega t$ and the current

$I = \dfrac{dq}{dt} = \omega C V_0\cos\omega t = I_0\sin\!\left(\omega t + \dfrac{\pi}{2}\right)$

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In a purely capacitive AC circuit the current leads the voltage by π/2 (equivalently, voltage lags current by 90°).

For a capacitor of capacitance C connected to an alternating voltage $v = v_0\sin(\omega t)$, the charge on the capacitor is $q = Cv = Cv_0\sin(\omega t)$. The current is the rate of change of charge:

$i = \frac{dq}{dt} = C v_0 \omega \cos(\omega t) = i_0 \sin\left(\omega t + \frac{\pi}{2}\right)$ …

In a purely capacitive AC circuit the current leads the voltage by 90°, so cos φ = cos 90° = 0.

Power factor is defined as $\cos\phi$, where $\phi$ is the phase difference between voltage and current. In a purely capacitive circuit, $\phi = 90°$, so $\cos\phi = 0$. This means the average power consumed, $P_{avg}=V_{rms}I_{rms}\cos\phi$, is zero — …

In a purely capacitive AC circuit the current leads the voltage by $90^\circ$, so the power factor $\cos\phi = \cos 90^\circ = 0$.

Power factor is defined as $\cos\phi$, where $\phi$ is the phase angle between the applied voltage and the current. In a pure capacitor, current leads voltage by exactly $\phi=90^\circ$ (no resistance to dissipate energy — the capacitor only stores and returns energy each half-c …

A capacitor blocks steady direct current once fully charged, whereas a resistor and an ideal inductor both allow d.c. to flow continuously.

A resistor obeys Ohm's law $V=IR$ for both d.c. and a.c.; it offers a fixed opposition to current and allows a steady direct current to flow continuously as long as a potential difference is maintained across it.

An inductor opposes only a change in current, through the induced back-emf $e=-L\dfrac{dI}{dt}$. For a steady (unchanging) direct current, $\dfrac{dI}{dt}=0$, so the induced emf is zero and the inductor offers no opposition -- d.c. flows through it freely once the current has settled to a constant value (it behaves essentially like a plain conductor).

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A capacitor does not allow direct current to flow through it in the steady state.

In a d.c. circuit, once a capacitor is charged to the applied voltage, no further charge flows onto its plates and the current becomes zero — the capacitor behaves like an open circuit for steady d.c. An inductor, by contrast, offers zero opposition to steady d.c. (since $V=L\,dI/dt=0$ for constant current) and behaves like a plain wire. A resistor always allows d.c. to pass, since Ohm's law …