Alternating Current and Electromagnetic Induction

Understanding Alternating Current (AC) and Electromagnetic Induction

Welcome to this comprehensive module on alternating current and electromagnetic induction. This course is designed for students of physics and engineering who want to deepen their grasp of AC fundamentals, power calculations, transformer operation, wave phenomena, and the underlying principles of electromagnetic induction. Each section expands on the concepts tested in the quiz, providing clear explanations, real‑world examples, and useful formulas.

1. Frequency of an AC Source

When a generator produces a sinusoidal voltage of the form e(t) = E_{0}\cos(\omega t), the angular frequency ω (in radians per second) is related to the ordinary frequency f (in hertz) by the simple relation:

• ω = 2πf

In the quiz, the voltage is e = 220\sqrt{2}\cos(100π t) V. Here ω = 100π rad/s, so:

• f = ω / (2π) = (100π) / (2π) = 50 Hz

Thus the generator supplies a standard mains frequency of 50 Hz, common in many countries.

2. Average Power Dissipated in a Resistor

For a sinusoidal current i(t) = I_{0}\cos(ωt) flowing through a resistor R, the instantaneous power is p(t) = i^{2}(t)R. The average (or real) power over a full cycle is given by:

• P_{avg} = I_{rms}^{2} R = \frac{I_{0}^{2}}{2} R

In the example, i = 2\cos(100π t) A, so I_{0}=2 A. The RMS current is I_{rms}=I_{0}/\sqrt{2}=\sqrt{2} A. Therefore:

• P = (\sqrt{2})^{2} × 10 Ω = 2 × 10 Ω = 20 W

This demonstrates that the average power is not simply the product of peak values, but depends on RMS quantities.

3. Current Drawn by a Lamp

A lamp rated at 220 V and 75 W operates on an AC supply. The current drawn can be found using the power formula for resistive loads:

• I = \frac{P}{V}

Substituting the given values:

• I = 75 W / 220 V ≈ 0.34 A

Because the lamp is essentially a resistor, the RMS current measured by an ammeter is 0.34 A. This illustrates the practical use of the power‑voltage‑current relationship in everyday appliances.

4. Transformer Turns Ratio

Ideal transformers obey the turns ratio relationship:

• \frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}

where V_{p} and V_{s} are the primary and secondary RMS voltages, and N_{p}, N_{s} are the corresponding turn counts. The total number of turns is given as 900, and the secondary voltage is 1000 V rms while the primary voltage is 200 V rms.

• First find the ratio: Vp/Vs = 200/1000 = 1/5.
• Thus Np/Ns = 1/5 → Np = Ns/5.
• Since Np + Ns = 900, substitute Ns = 5Np → Np + 5Np = 900 → 6Np = 900 → Np = 150 turns.

The primary winding therefore contains 150 turns, and the secondary contains 750 turns. This section reinforces how voltage transformation is directly linked to coil geometry.

5. Fundamental Frequency of a Vibrating String

For a string fixed at both ends, the fundamental (first harmonic) frequency is determined by the wave speed v and the length L:

• f_{1} = \frac{v}{2L}

Given v = 300 m/s and L = 0.60 m:

• f₁ = 300 / (2 × 0.60) = 300 / 1.2 = 250 Hz

This frequency is within the audible range and is the pitch you would hear when the string is plucked.

6. Wavelength of an Electromagnetic Wave

Electromagnetic (EM) waves travel at the speed of light c ≈ 3.0×10⁸ m/s in free space. The relationship between wavelength λ, frequency f, and speed is:

• λ = \frac{c}{f}

For a frequency of 95.6 MHz (which is 95.6×10⁶ Hz):

• λ = 3.0×10⁸ m/s ÷ 95.6×10⁶ Hz ≈ 3.14 m

The result, about 3.1 m, is typical for VHF radio transmissions. Understanding this link is essential for antenna design and wireless communication.

7. Determining the Period from a Graph

The period T of a periodic waveform is the time required for one complete cycle. If a graph shows that the waveform repeats every 0.04 s, then:

• T = 0.04 s

Correspondingly, the frequency would be f = 1/T = 25 Hz. Recognizing the period directly from a time‑domain plot is a valuable skill for analyzing oscilloscopes and sensor outputs.

8. Electromagnetic Induction – The Core Principle

When a magnet rotates near a coil, the magnetic flux through the coil changes with time. According to Faraday’s Law of Electromagnetic Induction:

• ε = -\frac{d\Phi_{B}}{dt}

where ε is the induced emf and Φ_B is the magnetic flux. This changing flux generates an alternating voltage, which drives current in the coil – the operating principle of a bicycle dynamo, generators, and many other devices.

9. Key Formulas Summary

• Frequency from angular frequency: f = ω / (2π)
• RMS values for sinusoidal waveforms: I_{rms}=I_0/√2, V_{rms}=V_0/√2
• Average power in a resistor: P = I_{rms}² R = V_{rms}² / R
• Transformer turns ratio: V_p/V_s = N_p/N_s
• Fundamental frequency of a string: f = v / (2L)
• EM wave wavelength: λ = c / f
• Period and frequency: T = 1/f
• Faraday’s law: ε = -dΦ_B/dt

10. Applying the Concepts – Practice Problems

To reinforce learning, try solving these variations:

• 1. A generator outputs e = 150\sqrt{2}\cos(200π t) V. What is its frequency?
• 2. A 5 Ω resistor carries a current i = 3\cos(120π t) A. Compute the average power dissipated.
• 3. An ideal transformer has a primary voltage of 120 V rms and a turns ratio of 1:4. What secondary voltage is produced?
• 4. A radio station broadcasts at 2.5 MHz. Find the wavelength of its signal.

Working through these will solidify the relationships between voltage, current, frequency, and physical dimensions.

11. SEO‑Friendly Tips for Learning AC Topics

When searching for resources on alternating current, use specific keywords such as "RMS voltage calculation", "transformer turns ratio example", or "fundamental frequency of a string". Adding terms like "step‑by‑step" or "practice problems" improves the relevance of results. Additionally, bookmarking reputable educational sites (e.g., university physics departments) ensures access to accurate formulas and diagrams.

By mastering the concepts outlined above, you will be well‑equipped to tackle both theoretical problems and practical engineering tasks involving alternating current and electromagnetic induction.