Alternating Current (AC): CBSE Class 12 Physics Guide

Welcome! You've likely spent a lot of time studying Direct Current (DC), where electricity flows in one constant direction. Now, get ready to explore its fascinating counterpart: Alternating Current (AC). This is the type of electricity that powers our homes, schools, and industries. Why AC? Its ability to be easily transformed to high or low voltages makes it incredibly efficient for long-distance power transmission. In this chapter, we'll dive deep into the world of AC. We will start with the fundamental definitions like peak and RMS values, understand how AC behaves when flowing through resistors, inductors, and capacitors, and then combine them in the all-important series LCR circuit. You'll master concepts like impedance, phase difference, resonance, and the working principle of transformers. This chapter is not just theoretical; it's the physics behind the modern electrical world.

Fundamental Concepts of Alternating Current

Alternating Current (AC): An electric current which periodically reverses its direction and changes its magnitude continuously with time. It is typically represented by a sine wave: I = I₀ sin(ωt).
Peak Value (I₀, V₀): The maximum value or amplitude of the alternating current or voltage during a cycle. It's the highest point on the AC waveform.
RMS Value (I_rms, V_rms): The Root Mean Square or 'effective' value of AC. It is the equivalent DC value that would produce the same heating effect in a resistor. For a sinusoidal AC, V_rms = V₀/√2 and I_rms = I₀/√2. Standard domestic supply voltage (e.g., 220V) is an RMS value.
Frequency (f): The number of complete cycles of the alternating current or voltage that occur in one second. It is measured in Hertz (Hz). In India, the standard AC frequency is 50 Hz.
Phase (ωt + φ): The argument of the sine or cosine function representing the AC quantity. It specifies the state of the wave at any given time t. The initial phase is φ.

AC Circuits with Resistors, Inductors, and Capacitors

Understanding how individual components behave in an AC circuit is key.

Purely Resistive Circuit: When an AC voltage V = V₀ sin(ωt) is applied across a resistor (R), the current is I = I₀ sin(ωt), where I₀ = V₀/R. The voltage and current are in the same phase (phase difference φ = 0). The opposition to current flow is simply the resistance R.

Purely Inductive Circuit: For an inductor (L), the voltage leads the current by a phase angle of π/2 (90°). The inductor offers opposition to the flow of AC, known as Inductive Reactance (X_L), given by X_L = ωL = 2πfL. It is directly proportional to the frequency; higher frequency means higher opposition.

Purely Capacitive Circuit: For a capacitor (C), the current leads the voltage by a phase angle of π/2 (90°). The opposition offered by the capacitor is called Capacitive Reactance (X_C), given by X_C = 1/(ωC) = 1/(2πfC). It is inversely proportional to the frequency; higher frequency means lower opposition.

These individual behaviors combine in more complex circuits like the series LCR circuit, where the total opposition, called Impedance (Z), depends on R, X_L, and X_C.

Worked Examples: Solving AC Circuit Problems

Example 1: Analyzing an AC Voltage Equation An alternating voltage is given by the equation v(t) = 283 sin(314t), where v is in volts and t is in seconds. Find (a) the peak voltage, (b) the RMS voltage, and (c) the frequency of the source. Solution: Step 1: Compare with the standard equation. The standard equation for an alternating voltage is v(t) = V₀ sin(ωt). Step 2: Identify Peak Voltage (V₀) and Angular Frequency (ω). By comparing the given equation with the standard one, we get: Peak Voltage, V₀ = 283 V. Angular Frequency, ω = 314 rad/s. Step 3: Calculate RMS Voltage (V_rms). The RMS voltage is related to the peak voltage by V_rms = V₀ / √2. V_rms = 283 / √2 ≈ 283 / 1.414 ≈ 200.1 V. So, the RMS voltage is approximately 200 V. Step 4: Calculate Frequency (f). The angular frequency ω is related to frequency f by ω = 2πf. f = ω / (2π) = 314 / (2 * 3.14) = 314 / 6.28 = 50 Hz. Final Answer: (a) Peak voltage is 283 V, (b) RMS voltage is approx. 200 V, and (c) frequency is 50 Hz.
Example 2: Calculating Impedance in a Series LCR Circuit A series LCR circuit has a resistor R = 20 Ω, an inductor L = 1.5 H, and a capacitor C = 35 µF. It is connected to a variable frequency 200 V AC supply. Calculate the impedance of the circuit when the frequency of the supply is 50 Hz. Solution: Step 1: List the given values. R = 20 Ω, L = 1.5 H, C = 35 µF = 35 × 10⁻⁶ F, V = 200 V, f = 50 Hz. Step 2: Calculate the angular frequency (ω). ω = 2πf = 2 3.14 50 = 314 rad/s. Step 3: Calculate the inductive reactance (X_L). X_L = ωL = 314 1.5 = 471 Ω. Step 4: Calculate the capacitive reactance (X_C). X_C = 1 / (ωC) = 1 / (314 35 × 10⁻⁶) = 1 / (0.01099) ≈ 91 Ω. * Step 5: Calculate the impedance (Z). The formula for impedance in a series LCR circuit is Z = √[R² + (X_L - X_C)²]. Z = √[20² + (471 - 91)²] Z = √[400 + (380)²] Z = √[400 + 144400] = √144800 ≈ 380.5 Ω. Final Answer: The impedance of the circuit at 50 Hz is approximately 380.5 Ω.

