Different types of AC circuits

13.5 Different Types of AC Circuits

In this section the voltage-current relations are derived for individual circuit elements and then for the series combination, all carrying a sinusoidal current. The capacitor and the inductor are assumed ideal unless otherwise stated.

(a) AC Voltage Applied to a Resistor

A resistor of resistance $R$ is connected across a source of alternating emf whose instantaneous value is

$$e = e_0\sin\omega t \qquad \ldots(1)$$

The instantaneous emf equals the instantaneous potential drop across the resistance, so

$$e = iR \qquad \ldots(2)$$

From (1) and (2),

$$iR = e_0\sin\omega t \qquad \Rightarrow \qquad i = \dfrac{e_0}{R}\sin\omega t = i_0\sin\omega t, \qquad i_0 = \dfrac{e_0}{R}$$

Comparing $i_0 = e_0/R$ with Ohm's law, resistors behave in exactly the same way for AC and for DC. The behaviour of $R$ in DC and AC circuits is the same, and $R$ can reduce DC as well as AC equally effectively.

Comparing the two boxed equations, there is zero phase difference between the instantaneous alternating current and the instantaneous alternating emf in a resistive circuit — they are in phase. Both $e$ and $i$ reach zero, minimum and maximum values at the same instants.

Phasor diagram: since $i$ and $e$ are in phase, both phasors point in the same direction, making an angle $\omega t$ with OX. Their projections on the vertical axis give the instantaneous values. The phase angle between them is zero.

(b) AC Voltage Applied to an Inductor

A pure inductor $L$ of negligible resistance is connected across $e = e_0\sin\omega t$. When the key is closed, the current begins to grow, the flux linked with the coil changes, and an induced emf is produced which opposes the applied emf. By Lenz's law

$$e' = -L\dfrac{di}{dt}$$

To maintain the current, the applied emf must be equal and opposite to the induced emf. By Kirchhoff's voltage law, with no resistance in the circuit,

$$e = -e' = L\dfrac{di}{dt} \qquad \Rightarrow \qquad di = \dfrac{e}{L}dt = \dfrac{e_0}{L}\sin\omega t\,dt$$

Integrating both sides,

$$i = \dfrac{e_0}{L}\int\sin\omega t\,dt = -\dfrac{e_0}{\omega L}\cos\omega t + \text{constant}$$

The constant of integration is time independent and has the dimensions of current. As the emf oscillates about zero, $i$ also oscillates about zero, so there cannot be any time-independent component of current — the constant is zero. Writing $-\cos\omega t = \sin(\omega t - \pi/2)$,

$$i = \dfrac{e_0}{\omega L}\sin\!\left(\omega t - \dfrac{\pi}{2}\right) = i_0\sin\!\left(\omega t - \dfrac{\pi}{2}\right), \qquad i_0 = \dfrac{e_0}{\omega L}$$

Comparing with $e = e_0\sin\omega t$, the alternating current lags behind the alternating emf by a phase angle of $\pi/2$ radian (90°). Equivalently, the voltage across $L$ leads the current by 90° — which is the standard board question "show that in an AC circuit containing a pure inductor the voltage is ahead of the current by $\pi/2$ in phase".

Phasor diagram: the phasor representing the peak emf $e_0$ makes an angle $\omega t$ anticlockwise from the horizontal axis; since the current lags by 90°, the phasor representing $i_0$ is turned 90° clockwise from $e_0$.

Graph: plot $e$ and $i$ against $\omega t$. The current curve is the emf curve shifted to the right by a quarter cycle — the current reaches its peak a quarter period after the emf does.

Inductive reactance

The opposing nature of an inductor to the flow of alternating current is called inductive reactance.

