Unit 4
Electricity and Magnetism
Class 12 Physics • Chapter 19
Alternating Currents
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NEB/CDC syllabus scope:
Chapter 19 is a 6-teaching-hour Electricity and Magnetism chapter. It covers peak and RMS values of AC current and voltage; AC through a resistor, capacitor and inductor; phasor diagrams; series circuits containing resistance, capacitance and inductance; series resonance and quality factor; and power in AC circuits including power factor.
1. Introduction to Alternating Current
Alternating Current (AC)
An alternating current is an electric current whose magnitude changes continuously with time and whose direction reverses periodically.
An alternating voltage similarly changes magnitude and polarity with time. The most important practical AC waveform is the sinusoidal waveform.
| Direct Current (DC) | Alternating Current (AC) |
|---|---|
| Flows in one direction | Reverses direction periodically |
| Ideal DC magnitude is constant | Magnitude usually changes with time |
| Battery is a common source | Alternator / power generator is a common source |
| Frequency of ideal DC = 0 | AC has non-zero frequency |
Diagram 1: Direct current and sinusoidal alternating current
2. Sinusoidal Alternating Current and Voltage
A sinusoidal AC current may be written as:
i = I₀ sinωt
Similarly, sinusoidal voltage:
v = V₀ sinωt
where:
- I₀ = peak/current amplitude
- V₀ = peak voltage
- ω = angular frequency
- f = frequency
- T = time period
ω = 2πf
f = 1/T
Diagram 2: Peak value and time period of sinusoidal AC
3. Peak and RMS Values of AC
3.1 Peak Value
The maximum magnitude reached by alternating current or voltage is called its peak value or amplitude.
3.2 RMS Value
RMS value
The RMS value of an alternating current is the value of steady DC that would produce the same average heating effect in a resistor.
3.3 Derivation of RMS Current
For i = I₀ sinωt:
i² = I₀² sin²ωtThe mean value of sin²ωt over one complete cycle is 1/2:
⟨i²⟩ = I₀²/2Therefore:
Irms = I₀/√2 ≈ 0.707I₀
Vrms = V₀/√2 ≈ 0.707V₀
Household AC rating
An AC supply voltage is normally specified by its RMS value, not its peak value.
3.4 Average Value over a Half-Cycle
Although the average of a sinusoidal current over a complete cycle is zero, the average magnitude over one half-cycle is:
Iavg = 2I₀/π ≈ 0.637I₀
4. AC Through a Pure Resistor
For a resistor R connected to v = V₀ sinωt:
i = v/R = (V₀/R)sinωtTherefore:
i = I₀ sinωt
I₀ = V₀/R
Voltage and current reach maximum, zero and minimum values at the same instant.
Pure resistor phase relation
Current and voltage are in phase: φ = 0°.
Diagram 3: Resistor — voltage and current are in phase
5. AC Through a Pure Inductor
For a pure inductance L:
v = L(di/dt)If v = V₀ sinωt, integration gives a current that lags the voltage by 90°:
i = I₀ sin(ωt − π/2)
5.1 Inductive Reactance
Inductive reactance
The opposition offered by an inductor to alternating current.
XL = ωL = 2πfL
Irms = Vrms/XL
Pure inductor memory line
Current lags voltage by 90°. As frequency increases, XL increases.
Diagram 4: Inductor waveform and phasor relation
6. AC Through a Pure Capacitor
For a capacitor C:
q = Cv i = dq/dt = C(dv/dt)If v = V₀ sinωt:
i = I₀ sin(ωt + π/2)
6.1 Capacitive Reactance
Capacitive reactance
The opposition offered by a capacitor to alternating current.
XC = 1/(ωC) = 1/(2πfC)
Irms = Vrms/XC
Pure capacitor memory line
Current leads voltage by 90°. As frequency increases, XC decreases.
Diagram 5: Capacitor waveform and phasor relation
7. Phasor Diagrams
Phasor
A rotating vector representation used to show the magnitude and phase relationship of sinusoidal AC quantities having the same frequency.
| Element | Phase relation | Reactance/Resistance |
|---|---|---|
| Resistor R | V and I in phase | R |
| Inductor L | V leads I by 90° | XL = ωL |
| Capacitor C | I leads V by 90° | XC = 1/ωC |
Diagram 6: Basic R, L and C voltage phasors
8. Series RL Circuit
In a series RL circuit, the same current flows through R and L.
