Chapter 7: Alternating Current
Welcome to the dynamic world of Alternating Current (AC). Unlike Direct Current (DC), which flows in one direction, AC periodically reverses, creating the oscillating backbone of modern electrical power. This simple property is responsible for powering nearly everything in our lives. This guide delves into the core concepts of AC, its behavior in circuits, and the ingenious devices it enables.
⚡ The War of Currents
In the late 1880s, a fierce rivalry known as the “War of Currents” raged between Thomas Edison, who championed DC, and Nikola Tesla & George Westinghouse, who advocated for AC. AC’s ability to be easily transformed to high voltages for efficient long-distance transmission ultimately won, shaping the electrical grid we use today.
AC Fundamentals: Sinusoids and RMS Value
Peak vs. RMS Value
The voltage and current in AC circuits are constantly changing. The maximum value is called the **peak value** ($v_m, i_m$). However, for practical purposes like calculating power, we use the **Root Mean Square (RMS)** value. The RMS value is the equivalent DC value that would dissipate the same amount of heat in a resistor.
Standard household voltage (e.g., 220V in India) is the RMS value.
Practice Problems: Basics & RMS Values
Problem 1: The peak voltage of an AC supply is 300 V. What is the RMS voltage?
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Given: Peak voltage \(v_m = 300\) V.
Formula: \(V_{rms} = \frac{v_m}{\sqrt{2}}\)
Solution: \(V_{rms} = \frac{300}{\sqrt{2}} \approx \frac{300}{1.414} \approx 212.1\) V.
Problem 2: The RMS value of current in an AC circuit is 10 A. What is the peak current?
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Given: RMS current \(I_{rms} = 10\) A.
Formula: \(i_m = I_{rms} \times \sqrt{2}\)
Solution: \(i_m = 10 \times \sqrt{2} \approx 10 \times 1.414 \approx 14.14\) A.
Problem 3: An AC voltage is given by \(v = 140 \sin(314t)\). Find the RMS voltage and frequency.
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Comparing with \(v = v_m \sin(\omega t)\), we get \(v_m = 140\) V and \(\omega = 314\) rad/s.
RMS Voltage: \(V_{rms} = \frac{140}{\sqrt{2}} \approx 99\) V.
Frequency: \(\omega = 2\pi f \Rightarrow f = \frac{\omega}{2\pi} = \frac{314}{2 \times 3.14} = 50\) Hz.
Problem 4: A light bulb is rated at 100 W for a 220 V supply. Find the resistance of the bulb.
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The rated voltage is the RMS voltage. Formula: \(P = \frac{V_{rms}^2}{R}\).
Resistance \(R = \frac{V_{rms}^2}{P} = \frac{(220)^2}{100} = \frac{48400}{100} = 484\) Ω.
Phasors: Visualizing AC
Since AC voltages and currents are sinusoids, analyzing complex circuits can be tricky. **Phasors** simplify this. A phasor is a rotating vector whose length represents the peak amplitude ($v_m$ or $i_m$) and whose angle represents the phase. Its projection on the vertical axis gives the instantaneous value.
Why Phasors?
Phasors turn complex trigonometric problems into simple vector addition. By “freezing” the rotating phasors at t=0, we can see the phase relationships between voltage and current in different components at a glance, making circuit analysis much more intuitive.
AC Through R, L, and C
The behavior of AC depends entirely on the component it flows through. The relationship between voltage and current changes dramatically for resistors, inductors, and capacitors.
Practice Problems: Reactance & Individual Components
Problem 5: A 44 mH inductor is connected to 220 V, 50 Hz AC. Find the inductive reactance.
Show Solution
Given: L = 44 mH = \(44 \times 10^{-3}\) H, f = 50 Hz.
Formula: \(X_L = 2\pi fL\)
Solution: \(X_L = 2 \times 3.14 \times 50 \times 44 \times 10^{-3} \approx 13.82\) Ω.
Problem 6: What is the capacitive reactance of a 60 µF capacitor connected to a 110 V, 60 Hz supply?
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Given: C = 60 µF = \(60 \times 10^{-6}\) F, f = 60 Hz.
Formula: \(X_C = \frac{1}{2\pi fC}\)
Solution: \(X_C = \frac{1}{2 \times 3.14 \times 60 \times 60 \times 10^{-6}} \approx 44.2\) Ω.
Problem 7: A 100 Ω resistor is connected to a 220 V, 50 Hz AC supply. Find the RMS current.
