Chapter Notes

Introduction

So far, we have studied direct current (dc), which flows in a single direction. However, the electricity in our homes and offices is alternating current (ac). In ac, the voltage and current vary with time like a sine wave, constantly changing direction.

Why is AC so common?

• Easy Voltage Conversion: AC voltages can be easily increased or decreased using a device called a transformer. This is crucial for efficient power transmission.
• Economical Transmission: Electrical energy can be sent over long distances with less energy loss by using high-voltage AC.
• Useful Properties: AC circuits have special characteristics, like resonance, that are used in devices such as radios for tuning into specific stations.

AC Voltage Applied to a Resistor

Let's consider a simple circuit with a resistor connected to an AC voltage source. The symbol for an AC source is a circle with a sine wave inside ($\Theta$).

The voltage from the source changes sinusoidally over time and can be described by the equation: $v = v_m \sin \omega t$ Here, $v$ is the instantaneous voltage, $v_m$ is the amplitude or peak voltage, and $\omega$ is the angular frequency.

Using Kirchhoff's loop rule, we can find the current ($i$) through the resistor ($R$): $v_m \sin \omega t = iR$ $i = \frac{v_m}{R} \sin \omega t$

This can be written as: $i = i_m \sin \omega t$ where $i_m = \frac{v_m}{R}$ is the current amplitude or peak current. This relationship is Ohm's law, which applies to both DC and AC circuits for resistors.

An important observation is that the equations for voltage and current both have the same sine function ($\sin \omega t$). This means that the voltage and current rise and fall together. They reach their maximum, minimum, and zero values at the same time. We say that the voltage and current are in phase.

Power in a Resistive AC Circuit

Although the current alternates and its average value over a full cycle is zero, it still dissipates energy as heat in the resistor. This is because Joule heating depends on $i^2$, which is always positive.

The instantaneous power ($p$) dissipated in the resistor is: $p = i^2 R = (i_m \sin \omega t)^2 R = i_m^2 R \sin^2 \omega t$

The average power ($\bar{p}$ or $P$) over one complete cycle is: $\bar{p} = \langle i_m^2 R \sin^2 \omega t \rangle$ Since $i_m$ and $R$ are constants, and the average value of $\sin^2 \omega t$ over a cycle is $1/2$, the average power is: $\bar{p} = i_m^2 R \langle \sin^2 \omega t \rangle = \frac{1}{2} i_m^2 R$

Root Mean Square (RMS) Values

To make AC power calculations look like DC power calculations ($P = I^2R$), we use a special value for current and voltage called the root mean square (rms) value.

The rms current, denoted by $I$, is defined as: $I = \sqrt{\overline{i^2}} = \frac{i_m}{\sqrt{2}} \approx 0.707 i_m$ The rms voltage, denoted by $V$, is defined as: $V = \frac{v_m}{\sqrt{2}} \approx 0.707 v_m$

Using these rms values, the average power can be written in the familiar forms: $P = I^2 R = IV = \frac{V^2}{R}$ Also, the AC version of Ohm's law using rms values is: $V = IR$

Given

• Power, $P = 100 \text{ W}$
• RMS Voltage, $V = 220 \text{ V}$

To Find

(a) Resistance, $R$ (b) Peak voltage, $v_m$ (c) RMS current, $I$

Formula

$R = \frac{V^2}{P}$ $v_m = \sqrt{2} V$ $I = \frac{P}{V}$

Solution

(a) Resistance of the bulb $R = \frac{(220 \text{ V})^2}{100 \text{ W}} = \frac{48400}{100} \Omega = 484 \Omega$ Answer for part (a) = $484 \Omega$

(b) Peak voltage of the source $v_m = \sqrt{2} \times 220 \text{ V} \approx 1.414 \times 220 \text{ V} = 311 \text{ V}$ Answer for part (b) = $311 \text{ V}$

(c) RMS current through the bulb $I = \frac{100 \text{ W}}{220 \text{ V}} = 0.454 \text{ A}$ Answer for part (c) = $0.454 \text{ A}$

Representation of AC Current and Voltage by Rotating Vectors - Phasors

To visualize the phase relationship between voltage and current in AC circuits, we use a concept called phasors.