- VR = IR is in phase with I.
- VL = IXL leads I by 90°.
The total voltage is the vector sum:
V = √(VR² + VL²)
Z = √(R² + XL²)
tanφ = XL/R
The current lags the supply voltage by φ.
Diagram 7: RL voltage triangle and phase angle
9. Series RC Circuit
In a series RC circuit:
- VR = IR is in phase with I.
- VC = IXC lags I by 90°.
V = √(VR² + VC²)
Z = √(R² + XC²)
tanφ = −XC/R
The current leads the supply voltage.
10. Series RLC Circuit
In a series RLC circuit, current is common to all elements.
- VR = IR
- VL = IXL
- VC = IXC
Because VL and VC are 180° opposite in the phasor diagram, the net reactive voltage is VL − VC.
V = √[VR² + (VL − VC)²]
Diagram 8: General series RLC voltage phasors
11. Impedance and Phase Angle of a Series RLC Circuit
Impedance, Z
The total opposition offered by an AC circuit, combining resistance and reactance.
Z = √[R² + (XL − XC)²]
I = V/Z
11.1 Phase Angle
tanφ = (XL − XC)/R
| Condition | Nature | Phase relation |
|---|---|---|
| XL > XC | Net inductive | Current lags voltage |
| XL < XC | Net capacitive | Current leads voltage |
| XL = XC | Resonance | Current and voltage in phase |
Diagram 9: Impedance triangle for a net-inductive RLC circuit
12. Series Resonance in an RLC Circuit
Series resonance
The condition in a series RLC circuit at which inductive reactance equals capacitive reactance.
XL = XC
Thus:
ω₀L = 1/(ω₀C)
ω₀ = 1/√(LC)
f₀ = 1/(2π√LC)
12.1 Conditions at Resonance
- XL = XC.
- Net reactance = 0.
- Z = R, the minimum possible series impedance.
- Current is maximum: I = V/R.
- φ = 0.
- Voltage and current are in phase.
- Power factor = 1.
- Average power is maximum for fixed applied V and R.
Diagram 10: Resonance current peak
13. Quality Factor of a Series Resonant Circuit
Quality factor, Q
A dimensionless measure of the sharpness/selectivity of resonance in a resonant circuit.
For a series RLC circuit:
Q = ω₀L/R
Q = 1/(ω₀CR)
In terms of resonance frequency and bandwidth:
Q = f₀/Δf
where Δf = f₂ − f₁ is the bandwidth between half-power frequencies.
Interpretation
Higher Q means a narrower, sharper resonance curve and greater frequency selectivity.
Diagram 11: Higher Q gives sharper resonance
14. Power in AC Circuits
Let:
v = V₀ sinωt i = I₀ sin(ωt − φ)Instantaneous power:
p = viOn averaging over one complete cycle:
Pavg = VrmsIrmscosφ
14.1 Pure Resistor
φ = 0, so cosφ = 1:
P = VI14.2 Pure Inductor or Pure Capacitor
|φ| = 90°, so cosφ = 0:
Pavg = 0Ideal inductors and capacitors alternately store and return energy; they do not consume net average power over a complete cycle.
Diagram 12: Real power in a purely resistive AC load
15. Power Factor
Power factor
The cosine of the phase angle between voltage and current.
Power factor = cosφ = R/Z
Thus:
P = VI cosφ
15.1 Significance
- Unity power factor means voltage and current are in phase.
- Low power factor requires a larger current to deliver the same real power at the same voltage.
- Higher current causes greater I²R losses in conductors.
- Inductive loads generally have lagging power factor.
- Capacitive loads generally have leading power factor.
At series resonance
φ = 0 and cosφ = 1; therefore the power factor is unity.
16. Choke Coil — Useful AC Application
A choke coil is an inductor designed to provide appreciable inductive reactance in an AC circuit while having comparatively low resistance.
XL = 2πfL
It can limit AC current with less real power loss than an equivalent purely resistive current-limiting element, because an ideal inductor has zero average power consumption.
Real devices
A practical choke has winding resistance and core losses, so its power consumption is not exactly zero.