Show Solution
Given: R = 100 Ω, V_rms = 220 V.
Formula: \(I_{rms} = \frac{V_{rms}}{R}\)
Solution: \(I_{rms} = \frac{220}{100} = 2.2\) A.
Problem 8: Find the RMS current through the inductor from Problem 5.
Show Solution
From Problem 5, \(X_L \approx 13.82\) Ω. Given V_rms = 220 V.
Formula: \(I_{rms} = \frac{V_{rms}}{X_L}\)
Solution: \(I_{rms} = \frac{220}{13.82} \approx 15.92\) A.
Problem 9: Find the RMS current through the capacitor from Problem 6.
Show Solution
From Problem 6, \(X_C \approx 44.2\) Ω. Given V_rms = 110 V.
Formula: \(I_{rms} = \frac{V_{rms}}{X_C}\)
Solution: \(I_{rms} = \frac{110}{44.2} \approx 2.49\) A.
Problem 10: How does the inductive reactance change if the frequency is doubled?
Show Solution
Since \(X_L = \omega L = 2\pi fL\), \(X_L\) is directly proportional to \(f\).
If the frequency is doubled, the inductive reactance also doubles.
Problem 11: How does capacitive reactance change if the frequency is doubled?
Show Solution
Since \(X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}\), \(X_C\) is inversely proportional to \(f\).
If the frequency is doubled, the capacitive reactance is halved.
The Series LCR Circuit
When a resistor, inductor, and capacitor are connected in series, their individual effects combine. The total opposition is **Impedance (Z)**, and the phase angle ($\phi$) depends on the relative values of $X_L$ and $X_C$.
Phasor Diagram & Impedance
The voltage across the inductor ($V_L$) and capacitor ($V_C$) are 180° out of phase. We find the net reactive voltage ($V_L – V_C$) and combine it with the resistor voltage ($V_R$) using the Pythagorean theorem to find the total voltage and impedance.
Practice Problems: LCR Circuits & Impedance
Problem 12: A series circuit has R=10Ω, XL=20Ω, XC=12Ω. Calculate the impedance.
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Formula: \(Z = \sqrt{R^2 + (X_L – X_C)^2}\)
Solution: \(Z = \sqrt{10^2 + (20 – 12)^2} = \sqrt{100 + 8^2} = \sqrt{100+64} = \sqrt{164} \approx 12.8\) Ω.
Problem 13: In the circuit from Problem 12, find the phase angle \(\phi\).
Show Solution
Formula: \(\tan \phi = \frac{X_L – X_C}{R}\)
Solution: \(\tan \phi = \frac{20 – 12}{10} = \frac{8}{10} = 0.8\). So, \(\phi = \arctan(0.8) \approx 38.66°\). Since \(X_L > X_C\), the circuit is inductive and voltage leads current.
Problem 14: A 100V AC source is connected to the circuit in Problem 12. Find the RMS current.
Show Solution
From Problem 12, \(Z \approx 12.8\) Ω. Given V_rms = 100 V.
Formula: \(I_{rms} = \frac{V_{rms}}{Z}\)
Solution: \(I_{rms} = \frac{100}{12.8} \approx 7.81\) A.
Problem 15: A circuit has R=30Ω, L=0.1H, C=100µF connected to a 200V, 50Hz supply. Find \(X_L, X_C\), and Z.
Show Solution
\(X_L = 2\pi fL = 2 \times 3.14 \times 50 \times 0.1 = 31.4\) Ω.
\(X_C = \frac{1}{2\pi fC} = \frac{1}{2 \times 3.14 \times 50 \times 100 \times 10^{-6}} \approx 31.83\) Ω.
\(Z = \sqrt{30^2 + (31.4 – 31.83)^2} = \sqrt{900 + (-0.43)^2} = \sqrt{900.18} \approx 30.003\) Ω.
Problem 16: For the circuit in Problem 15, is it inductive or capacitive? What is the phase angle?
Show Solution
Since \(X_C > X_L\) (31.83Ω > 31.4Ω), the circuit is predominantly **capacitive**. Current leads voltage.
\(\tan \phi = \frac{31.4 – 31.83}{30} = \frac{-0.43}{30} \approx -0.0143\). So, \(\phi = \arctan(-0.0143) \approx -0.82°\). The negative sign confirms it’s capacitive.
Resonance: The Sweet Spot
In a series LCR circuit, there is a specific frequency where the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$). This is the **resonant frequency** ($\omega_0$). At this point, the impedance is at its minimum (Z=R), and the current is maximum.