Exam Traps and Key Points

A very common mistake in exams is using peak voltage (V₀) or peak current (I₀) for power calculations. Always use the RMS values! Remember, the average power dissipated is P_avg = V_rms I_rms cos(φ). The domestic supply voltage (e.g., 220V in India) is always the RMS value, not the peak value.

Another key point is the phase relationship mnemonic 'ELI the ICE man'. For an Inductor (L), Voltage (E) leads Current (I). For a Capacitor (C), Current (I) leads Voltage (E). This is crucial for correctly drawing phasor diagrams and solving LCR circuit problems.

Practice Questions with Solutions

Q: A series LCR circuit with R = 10 Ω, L = 50 mH, and C = 200 μF is connected to a variable frequency AC source. At what angular frequency will the circuit exhibit resonance? What is the current at resonance if the applied RMS voltage is 100 V? A: Step 1: The condition for resonance in a series LCR circuit is when the inductive reactance (X_L) equals the capacitive reactance (X_C). The resonant angular frequency (ω₀) is given by the formula ω₀ = 1/√(LC). Step 2: Substitute the given values for L and C. L = 50 mH = 50 × 10⁻³ H and C = 200 μF = 200 × 10⁻⁶ F. ω₀ = 1 / √( (50 × 10⁻³) * (200 × 10⁻⁶) ) = 1 / √(10000 × 10⁻⁹) = 1 / √(10⁻⁵) = 1 / (10⁻²⁵) = √100000 = 100√10 ≈ 316.2 rad/s. Step 3: At resonance, X_L = X_C, so the impedance Z becomes minimum and is equal to the resistance R. Z = R = 10 Ω. Step 4: The current at resonance is given by I_rms = V_rms / Z = V_rms / R. I_rms = 100 V / 10 Ω = 10 A. Final answer: The resonant angular frequency is approximately 316.2 rad/s, and the current at resonance is 10 A.
Q: A coil has an inductance of 1 H. At what frequency will it have a reactance of 3140 Ω? A: Step 1: The formula for inductive reactance (X_L) is X_L = 2πfL. Step 2: We are given X_L = 3140 Ω and L = 1 H. We need to find the frequency f. Step 3: Rearrange the formula to solve for f: f = X_L / (2πL). Step 4: Substitute the values: f = 3140 / (2 3.14 1) = 3140 / 6.28 = 500 Hz. Final answer: The coil will have a reactance of 3140 Ω at a frequency of 500 Hz.
Q: A 100 W bulb is connected to a 220 V, 50 Hz AC mains. Calculate (a) the resistance of the bulb and (b) the RMS current flowing through it. A: Step 1: The power of a purely resistive device like a bulb is given by P = V_rms² / R. Step 2: We are given P = 100 W and V_rms = 220 V. We can rearrange the formula to find the resistance R. R = V_rms² / P = (220)² / 100 = 48400 / 100 = 484 Ω. Step 3: The RMS current (I_rms) can be calculated using Ohm's law for AC circuits: V_rms = I_rms * R. Step 4: Rearrange to find I_rms: I_rms = V_rms / R = 220 V / 484 Ω ≈ 0.45 A. Final answer: (a) The resistance of the bulb is 484 Ω. (b) The RMS current is approximately 0.45 A.
Q: A step-down transformer converts a 2200 V line to 220 V. The primary coil has 5000 turns. If the efficiency is 90% and the output power is 8 kW, find the number of turns in the secondary coil and the input power. A: Step 1: Use the transformer equation for voltages and turns: V_s / V_p = N_s / N_p. Here, V_p = 2200 V, V_s = 220 V, and N_p = 5000. Step 2: Solve for the number of turns in the secondary coil (N_s). N_s = N_p (V_s / V_p) = 5000 (220 / 2200) = 5000 * (1 / 10) = 500 turns. Step 3: The efficiency (η) of a transformer is the ratio of output power (P_out) to input power (P_in). η = P_out / P_in. Step 4: We are given η = 90% = 0.9 and P_out = 8 kW = 8000 W. We need to find P_in. P_in = P_out / η = 8000 W / 0.9 ≈ 8888.9 W. Final answer: The number of turns in the secondary coil is 500, and the input power is approximately 8.89 kW.

Frequently Asked Questions

Why is AC preferred over DC for long-distance power transmission?

AC voltage can be easily and efficiently stepped up to very high levels using transformers. This reduces the current for the same power (since P = VI), which in turn minimizes power loss (P_loss = I²R) in the transmission lines. The voltage is then stepped down for safe use at the destination.

What is a phasor diagram?

A phasor diagram is a graphical method to represent alternating quantities like voltage and current. Phasors are vectors that rotate counter-clockwise with the source's angular frequency. Their length represents the peak value (or RMS value), and the angle between them represents the phase difference.

What is the power factor and why is it important?

The power factor (cos φ) is the ratio of the true power (power dissipated) to the apparent power (product of RMS voltage and current) in an AC circuit. A power factor close to 1 is desirable because it means the circuit is using power efficiently. A low power factor implies a large current is drawn to deliver the same true power, leading to higher energy losses.