A phasor is a vector that rotates counter-clockwise about the origin with an angular speed $\omega$.

• The length of the phasor represents the amplitude (peak value) of the quantity (e.g., $v_m$ or $i_m$).
• The vertical projection of the phasor at any instant gives the instantaneous value of the quantity (e.g., $v = v_m \sin \omega t$).

For a resistor, the voltage and current are in phase. This means their phasors, $\mathbf{V}$ and $\mathbf{I}$, always point in the same direction as they rotate. The phase angle between them is zero.

AC Voltage Applied to an Inductor

Now, let's connect the AC source to a pure inductor ($L$), assuming it has negligible resistance.

According to Kirchhoff's loop rule, the applied voltage must be equal to the self-induced emf in the inductor: $v - L \frac{di}{dt} = 0$ $v_m \sin \omega t = L \frac{di}{dt}$

To find the current $i$, we integrate this equation with respect to time: $i = \int \frac{v_m}{L} \sin(\omega t) dt = -\frac{v_m}{\omega L} \cos(\omega t)$ Using the trigonometric identity $-\cos(\theta) = \sin(\theta - \pi/2)$, we get: $i = \frac{v_m}{\omega L} \sin(\omega t - \frac{\pi}{2})$

This can be written as: $i = i_m \sin(\omega t - \frac{\pi}{2})$ Comparing this to the voltage equation $v = v_m \sin \omega t$, we see a phase difference of $-\pi/2$. This means that in a purely inductive circuit, the current lags the voltage by $\pi/2$ radians (or 90 degrees).

Inductive Reactance

The term $\omega L$ in the current equation acts like resistance, as it limits the current. This opposition to current flow is called inductive reactance ($X_L$). $X_L = \omega L$ The unit of inductive reactance is the ohm ($\Omega$). The current amplitude is then given by: $i_m = \frac{v_m}{X_L}$ Inductive reactance is directly proportional to the angular frequency ($\omega$) and the inductance ($L$). This means it offers more opposition to higher frequency currents.

Power in an Inductive Circuit

The instantaneous power supplied to the inductor is: $p_L = iv = (-i_m \cos \omega t)(v_m \sin \omega t) = -\frac{i_m v_m}{2} \sin(2\omega t)$ The average power over a complete cycle is zero because the average of $\sin(2\omega t)$ over a cycle is zero. $P_L = \langle -\frac{i_m v_m}{2} \sin(2\omega t) \rangle = 0$ Therefore, a pure inductor does not consume any power. It stores energy in its magnetic field during one part of the cycle and returns it to the source in another part.

Given

• Inductance, $L = 25.0 \text{ mH} = 25.0 \times 10^{-3} \text{ H}$
• RMS Voltage, $V = 220 \text{ V}$
• Frequency, $\nu = 50 \text{ Hz}$

To Find

• Inductive reactance, $X_L$
• RMS current, $I$

Formula

$X_L = 2\pi\nu L$ $I = \frac{V}{X_L}$

Solution

First, calculate the inductive reactance: $X_L = 2 \times 3.14 \times 50 \text{ Hz} \times (25 \times 10^{-3} \text{ H}) = 7.85 \Omega$

Next, calculate the rms current: $I = \frac{220 \text{ V}}{7.85 \Omega} = 28 \text{ A}$

Final Answer The inductive reactance is $7.85 \Omega$ and the rms current is $28 \text{ A}$.

AC Voltage Applied to a Capacitor

Consider an AC source connected to a capacitor ($C$), forming a purely capacitive circuit.

The instantaneous voltage across the capacitor is $v = q/C$, where $q$ is the charge. By Kirchhoff's loop rule, this must equal the source voltage: $v_m \sin \omega t = \frac{q}{C}$ The current is the rate of change of charge, $i = dq/dt$. So we first solve for $q$ and then differentiate: $q = v_m C \sin \omega t$ $i = \frac{d}{dt}(v_m C \sin \omega t) = \omega C v_m \cos(\omega t)$ Using the identity $\cos(\theta) = \sin(\theta + \pi/2)$, we get: $i = i_m \sin(\omega t + \frac{\pi}{2})$ Comparing this to the voltage equation $v = v_m \sin \omega t$, we see a phase difference of $+\pi/2$. This means that in a purely capacitive circuit, the current leads the voltage by $\pi/2$ radians (or 90 degrees).