The Tuning Circuit
Resonance is the principle behind tuning a radio or TV. The circuit is “tuned” by changing the capacitance until its resonant frequency matches the frequency of the desired station’s signal. This maximizes the current for that specific frequency while suppressing others.
At Resonance: \(X_L = X_C\)
$$\omega_0 = \frac{1}{\sqrt{LC}} \quad \text{or} \quad f_0 = \frac{1}{2\pi\sqrt{LC}}$$Practice Problems: Resonance
Problem 17: A series LCR circuit has L = 8H, C = 0.5µF, and R = 100Ω. Find the resonant frequency.
Show Solution
Formula: \(\omega_0 = \frac{1}{\sqrt{LC}}\)
Solution: \(\omega_0 = \frac{1}{\sqrt{8 \times 0.5 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-6}}} = \frac{1}{2 \times 10^{-3}} = 500\) rad/s.
Problem 18: What is the impedance of the circuit in Problem 17 at resonance?
Show Solution
At resonance, \(X_L = X_C\). Therefore, the impedance Z is at its minimum and is equal to the resistance R.
Solution: \(Z = R = 100\) Ω.
Problem 19: If the circuit in Problem 17 is connected to a 200V AC supply at resonant frequency, what is the current?
Show Solution
At resonance, Z = R = 100Ω. Given V_rms = 200V.
Solution: \(I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{100} = 2\) A. This is the maximum current the circuit can have.
Problem 20: A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 µH, what must be the range of its variable capacitor?
Show Solution
Formula: \(f_0 = \frac{1}{2\pi\sqrt{LC}} \Rightarrow C = \frac{1}{4\pi^2 f_0^2 L}\)
For \(f_1 = 800\) kHz = \(8 \times 10^5\) Hz:
\(C_1 = \frac{1}{4(3.14)^2 (8 \times 10^5)^2 (200 \times 10^{-6})} \approx 198\) pF.
For \(f_2 = 1200\) kHz = \(12 \times 10^5\) Hz:
\(C_2 = \frac{1}{4(3.14)^2 (12 \times 10^5)^2 (200 \times 10^{-6})} \approx 88\) pF.
The required range of the capacitor is from **88 pF to 198 pF**.
Power in AC Circuits
In DC, power is simply P=VI. In AC, since voltage and current are not always in phase, the power delivered to the circuit is more complex. The average power depends on the phase difference $\phi$.
Power Factor
The term $\cos\phi$ is called the **power factor**. It represents the fraction of the total apparent power that does useful work. A power factor of 1 (for a purely resistive circuit) is most efficient, while a power factor of 0 (for purely reactive circuits) means no net power is consumed.
The component of current \(I_{rms}\sin\phi\), which is 90° out of phase with voltage, is called **wattless current** as it consumes no power.
Practice Problems: Power Factor & Average Power
Problem 21: A series LCR circuit with R=10Ω and Z=20Ω is connected to a 200V source. Find the power factor.
Show Solution
Formula: \(\cos\phi = \frac{R}{Z}\)
Solution: \(\cos\phi = \frac{10}{20} = 0.5\). The power factor is 0.5.
Problem 22: For the circuit in Problem 21, find the average power consumed.
Show Solution
First, find the current: \(I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{20} = 10\) A.
Formula: \(P_{avg} = V_{rms} I_{rms} \cos\phi\).
Solution: \(P_{avg} = 200 \times 10 \times 0.5 = 1000\) W.
Alternatively: \(P_{avg} = I_{rms}^2 R = (10)^2 \times 10 = 100 \times 10 = 1000\) W.
Problem 23: A choke coil and a bulb are connected in series to a DC source. The bulb glows brightly. When connected to an AC source of the same voltage, it glows dimly. Why?
Show Solution
A choke coil is an inductor. For DC, its resistance is very low, so it offers little opposition. For AC, it has a large inductive reactance (\(X_L = 2\pi fL\)) in addition to its resistance. This increases the total impedance of the circuit, reducing the current and causing the bulb to glow dimly.
Problem 24: What is the power consumed in a purely inductive or purely capacitive circuit?
Show Solution
In a purely inductive or capacitive circuit, the phase angle \(\phi\) is \(\pm \frac{\pi}{2}\) (or \(\pm 90°\)).
The power factor is \(\cos(\pm \frac{\pi}{2}) = 0\).
Therefore, the average power consumed is \(P_{avg} = V_{rms} I_{rms} \times 0 = 0\). No power is consumed.