Comparing $i_0 = e_0/\omega L$ with Ohm's law $i_0 = e_0/R$, the quantity $\omega L$ is the effective resistance offered by the inductance $L$. It is denoted $X_L$:

$$X_L = \omega L = 2\pi fL \qquad (\because \omega = 2\pi f)$$

• Its function is similar to that of resistance in a purely resistive circuit.
• It is directly proportional to the inductance $L$ and to the frequency $f$ of the AC.
• Its dimensions are the same as those of resistance and its SI unit is the ohm ($\Omega$).
• In a DC circuit $f = 0$, so $X_L = 0$: a pure inductor offers zero resistance to DC and cannot reduce it. An inductor passes DC and blocks AC of very high frequency.
• In an inductive circuit the self-induced emf opposes the growth as well as the decay of current.

(c) AC Voltage Applied to a Capacitor

A capacitor of capacitance $C$ is connected across $e = e_0\sin\omega t$. The current transfers charge to the plates, producing a potential difference between them; as the current reverses each half cycle, the capacitor is alternately charged and discharged.

If $q$ is the charge at an instant $t$, then $V = q/C$, i.e. $q = CV = Ce$ since the capacitor voltage equals the source emf at every instant. Hence

$$i = \dfrac{dq}{dt} = \dfrac{d}{dt}(Ce) = C\dfrac{d}{dt}(e_0\sin\omega t) = \omega Ce_0\cos\omega t$$

$$\therefore\ i = \dfrac{e_0}{1/\omega C}\cos\omega t = \dfrac{e_0}{1/\omega C}\sin\!\left(\omega t + \dfrac{\pi}{2}\right)$$

The current is maximum when $\sin(\omega t + \pi/2) = 1$, so the peak current is

$$i_0 = \dfrac{e_0}{1/\omega C} = \omega Ce_0 \qquad \Rightarrow \qquad i = i_0\sin\!\left(\omega t + \dfrac{\pi}{2}\right)$$

Comparing with $e = e_0\sin\omega t$, in an AC circuit containing a capacitor only the alternating current leads the alternating emf by a phase angle of $\pi/2$ radian.

Phasor diagram: the emf phasor makes an angle $\omega t$ anticlockwise with the horizontal axis; since the current leads by 90°, the phasor representing $i_0$ is turned 90° anticlockwise with respect to $e_0$. The projections of both phasors on the vertical axis give the instantaneous values.

Graph: plot $e$ and $i$ against $\omega t$. The current curve is the emf curve shifted to the left by a quarter cycle.

Capacitive reactance

The capacitive reactance of a capacitor is defined as the ratio of the rms voltage (emf) across the capacitor to the corresponding rms current.

Comparing $i_0 = e_0/(1/\omega C)$ with Ohm's law, the quantity $1/\omega C$ is the effective resistance offered by the capacitor, denoted $X_C$:

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

• Its function in a purely capacitive circuit is to limit the amplitude of the current, just as resistance does in a resistive circuit.
• $X_C$ varies inversely as the frequency of the AC and inversely as the capacitance.
• Its dimensions are the same as those of resistance and its SI unit is the ohm ($\Omega$).
• In a DC circuit $f = 0$, so $X_C = \infty$: a capacitor blocks DC and acts as an open circuit, while it passes AC of high frequency.

Comparison of Resistance and Reactance

(d) Series LCR Circuit — Derivation of the Impedance

A pure resistor $R$, a pure inductance $L$ and an ideal capacitor of capacitance $C$ are connected in series across a source of alternating emf. Because they are in series, the current at any instant through all three elements has the same amplitude and phase; let it be

$$i = i_0\sin\omega t$$

The voltage across each element bears a different phase relationship with this current:

$$e_R = iR, \qquad e_L = iX_L, \qquad e_C = iX_C \qquad \ldots(1)$$

Phasor diagram, drawn step by step:

• $e_R$ is in phase with the current, so the vector $e_R$ is drawn in the same direction as $i$, along the positive X-axis — call it $OA$.
• $e_L$ leads the current by $\pi/2$ radian (90°), so the vector $e_L$ is drawn along the positive Y-axis — call it $OB$.
• $e_C$ lags the current by $\pi/2$ radian (90°), so the vector $e_C$ is drawn along the negative Y-axis, $OY'$ — call it $OC$.
• The voltages across $L$ and $C$ have a phase difference of 180°, so the net reactive voltage is $(e_L - e_C)$. Taking $e_L > e_C$, this net reactive emf is in the direction of $e_L$; call it $OB'$.
• The resultant of $OA$ and $OB'$ is the diagonal $OK$ of the rectangle $OAKB'$, and it represents the applied emf $e$.