Capacitive Reactance

The term $1/(\omega C)$ acts as the opposition to current flow in a capacitive circuit. This is called capacitive reactance ($X_C$). $X_C = \frac{1}{\omega C}$ The unit of capacitive reactance is also the ohm ($\Omega$). The current amplitude is given by: $i_m = \frac{v_m}{X_C}$ Capacitive reactance is inversely proportional to the angular frequency ($\omega$) and the capacitance ($C$). It offers less opposition to higher frequency currents.

Power in a Capacitive Circuit

The instantaneous power supplied to the capacitor is: $p_c = iv = (i_m \cos \omega t)(v_m \sin \omega t) = \frac{i_m v_m}{2} \sin(2\omega t)$ Just like the inductor, the average power over a complete cycle is zero. $P_C = \langle \frac{i_m v_m}{2} \sin(2\omega t) \rangle = 0$ A pure capacitor does not consume power. It stores energy in its electric field and returns it to the source.

Solution

• DC Connection: When connected to a DC source, the capacitor will charge up. Once fully charged, it blocks the flow of current. Therefore, the lamp will not glow. Reducing the capacitance will not change this outcome.
• AC Connection: When connected to an AC source, the capacitor offers a capacitive reactance ($X_C = 1/\omega C$) and allows current to flow. The capacitor charges and discharges continuously, so the lamp will shine. If the capacitance ($C$) is reduced, the capacitive reactance ($X_C$) will increase. This increased opposition to current will cause the current to decrease, and the lamp will shine less brightly.

Given

• Capacitance, $C = 15.0 \mu\text{F} = 15.0 \times 10^{-6} \text{ F}$
• RMS Voltage, $V = 220 \text{ V}$
• Frequency, $\nu = 50 \text{ Hz}$

To Find

• Capacitive reactance, $X_C$
• RMS current, $I$
• Peak current, $i_m$
• Effect of doubling the frequency

Formula

$X_C = \frac{1}{2\pi\nu C}$ $I = \frac{V}{X_C}$ $i_m = \sqrt{2} I$

Solution

First, calculate the capacitive reactance: $X_C = \frac{1}{2\pi(50 \text{ Hz})(15.0 \times 10^{-6} \text{ F})} = 212 \Omega$

Next, calculate the rms current: $I = \frac{220 \text{ V}}{212 \Omega} = 1.04 \text{ A}$

Then, calculate the peak current: $i_m = \sqrt{2} \times 1.04 \text{ A} = 1.47 \text{ A}$

Effect of Doubling Frequency: If the frequency is doubled ($\nu' = 2\nu$), the capacitive reactance becomes: $X_C' = \frac{1}{2\pi(2\nu)C} = \frac{1}{2} X_C$ The capacitive reactance is halved. Since current is inversely proportional to reactance ($I = V/X_C$), the current will be doubled.

Final Answer The capacitive reactance is $212 \Omega$, the rms current is $1.04 \text{ A}$, and the peak current is $1.47 \text{ A}$. If the frequency is doubled, the reactance is halved and the current is doubled.

Solution

Inserting an iron rod into the inductor increases the magnetic field inside it, which significantly increases the inductance ($L$) of the coil.

The inductive reactance is given by $X_L = \omega L$. Since $L$ increases, the inductive reactance $X_L$ also increases.

In the series circuit, this increased reactance means the inductor opposes the current more strongly. As a result, a larger portion of the source voltage drops across the inductor, leaving less voltage across the light bulb. With less voltage, the current through the bulb decreases, and its glow diminishes.

Final Answer (b) The glow of the light bulb decreases.

AC Voltage Applied to a Series LCR Circuit

Let's analyze a circuit containing a resistor ($R$), an inductor ($L$), and a capacitor ($C$) connected in series to an AC source with voltage $v = v_m \sin \omega t$.

