Alternating Current And Electromagnetic Waves NEET Notes and Important Questions by PW

 Alternating Current and Electromagnetic Waves can be challenging to revise because they combine numerical calculations with concepts involving phase relationships, circuit behaviour and electromagnetic wave properties. Questions on RMS values, LCR resonance, power factor and transformers require you to identify the correct formula, while electromagnetic wave questions test field directions, wave speed, energy and spectrum order.

PW’s Alternating Current and Electromagnetic Waves one-shot video brings together important concepts, formulas, tricks and Previous Year Questions (PYQs) to support your NEET Physics preparation. Use the chapter notes alongside the video to revise key equations, understand common question patterns and practise numerical problems.

Direct Current And Alternating Current

Direct Current

Direct current (DC) is unidirectional current. It continues to flow in the same direction, even if its magnitude changes with time.

A current may be:

• Constant and positive

• Constant and negative

• Variable but always on the same side of the time axis

If current changes from positive to negative without repeating periodically, it is neither ordinary DC nor pure AC.

Alternating Current

For pure alternating current, three conditions are essential:

1. Bidirectional: Current flows in both directions.

2. Periodic: The waveform repeats after a fixed time interval.

3. Symmetric: Positive and negative portions correspond to each other.

Common AC waveforms include sinusoidal, square, and triangular waves.

Memory Tip: Pure AC means Bidirectional + Periodic + Symmetric. All three conditions must be present.

AC–DC Comparison

AC is preferred for long-distance transmission and voltage transformation. DC is useful because it can be stored in batteries.

Skin Effect And Measurement Of AC

AC produces a changing magnetic field. The changing magnetic flux induces a back EMF inside the conductor, opposing current flow in its interior. Therefore, AC tends to flow near the surface. This phenomenon is called the skin effect.

Multiple-strand wires are useful for AC because they provide a larger total surface area.

Memory Tip: Multiple strands are especially important for AC because AC flows near the surface.

A moving-coil galvanometer measures DC. Its torque remains in one direction for DC, producing steady deflection. With AC, the torque reverses repeatedly, and the coil’s inertia prevents steady deflection.

A hot-wire ammeter measures both AC and DC. It works on the heating effect of current, which depends on the square of the current.

Average And RMS Current

For discrete current values:

Average current = Sum of current values / Number of values

For a continuous current:

Average current = Integral of I dt / Integral of dt

For a symmetric AC waveform, the average over a complete cycle is zero because positive and negative areas cancel.

For sinusoidal current:

• Full-cycle average = 0

• Conventional positive half-cycle average = 2I0 / pi

• RMS current = I0 / square root of 2

Here, I0 is the peak current.

The RMS method means:

1. Square the quantity.

2. Find its mean.

3. Take the root.

Memory Tip: RMS means Square → Mean → Root.

For a current containing DC and AC components:

i = IDC + I0 sin omega t

RMS current = square root of [IDC² + I0²/2]

For two sinusoidal AC components:

RMS current = square root of [(I1² + I2²)/2]

Sinusoidal AC Equation

The standard form is:

i = I0 sin(omega t + phi)

For:

i = 100 sin(50 pi t + pi/6)

• Peak current, I0 = 100 A

• Angular frequency, omega = 50 pi rad/s

• Frequency, f = 25 Hz

• Initial phase = 30 degrees

• Peak-to-peak current = 200 A

• Time period, T = 1/25 s

• RMS current = 100 / square root of 2 A

At t = 0:

i = 100 sin 30 degrees = 50 A

The current reaches its maximum when the phase becomes 90 degrees. The phase change is 60 degrees, so:

Time taken = T × 60/360 = T/6

Memory Tip: Always calculate time from the initial phase to the required final phase. Do not assume that every sinusoidal wave starts from zero.

If an AC source is given directly as “220 V, 50 Hz,” 220 V means RMS voltage. If the source is given as an equation, the coefficient generally represents peak voltage.

Pure Resistance, Inductance, And Capacitance

Pure Resistor

For a resistor:

i = I0 sin ωt

Voltage and current are in the same phase.

• Phase difference = 0 degrees

• Impedance, Z = R

• Average power = Vrms Irms

• Power is dissipated as heat

Pure Inductor

Inductive reactance is:

XL = omega L = 2 pi f L

The current is:

I = V / XL

In a pure inductor:

• Voltage leads current by 90 degrees.

• Current lags voltage by 90 degrees.

• Average power loss is zero.

• It behaves as a wire in steady-state DC.

Increasing frequency or inductance increases XL and decreases current.

Pure Capacitor

Capacitive reactance is:

XC = 1 / (omega C) = 1 / (2 pi f C)

In a pure capacitor:

• Current leads voltage by 90 degrees.

• Average power loss is zero.

• It behaves as an open circuit in steady-state DC.

Increasing frequency or capacitance decreases XC and increases current.

Memory Tip: In a resistor, voltage and current stay together. In an inductor, voltage leads. In a capacitor, current leads.

Series LCR Circuit

The impedance of a series LCR circuit is:

Z = square root of [R² + (XL − XC)²]

The average power is:

P = Vrms Irms cos phi

The power factor is:

cos phi = R/Z

Resonance

Resonance occurs when:

XL = XC

The resonant angular frequency is:

omega0 = 1 / square root of (LC)

The resonant frequency is:

f0 = 1 / [2 pi square root of (LC)]

At resonance:

• Net reactance is zero.

• Impedance is minimum: Z = R

• Current is maximum.

• Phase difference is zero.

• Power factor is one.

• Average power is maximum.

The bandwidth of a series LCR circuit is:

Delta omega = R/L

The quality factor is:

Q = omega0 / Delta omega = omega0 L/R

A narrow resonance curve indicates smaller bandwidth and higher quality factor.