The AC Generator (Dynamo)
The AC generator is a device that converts mechanical energy into electrical energy based on the principle of **electromagnetic induction**. When a conducting coil is rotated in a uniform magnetic field, the magnetic flux through it changes, inducing an alternating EMF.
Working Principle
As the coil rotates, the angle $\theta$ between the magnetic field and the area vector of the coil changes. The induced EMF is proportional to the rate of change of magnetic flux ($\Phi_B = NBA\cos\theta$). This sinusoidal change in flux results in a sinusoidal EMF output.
Practice Problems: AC Generator
Problem 25: A rectangular coil of 100 turns, area 0.1 m², rotates at 50 rad/s in a magnetic field of 0.5 T. What is the peak EMF?
Show Solution
Given: N=100, A=0.1m², ω=50 rad/s, B=0.5 T.
Formula: \(\mathcal{E}_{max} = NBA\omega\)
Solution: \(\mathcal{E}_{max} = 100 \times 0.5 \times 0.1 \times 50 = 250\) V.
Problem 26: For the generator in Problem 25, what is the instantaneous EMF when the plane of the coil is parallel to the magnetic field?
Show Solution
When the plane of the coil is parallel to the field, the angle \(\theta\) between the area vector and the field is 90° or \(\frac{\pi}{2}\).
The EMF is given by \(\mathcal{E} = \mathcal{E}_{max} \sin(\omega t)\). At this orientation, \(\sin(\omega t) = \sin(90°) = 1\).
The EMF is at its maximum value, \(\mathcal{E} = \mathcal{E}_{max} = 250\) V.
Problem 27: At what orientation of the coil is the induced EMF zero?
Show Solution
The EMF \(\mathcal{E} = \mathcal{E}_{max} \sin(\omega t)\) is zero when \(\sin(\omega t) = 0\). This occurs when \(\omega t = 0°\) or \(180°\).
This corresponds to the orientation where the plane of the coil is **perpendicular** to the magnetic field lines. At this point, the magnetic flux is maximum, but its rate of change is momentarily zero.
Transformers
A transformer is a passive electrical device that transfers electrical energy from one circuit to another through electromagnetic induction. It is used to “step-up” or “step-down” AC voltages without changing the frequency.
Mutual Induction at Work
An alternating current in the **primary coil** creates a changing magnetic flux in the iron core. This changing flux is linked to the **secondary coil**, inducing an alternating EMF in it. The ratio of the voltages is determined by the ratio of the number of turns in the coils.
Transformer Equation (Ideal):
$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$$If $N_s > N_p$: Step-up transformer ($V_s > V_p$)
If $N_s < N_p$: Step-down transformer ($V_s < V_p$)
Practice Problems: Transformers
Problem 28: A step-down transformer has 2000 turns on the primary and 100 on the secondary. The primary is connected to a 2400 V AC supply. What is the secondary voltage?
Show Solution
Given: Np=2000, Ns=100, Vp=2400 V.
Formula: \(\frac{V_s}{V_p} = \frac{N_s}{N_p}\)
Solution: \(V_s = V_p \times \frac{N_s}{N_p} = 2400 \times \frac{100}{2000} = 120\) V.
Problem 29: In an ideal transformer, the primary coil has 500 turns and is connected to a 120 V supply. The secondary has 2000 turns. If the primary draws 2 A, what is the current in the secondary?
Show Solution
Given: Np=500, Ns=2000, Ip=2 A.
Formula: \(\frac{I_s}{I_p} = \frac{N_p}{N_s}\)
Solution: \(I_s = I_p \times \frac{N_p}{N_s} = 2 \times \frac{500}{2000} = 0.5\) A.
Problem 30: A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 A, calculate the efficiency of the transformer.
Show Solution
Efficiency \(\eta = \frac{\text{Output Power}}{\text{Input Power}}\)
Output Power (Power of lamp) = 100 W.
Input Power = \(V_p \times I_p = 220 \text{ V} \times 0.5 \text{ A} = 110\) W.
Solution: \(\eta = \frac{100}{110} \approx 0.909\). The efficiency is about 90.9%.
Problem 31: Why is the core of a transformer laminated?
Show Solution
The changing magnetic flux in the iron core induces large circulating currents called **eddy currents**. These currents produce unwanted heat (\(I^2R\) loss) and reduce the transformer’s efficiency. By laminating the core (using thin, insulated sheets of iron), the path for these eddy currents is broken, significantly reducing their magnitude and minimizing energy loss.