Since all components are in series, the current $i$ is the same through each one at any instant. Let the current be $i = i_m \sin(\omega t + \phi)$, where $\phi$ is the phase difference between the source voltage and the circuit current.

Phasor-Diagram Solution

We can find the overall behavior of the circuit by combining the phasors for the voltage across each component.

• Resistor Voltage ($V_R$): In phase with the current $I$. Its phasor $\mathbf{V_R}$ is parallel to $\mathbf{I}$. Amplitude: $v_{Rm} = i_m R$.
• Inductor Voltage ($V_L$): Leads the current $I$ by $\pi/2$. Its phasor $\mathbf{V_L}$ is $90^\circ$ ahead of $\mathbf{I}$. Amplitude: $v_{Lm} = i_m X_L$.
• Capacitor Voltage ($V_C$): Lags the current $I$ by $\pi/2$. Its phasor $\mathbf{V_C}$ is $90^\circ$ behind $\mathbf{I}$. Amplitude: $v_{Cm} = i_m X_C$.

The phasors $\mathbf{V_L}$ and $\mathbf{V_C}$ are in opposite directions. Their resultant phasor has a magnitude of $|v_{Cm} - v_{Lm}|$. The total source voltage phasor $\mathbf{V}$ is the vector sum of $\mathbf{V_R}$ and the resultant of $\mathbf{V_L}$ and $\mathbf{V_C}$.

Using the Pythagorean theorem on the phasor diagram: $v_m^2 = v_{Rm}^2 + (v_{Cm} - v_{Lm})^2$ Substituting the amplitudes: $v_m^2 = (i_m R)^2 + (i_m X_C - i_m X_L)^2 = i_m^2 [R^2 + (X_C - X_L)^2]$

Solving for the current amplitude $i_m$: $i_m = \frac{v_m}{\sqrt{R^2 + (X_C - X_L)^2}}$

Impedance

The term in the denominator represents the total opposition to current flow in an LCR circuit and is called impedance ($Z$). $Z = \sqrt{R^2 + (X_C - X_L)^2}$ Impedance is measured in ohms ($\Omega$). The current amplitude is simply: $i_m = \frac{v_m}{Z}$

The phase angle $\phi$ between the source voltage and the current can be found from the impedance diagram: $\tan \phi = \frac{X_C - X_L}{R}$

• If $X_C > X_L$, $\phi$ is positive, the circuit is predominantly capacitive, and the current leads the voltage.
• If $X_L > X_C$, $\phi$ is negative, the circuit is predominantly inductive, and the current lags the voltage.

Resonance

A key feature of LCR circuits is resonance. This occurs when the circuit oscillates with maximum amplitude at a specific frequency.

The current amplitude is $i_m = v_m / Z$. It will be maximum when the impedance $Z$ is minimum. Looking at the impedance formula, $Z$ is minimum when the term $(X_C - X_L)$ is zero. The condition for resonance is: $X_C = X_L$ $\frac{1}{\omega_0 C} = \omega_0 L$

Solving for the frequency gives the resonant frequency ($\omega_0$): $\omega_0 = \frac{1}{\sqrt{LC}}$

At resonance:

• The impedance is at its minimum value, $Z = R$.
• The current amplitude is at its maximum value, $i_m = v_m/R$.
• The phase angle $\phi$ is zero, meaning the current and voltage are in phase.

Given

• Resistance, $R = 200 \Omega$
• Capacitance, $C = 15.0 \mu\text{F} = 15.0 \times 10^{-6} \text{ F}$
• Source Voltage, $V = 220 \text{ V}$
• Frequency, $\nu = 50 \text{ Hz}$

To Find

(a) Current in the circuit, $I$ (b) Voltage across resistor ($V_R$) and capacitor ($V_C$) and explain the sum.

