Q.A 30 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the rms value of the current in the circuit. OR A power transmission line feeds input power at 2200 V to a step down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 220 V ?

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 …