17. Worked Numericals
Example 1: Peak from RMS Voltage
An AC supply is 220 V RMS. Find its peak voltage.
V₀ = √2 Vrms V₀ = 1.414 × 220 ≈ 311 V
Example 2: RMS Current
Peak current I₀ = 8 A.
Irms = 8/√2 ≈ 5.66 A
Example 3: Inductive Reactance
L = 0.20 H and f = 50 Hz.
XL = 2πfL XL = 2π × 50 × 0.20 ≈ 62.8 Ω
Example 4: Capacitive Reactance
C = 20 μF and f = 50 Hz.
XC = 1/(2πfC) XC = 1/[2π × 50 × 20 × 10⁻⁶] ≈ 159 Ω
Example 5: Series RLC Impedance
R = 40 Ω, XL = 70 Ω and XC = 40 Ω.
Z = √[R² + (XL − XC)²] Z = √(40² + 30²) = 50 Ω
Example 6: Resonance Frequency
L = 0.10 H and C = 100 μF.
f₀ = 1/(2π√LC) f₀ = 1/[2π√(0.10 × 100 × 10⁻⁶)] ≈ 50.3 Hz
Example 7: Power Factor and Real Power
V = 230 V, I = 5 A, cosφ = 0.8.
P = VIcosφ P = 230 × 5 × 0.8 = 920 W
Example 8: Quality Factor
A resonant circuit has f₀ = 1000 Hz and bandwidth Δf = 50 Hz.
Q = f₀/Δf = 1000/50 = 2018. Complete Formula Sheet
| Topic | Formula |
|---|---|
| Sinusoidal current | i = I₀ sinωt |
| Sinusoidal voltage | v = V₀ sinωt |
| Angular frequency | ω = 2πf |
| Frequency and period | f = 1/T |
| RMS current | Irms = I₀/√2 |
| RMS voltage | Vrms = V₀/√2 |
| Half-cycle average current | Iavg = 2I₀/π |
| Inductive reactance | XL = ωL = 2πfL |
| Capacitive reactance | XC = 1/(ωC) = 1/(2πfC) |
| RL impedance | Z = √(R² + XL²) |
| RC impedance | Z = √(R² + XC²) |
| RLC impedance | Z = √[R² + (XL − XC)²] |
| Series RLC phase | tanφ = (XL − XC)/R |
| Resonance condition | XL = XC |
| Resonant angular frequency | ω₀ = 1/√LC |
| Resonant frequency | f₀ = 1/(2π√LC) |
| Quality factor | Q = ω₀L/R = 1/(ω₀CR) |
| Q from bandwidth | Q = f₀/Δf |
| Average AC power | P = VrmsIrmscosφ |
| Power factor | cosφ = R/Z |
19. Common Exam Mistakes
- Confusing peak and RMS values. For a sine wave, RMS is peak/√2.
- Writing the average of sinusoidal AC over a complete cycle as a positive number. It is zero.
- Using Irms = I₀/2 instead of I₀/√2.
- Confusing frequency f with angular frequency ω. Remember ω = 2πf.
- Writing XL = 1/ωL. Correct: XL = ωL.
- Writing XC = ωC. Correct: XC = 1/ωC.
- Forgetting units of reactance and impedance: ohm (Ω).
- Reversing phase rules. Inductor: current lags; capacitor: current leads.
- Adding VR, VL and VC arithmetically in an RLC AC circuit. They are phasors.
- Writing Z = R + XL − XC. Correct: Z = √[R² + (XL − XC)²].
- Forgetting the sign of XL − XC when deciding whether current leads or lags.
- Writing the resonance condition as XL + XC = 0. In magnitudes, resonance is XL = XC.
- At series resonance, writing maximum impedance. It is minimum: Z = R.
- At series resonance, writing minimum current. Current is maximum.
- Forgetting that power factor at resonance is unity.
- Writing AC power as VI in every circuit. Correct general formula: P = VIcosφ using RMS values.
- Writing non-zero average power for an ideal pure inductor or capacitor.
- Confusing Q factor with power factor.
- Using microfarad directly without converting μF to F in SI calculations.
- Using peak voltage with RMS current in the average power formula.
20. Important Exam Questions
Short-Answer Questions
- Define alternating current.
- Write the mathematical expression for sinusoidal AC.
- Define peak value and RMS value.