Formula

$X_C = \frac{1}{2\pi\nu C}$ $Z = \sqrt{R^2 + X_C^2}$ $I = \frac{V}{Z}$ $V_R = IR \quad \text{and} \quad V_C = IX_C$

Solution

(a) Calculate the circuit current

First, find the capacitive reactance: $X_C = \frac{1}{2 \times 3.14 \times 50 \times 15.0 \times 10^{-6}} = 212.3 \Omega$ Next, find the impedance: $Z = \sqrt{(200 \Omega)^2 + (212.3 \Omega)^2} = \sqrt{40000 + 45071.29} = \sqrt{85071.29} = 291.67 \Omega$ Now, find the current: $I = \frac{220 \text{ V}}{291.67 \Omega} = 0.755 \text{ A}$ Answer for part (a) = $0.755 \text{ A}$

(b) Calculate individual voltages and resolve the paradox

Voltage across the resistor: $V_R = (0.755 \text{ A})(200 \Omega) = 151 \text{ V}$ Voltage across the capacitor: $V_C = (0.755 \text{ A})(212.3 \Omega) = 160.3 \text{ V}$ The algebraic sum is $V_R + V_C = 151 \text{ V} + 160.3 \text{ V} = 311.3 \text{ V}$. This is greater than the source voltage of $220 \text{ V}$.

Resolving the Paradox: The paradox is resolved by remembering that these voltages are not in phase. $V_R$ is in phase with the current, while $V_C$ lags the current by $90^\circ$. Therefore, they must be added as vectors (phasors) using the Pythagorean theorem: $V_{total} = \sqrt{V_R^2 + V_C^2} = \sqrt{(151)^2 + (160.3)^2} = \sqrt{22801 + 25696} = \sqrt{48497} \approx 220 \text{ V}$ This vector sum equals the source voltage, as expected.

Power in AC Circuit: The Power Factor

The instantaneous power in a series LCR circuit is $p = vi$. The average power ($P$) dissipated over one cycle is: $P = \frac{v_m i_m}{2} \cos \phi$ Using rms values, this becomes: $P = VI \cos \phi$ The term $\cos \phi$ is called the power factor. It determines how much of the supplied power is actually dissipated as heat.

Here are the key cases for the power factor:

• Case (i) Purely Resistive Circuit: $X_C = X_L = 0$, so $\phi = 0$. The power factor $\cos \phi = 1$. Power dissipation is maximum ($P = VI$).
• Case (ii) Purely Inductive or Capacitive Circuit: The phase angle $\phi$ is $\pm \pi/2$. The power factor $\cos \phi = 0$. No power is dissipated. The current that flows without dissipating power is called wattless current.
• Case (iii) LCR Series Circuit: Power is only dissipated in the resistor. The power factor is $\cos \phi = R/Z$.
• Case (iv) LCR Circuit at Resonance: $X_C = X_L$, so $\phi = 0$ and $\cos \phi = 1$. Power dissipation is maximum, given by $P = I^2R$.

Solution

(a) Low power factor and transmission loss The power delivered to a user is $P = IV \cos\phi$. To deliver a specific amount of power $P$ at a voltage $V$, the required current is $I = P / (V \cos\phi)$. If the power factor $\cos\phi$ is low (close to 0), the current $I$ must be very high to deliver the same power. The power lost as heat in the transmission lines is given by $P_{loss} = I^2R_{line}$. A higher current $I$ leads to a much larger power loss ($I^2$ dependency). Therefore, a low power factor is inefficient for power transmission.

(b) Improving the power factor In many industrial circuits (e.g., with motors), the load is inductive, causing the current to lag the voltage. This results in a lagging (low) power factor. To improve it, we need to counteract the lagging "wattless" component of the current. This can be done by connecting a capacitor in parallel with the load. The capacitor draws a leading current, which cancels out the lagging current from the inductive load. This brings the total current more in phase with the voltage, increasing the power factor closer to 1 and improving efficiency.

