AC Voltage Applied to a Capacitor

AC voltage applied to a capacitor

Diagram in words

• A capacitor $C$ connected directly across an ac source; a single loop with no resistance. On dc, the capacitor charges and the current stops; on ac it is charged and discharged alternately as the current reverses each half cycle, so a current flows continuously in the circuit.

Derivation (what a 3- or 5-mark answer must show)

• Let $v = v_m\sin\omega t$. From the loop rule the voltage across the capacitor equals the source voltage, and $v = q/C$, so the instantaneous charge is

$$q = Cv = Cv_m\sin\omega t$$

• The current is the rate of flow of charge:

$$i = \dfrac{dq}{dt} = \dfrac{d}{dt}\left(Cv_m\sin\omega t\right) = \omega C v_m\cos\omega t$$

• Writing $\cos\omega t = \sin\left(\omega t + \dfrac{\pi}{2}\right)$ and $\omega C v_m = \dfrac{v_m}{1/\omega C}$:

$$i = \dfrac{v_m}{1/\omega C}\sin\left(\omega t + \dfrac{\pi}{2}\right) = i_m\sin\left(\omega t + \dfrac{\pi}{2}\right), \qquad i_m = \omega C v_m = \dfrac{v_m}{X_C}$$

• Conclusion to state: in a purely capacitive ac circuit the current leads the applied voltage in phase by $\pi/2$ (a quarter of a cycle).

Capacitive reactance

• The quantity that plays the part of resistance is the capacitive reactance

$$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi\nu C}$$

• Definition to quote: capacitive reactance is the opposition offered by a capacitor to the flow of alternating current through it. Its SI unit is the ohm ($\Omega$), the same as that of resistance, and its dimensions are $[\mathrm{M\,L^2\,T^{-3}\,A^{-2}}]$.
• It limits the current amplitude exactly as resistance does, but it is inversely proportional both to the frequency and to the capacitance.
• The impedance of a pure capacitor is $Z = X_C = \dfrac{1}{2\pi\nu C}$.
• For dc, $\nu = 0$ and $X_C\to\infty$: a capacitor blocks dc but passes ac.

Graphs to draw

• $v$ and $i$ against $\omega t$: two sinusoids of the same period $2\pi/\omega$; the current curve is shifted a quarter cycle to the left of the voltage curve, so $i$ reaches its peak $i_m$ a quarter period before $v$ reaches $v_m$.
• $X_C$ against $\nu$: a rectangular hyperbola in the first quadrant, very large as $\nu\to0$ and falling towards zero as $\nu$ increases.
• $I$ against $\nu$ for a pure capacitor: since $I = V/X_C = 2\pi\nu CV$, this is a straight line through the origin.
• Power against $\omega t$: a sinusoid of twice the frequency, $p = \dfrac{v_m i_m}{2}\sin2\omega t$, with equal areas above and below the axis in each cycle.
• Phasor diagram: the current phasor $\mathbf I$ of length $i_m$ drawn $90^\circ$ ahead of (anticlockwise from) the voltage phasor $\mathbf V$ of length $v_m$, both rotating anticlockwise with angular speed $\omega$.

Power in a pure capacitor

$$p = iv = i_m\cos\omega t\; v_m\sin\omega t = \dfrac{i_m v_m}{2}\sin2\omega t$$

$$P_{av} = \dfrac{i_m v_m}{2}\langle\sin2\omega t\rangle = 0$$

• Equivalently, $P_{av} = VI\cos\left(\dfrac{\pi}{2}\right) = 0$. Over a complete cycle the capacitor takes energy from the source while charging and returns exactly the same energy while discharging; it dissipates none.

AC voltage applied to an inductor

Diagram in words

• An inductor $L$ of negligible winding resistance connected across an ac source — a purely inductive circuit.

Derivation

• With no resistor in the loop, Kirchhoff's rule gives $v - L\dfrac{di}{dt} = 0$, the second term being the self-induced (back) emf, negative by Lenz's law. Hence

$$\dfrac{di}{dt} = \dfrac{v}{L} = \dfrac{v_m}{L}\sin\omega t$$

• Integrating, and setting the integration constant to zero because the source emf oscillates symmetrically about zero so the current must too:

$$i = -\dfrac{v_m}{\omega L}\cos\omega t = \dfrac{v_m}{\omega L}\sin\left(\omega t - \dfrac{\pi}{2}\right) = i_m\sin\left(\omega t - \dfrac{\pi}{2}\right), \qquad i_m = \dfrac{v_m}{\omega L}$$

• Conclusion: in a purely inductive circuit the current lags the applied voltage by $\pi/2$, reaching its maximum a quarter period $T/4 = (\pi/2)/\omega$ after the voltage does.

Inductive reactance

$$X_L = \omega L = 2\pi\nu L, \qquad i_m = \dfrac{v_m}{X_L}$$

• Inductive reactance is the opposition offered by an inductor to alternating current. SI unit: ohm ($\Omega$); dimensions $[\mathrm{M\,L^2\,T^{-3}\,A^{-2}}]$. It is directly proportional to $L$ and to $\nu$, so the $X_L$–$\nu$ graph is a straight line through the origin, and $X_L = 0$ for dc.

Graphs and power

• $v$ and $i$ against $\omega t$: the current curve is shifted a quarter cycle to the right of the voltage curve.
• Phasor diagram: $\mathbf I$ drawn $90^\circ$ behind $\mathbf V$.
• $X_L$ against $\nu$: straight line through the origin (contrast with the hyperbola for $X_C$).
• Instantaneous power $p = -\dfrac{v_m i_m}{2}\sin2\omega t$, so $P_{av} = 0$: an ideal inductor dissipates no power, storing energy in its magnetic field for a quarter cycle and returning it in the next.

Standard results and standard applications

• Lamp in series with a capacitor: no glow on dc (no steady current once charged); it glows on ac. Reducing $C$ raises $X_C$ and dims it; inserting a dielectric raises $C$, lowers $X_C$ and brightens it.
• Bulb in series with a coil: inserting an iron rod raises $L$ (the iron is magnetised, increasing the field and hence the inductance), so $X_L$ rises, more of the applied voltage appears across the inductor, and the bulb dims. Removing the iron, or reducing the number of turns, brightens it.
• Identification rule for a 1- or 5-mark question: if $i = i_0\sin(\omega t + \pi/2)$ for $v = v_0\sin\omega t$, the device is a capacitor; if $i = i_0\sin(\omega t - \pi/2)$, it is an inductor.
• Reading a reactance graph: at the marked point read off $X_C$ and $\nu$ and invert, $C = \dfrac{1}{2\pi\nu X_C}$; for an inductor, $L = \dfrac{X_L}{2\pi\nu}$.