From the right-angled triangle $OAK$,

$$OK = \sqrt{OA^2 + OB'^2}$$

$$\therefore\ e^2 = e_R^2 + (e_L - e_C)^2$$

Using (1),

$$e^2 = (iR)^2 + (iX_L - iX_C)^2 = i^2\left[R^2 + (X_L - X_C)^2\right]$$

$$\therefore\ e = i\sqrt{R^2 + (X_L - X_C)^2} = iZ$$

$$\therefore\ Z = \sqrt{R^2 + (X_L - X_C)^2}$$

The quantity $\sqrt{R^2+(X_L-X_C)^2}$ represents the effective opposition offered by the inductor, capacitor and resistor connected in series to the flow of AC. This total effective resistance of the LCR circuit is called the impedance of the circuit, denoted $Z$.

• Impedance can be defined as the ratio of the rms voltage to the rms value of the current. It is expressed in ohm ($\Omega$).
• The reciprocal of the impedance of an AC circuit is called admittance; its SI unit is ohm$^{-1}$ or siemens.

Phase Angle and the Three Cases

From the phasor diagram, the voltage leads the current by a phase angle $\phi$ given by

$$\tan\phi = \dfrac{AK}{OA} = \dfrac{e_L - e_C}{e_R} = \dfrac{i_0X_L - i_0X_C}{i_0R} = \dfrac{X_L - X_C}{R}$$

$$\therefore\ \phi = \tan^{-1}\!\left(\dfrac{X_L - X_C}{R}\right)$$

so that the alternating current in the LCR circuit is $i = i_0\sin(\omega t - \phi)$ when the emf is $e = e_0\sin\omega t$; the textbook writes the pair as $i = i_0\sin(\omega t + \phi)$ and $e = e_0\sin(\omega t + \phi)$ depending on which quantity is taken as reference.

• If $X_L = X_C$, then $\tan\phi = 0$; voltage and current are in phase and the AC circuit is non-inductive.
• If $X_L > X_C$, then $\tan\phi$ is positive and $\phi$ is positive; the voltage leads the current and the circuit is inductance dominated.
• If $X_L < X_C$, then $\tan\phi$ is negative and $\phi$ is negative; the voltage lags the current and the circuit is capacitance dominated.

Impedance Triangle

From the three phasors $e_R = i_0R$, $e_L = i_0X_L$ and $e_C = i_0X_C$, dividing throughout by $i_0$ gives a right-angled triangle: the base $OA$ represents the ohmic resistance $R$, the perpendicular $AK$ represents the net reactance $(X_L - X_C)$, and the hypotenuse $OK$ represents the impedance $Z$. The angle $\angle AOK = \phi$ is the phase angle by which the voltage leads the current, with $\tan\phi = (X_L-X_C)/R$ and $\cos\phi = R/Z$.

Standard Results

• Resistor: $i = i_0\sin\omega t$, $i_0 = e_0/R$, phase difference zero.
• Inductor: $i = i_0\sin(\omega t - \pi/2)$, $i_0 = e_0/X_L$, $X_L = \omega L = 2\pi fL$, current lags by 90°.
• Capacitor: $i = i_0\sin(\omega t + \pi/2)$, $i_0 = e_0/X_C$, $X_C = 1/\omega C = 1/2\pi fC$, current leads by 90°.
• Series LCR: $Z = \sqrt{R^2+(X_L-X_C)^2}$, $\tan\phi = (X_L-X_C)/R$, $i_{rms} = e_{rms}/Z$, $\Delta t = \phi/\omega$.