- Derive Irms = I₀/√2.
- What is the average value of sinusoidal AC over a complete cycle?
- State the phase relation between current and voltage in a pure resistor.
- Define inductive reactance and write its formula.
- How does XL vary with frequency?
- State the phase relation in a pure inductor.
- Define capacitive reactance and write its formula.
- How does XC vary with frequency?
- State the phase relation in a pure capacitor.
- What is a phasor?
- Define impedance.
- Write the impedance of series RL, RC and RLC circuits.
- Define series resonance.
- State the condition for series resonance.
- Derive the resonant frequency formula.
- What happens to impedance and current at resonance?
- Define quality factor.
- What is bandwidth?
- Define power factor.
- Write the average AC power formula.
- Why does an ideal inductor consume zero average power?
- What is a choke coil?
Long Questions / Derivations
- Derive the RMS value of sinusoidal alternating current.
- Discuss AC through a pure resistor with waveform and phasor diagram.
- Derive the expression for current and reactance of a pure inductor.
- Derive the expression for current and reactance of a pure capacitor.
- Explain phasor diagrams of RL and RC series circuits.
- Derive the impedance of a series RLC circuit.
- Derive tanφ = (XL − XC)/R.
- Explain series resonance and derive f₀ = 1/(2π√LC).
- Explain quality factor and resonance sharpness.
- Derive the average power in an AC circuit.
- Explain power factor and its significance.
Numerical Practice
- Convert peak current/voltage to RMS values and vice versa.
- Calculate XL from f and L.
- Calculate XC from f and C.
- Calculate impedance and current of RL/RC/RLC circuits.
- Calculate phase angle from R, XL and XC.
- Find resonance frequency from L and C.
- Find L or C from a given resonance condition.
- Calculate Q factor from circuit parameters.
- Calculate Q from resonance frequency and bandwidth.
- Calculate average power and power factor.
Exam Strategy
Master these six anchors: RMS derivation → R/L/C phase rules → RLC impedance → resonance → Q factor → P = VIcosφ. Most Chapter 19 derivations and numericals are combinations of these ideas.
21. One-Minute Revision
- Chapter 19: Alternating Currents — Electricity and Magnetism.
- Sinusoidal current: i = I₀sinωt.
- ω = 2πf and f = 1/T.
- Irms = I₀/√2.
- Vrms = V₀/√2.
- Average sinusoidal AC over a full cycle = 0.
- Resistor: V and I are in phase.
- Inductor: current lags voltage by 90°.
- Capacitor: current leads voltage by 90°.
- XL = 2πfL; increases with f.
- XC = 1/(2πfC); decreases with f.
- RL: Z = √(R² + XL²).
- RC: Z = √(R² + XC²).
- RLC: Z = √[R² + (XL − XC)²].
- tanφ = (XL − XC)/R.
- If XL > XC, current lags.
- If XL < XC, current leads.
- Resonance: XL = XC.
- At resonance: Z = R and current is maximum.
- f₀ = 1/(2π√LC).
- At resonance φ = 0 and power factor = 1.
- Q = ω₀L/R.
- Q = f₀/Δf.
- High Q means sharp resonance.
- Average AC power = VrmsIrmscosφ.
- Power factor = cosφ = R/Z.
- Pure ideal L or C consumes zero average real power.
22. Diagram Practice
Students should practice these labelled diagrams for the NEB examination:
- DC vs sinusoidal AC.
- Sinusoidal waveform showing peak and period.
- Pure resistor waveform/phasor.
- Pure inductor waveform/phasor.
- Pure capacitor waveform/phasor.
- Basic R, L and C phasor relations.
- RL phasor triangle.
- Series RLC phasor diagram.
- Impedance triangle.
- Series-resonance current curve.
- High-Q vs low-Q resonance curves.
- Power behavior in a resistive AC circuit.
Source handling: The original Nepal eNotes PDF remains embedded above using the verified Google Drive file. The source page identifies this resource as Unit 4, Electricity and Magnetism, Chapter 19 – Alternating Currents. The typed section follows the current NEB/CDC Chapter 19 syllabus and is designed as a searchable, responsive study companion. Where the PDF viewer does not expose page text, the typed section is a syllabus-aligned reconstruction and is not claimed to be a word-for-word transcription.
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