Given

• Peak voltage, $v_m = 283 \text{ V}$
• Frequency, $\nu = 50 \text{ Hz}$
• Resistance, $R = 3 \Omega$
• Inductance, $L = 25.48 \text{ mH} = 25.48 \times 10^{-3} \text{ H}$
• Capacitance, $C = 796 \mu\text{F} = 796 \times 10^{-6} \text{ F}$

To Find

(a) Impedance, $Z$ (b) Phase difference, $\phi$ (c) Power dissipated, $P$ (d) Power factor, $\cos \phi$

Formula

$X_L = 2\pi\nu L \quad \text{and} \quad X_C = \frac{1}{2\pi\nu C}$ $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $\phi = \tan^{-1}\left(\frac{X_C - X_L}{R}\right)$ $P = I^2 R \quad \text{where} \quad I = \frac{v_m}{\sqrt{2} Z}$ $\text{Power factor} = \cos \phi$

Solution

(a) Impedance of the circuit First, calculate reactances: $X_L = 2 \times 3.14 \times 50 \times 25.48 \times 10^{-3} = 8 \Omega$ $X_C = \frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}} = 4 \Omega$ Now, calculate impedance: $Z = \sqrt{3^2 + (8 - 4)^2} = \sqrt{9 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \Omega$ Answer for part (a) = $5 \Omega$

(b) Phase difference $\phi = \tan^{-1}\left(\frac{4 - 8}{3}\right) = \tan^{-1}\left(\frac{-4}{3}\right) = -53.1^\circ$ Since $\phi$ is negative, the current lags the voltage. Answer for part (b) = $-53.1^\circ$

(c) Power dissipated First, find the rms current $I$: $I = \frac{i_m}{\sqrt{2}} = \frac{v_m/Z}{\sqrt{2}} = \frac{283/5}{\sqrt{2}} = \frac{56.6}{\sqrt{2}} \approx 40 \text{ A}$ Now, calculate the power: $P = I^2 R = (40 \text{ A})^2 \times 3 \Omega = 1600 \times 3 = 4800 \text{ W}$ Answer for part (c) = $4800 \text{ W}$

(d) Power factor $\text{Power factor} = \cos \phi = \cos(-53.1^\circ) = 0.6$ Answer for part (d) = $0.6$

Transformers

A transformer is a device used to change an AC voltage from one value to another. It works on the principle of mutual induction.

Construction and Working

A transformer consists of two coils, the primary coil ($N_p$ turns) and the secondary coil ($N_s$ turns), wound on a soft-iron core.

1. An alternating voltage ($v_p$) is applied to the primary coil.
2. This creates an alternating current, which produces an alternating magnetic flux in the iron core.
3. The changing magnetic flux links with the secondary coil and induces an alternating emf ($\varepsilon_s$) in it.

For an ideal transformer (no energy loss):

• The induced emf in each coil is proportional to the number of turns.
• The voltage across the primary ($v_p$) equals its back emf, and the voltage across the secondary ($v_s$) equals its induced emf.

This leads to the primary relationship for a transformer: $\frac{v_s}{v_p} = \frac{N_s}{N_p}$

If the transformer is 100% efficient, the input power equals the output power: $p_{in} = p_{out} \implies i_p v_p = i_s v_s$ This gives the relationship for currents: $\frac{i_p}{i_s} = \frac{v_s}{v_p} = \frac{N_s}{N_p}$

Types of Transformers

• Step-up Transformer: The secondary coil has more turns than the primary ($N_s > N_p$). This increases the voltage ($V_s > V_p$) but decreases the current ($I_s < I_p$).
• Step-down Transformer: The secondary coil has fewer turns than the primary ($N_s < N_p$). This decreases the voltage ($V_s < V_p$) but increases the current ($I_s > I_p$).

Energy Losses in Real Transformers

Real transformers are not 100% efficient due to:

1. Flux Leakage: Not all magnetic flux from the primary links with the secondary. Minimized by winding coils over each other.
2. Resistance of Windings: The wires have resistance, causing heat loss ($I^2R$). Minimized by using thick wire.
3. Eddy Currents: The changing magnetic flux induces circulating currents (eddy currents) in the iron core, causing heat. Minimized by using a laminated (layered) core.
4. Hysteresis: Energy is lost as heat each time the magnetic field reverses the magnetization of the core. Minimized by using a material with low hysteresis loss.

Transformers are essential for long-distance power transmission. Power is generated, stepped-up to a very high voltage (which reduces the current and $I^2R$ losses), transmitted, and then stepped-down in stages before reaching our homes at a safe voltage like 240 V.