Transformer

A transformer works on the principle of mutual inductance and requires AC.

For an ideal transformer:

Voltage ratio = Secondary turns / Primary turns

Vs/Vp = Ns/Np

Power is conserved:

Vp Ip = Vs Is

Therefore, current is inversely proportional to the number of turns.

• Step-up transformer: Ns > Np, voltage increases, current decreases.

• Step-down transformer: Ns < Np, voltage decreases, current increases.

Practical transformers have losses due to flux leakage, winding resistance, eddy currents, and hysteresis.

Electromagnetic Waves And Displacement Current

When a capacitor is charging, conduction current flows through the wires, but no charge crosses the gap between its plates. Maxwell explained this using displacement current.

The displacement current between capacitor plates is:

ID = epsilon0 A dE/dt

It can also be written as:

ID = C dV/dt

At any instant during charging:

ID = Conduction current

A changing electric field produces a magnetic field. This completes the current path in a charging capacitor.

Properties Of Electromagnetic Waves

Electromagnetic waves:

• Are transverse and non-mechanical.

• Do not require a material medium.

• Carry energy, momentum, and information.

• Have electric and magnetic fields in the same phase.

• Have electric and magnetic fields perpendicular to each other.

• Travel in the direction of E cross B.

• Do not transport electric charge.

The speed in a medium is:

v = 1 / square root of (mu epsilon)

In vacuum:

c = 1 / square root of (mu0 epsilon0)

Also:

E/B = v

In vacuum:

E/B = c

The electric and magnetic energy densities are equal. The average electric field and magnetic field are zero, but their average energy densities are not zero.

Electromagnetic Spectrum And Energy

The spectrum order is:

Gamma Rays → X-Rays → Ultraviolet → Visible Light → Infrared → Microwaves → Radio Waves

Memory Tip: “Gadi XUV in My Range.”

From gamma rays to radio waves:

• Wavelength increases.

• Frequency decreases.

• Photon energy decreases.

Photon energy is:

E = hc/lambda

The visible-light wavelength range is approximately 400 nm to 700 nm.

The Poynting vector is:

S = E cross B / mu0

Its average magnitude represents intensity:

Intensity = Average energy density × Wave speed

Electromagnetic radiation also carries momentum:

• Absorbed radiation: p = E/c

• Reflected radiation: p = 2E/c

Watch PW’s Alternating Current and Electromagnetic Waves for NEET 2026: Concepts, Formulas, and Tricks to revisit AC circuits, phase differences, resonance, transformers and electromagnetic wave properties. Review the important formulas and tricks, then practise questions to strengthen your understanding and identify topics that need further revision.

Worked-Out Questions on Alternating Current and Electromagnetic Waves

1. A sinusoidal current has a peak value of 10 A. Calculate its RMS value.

Answer: RMS current = Peak current / √2 = 10 / √2 = 7.07 A.

2. An alternating current is given by i = 100 sin(50πt + π/6). Find its frequency and time period.

Answer: Angular frequency = 50π rad/s. Frequency = 50π / 2π = 25 Hz. Time period = 1 / 25 = 0.04 s.

3. Calculate the inductive reactance of a 0.2 H inductor connected to a 50 Hz AC supply.

Answer: Inductive reactance = 2πfL = 2π × 50 × 0.2 = 62.83 Ω.

4. A 100 μF capacitor is connected to a 50 Hz AC supply. Calculate its capacitive reactance.

Answer: Capacitive reactance = 1 / (2πfC) = 1 / (2π × 50 × 100 × 10⁻⁶) = 31.83 Ω.

5. A series LCR circuit has resistance 3 Ω, inductive reactance 8 Ω and capacitive reactance 4 Ω. Find its impedance.

Answer: Impedance = √[R² + (XL − XC)²] = √[3² + (8 − 4)²] = √25 = 5 Ω.

6. Find the resonant frequency of a circuit with inductance 0.1 H and capacitance 100 μF.

Answer: Resonant frequency = 1 / (2π√LC) = 1 / (2π√(0.1 × 100 × 10⁻⁶)) = 50.33 Hz approximately.

7. An ideal transformer has 500 primary turns and 2,000 secondary turns. If the primary voltage is 100 V, find the secondary voltage.

Answer: Secondary voltage = Primary voltage × (Secondary turns / Primary turns) = 100 × (2,000 / 500) = 400 V. It is a step-up transformer.

8. A capacitor has a capacitance of 5 μF. If the voltage across it changes at 200 V/s, calculate the displacement current.

Answer: Displacement current = Capacitance × Rate of change of voltage = 5 × 10⁻⁶ × 200 = 10⁻³ A = 1 mA.

9. An electromagnetic wave travels in a vacuum with an electric field amplitude of 300 V/m. Calculate its magnetic field amplitude.

Answer: Magnetic field amplitude = Electric field amplitude / Speed of light = 300 / (3 × 10⁸) = 1 × 10⁻⁶ T.

10. Calculate the energy of a photon with wavelength 600 nm. Use Planck’s constant = 6.626 × 10⁻³⁴ J·s and speed of light = 3 × 10⁸ m/s.

Answer: Photon energy = hc / λ = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (600 × 10⁻⁹) = 3.31 × 10⁻¹⁹ J approximately.

Alternating Current and Electromagnetic Waves become easier to revise when you connect formulas with their physical meaning. Practise RMS and reactance calculations, understand resonance and transformer relationships, and revise electromagnetic wave properties and the spectrum. PW’s one-shot video can help you revisit important concepts, formulas, tricks and PYQs as part of your NEET Physics preparation.

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