CLASS 12 PHYSICS · CHAPTER 7 · STUDY NOTES
Alternating Current
An in-depth study of alternating current (AC) circuits, covering resistive, inductive, and capacitive elements, the phenomenon of resonance in series LCR circuits, power factors, and the principles of transformers.
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Introduction to Alternating Current
Watch the direction change, then distinguish a cycle from its phase.A current that changes direction
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i = 2 sin θ A. At 0°, i = 0.00 A; a negative value means the conventional current reverses.
Why this works and model limits
The animated carriers show slow net drift, not their full thermal motion or how quickly an electrical signal propagates. They oscillate locally under AC; they do not travel from the power station to your device in each cycle. Phase θ = ωt and ω = 2πf.
- Alternating current
- A current whose direction reverses with time. A sinusoid is one important waveform, rather than the definition of all AC.
For sinusoidal AC, v = v_peak sin(ωt), with ω = 2πf. Frequency f counts cycles per second in hertz; phase ωt is an angle in radians. The period is T = 1/f.
Ordinary transformers make changing AC voltage levels convenient. Raising transmission voltage lowers current for a given delivered power and reduces line heating I²R. This explains an advantage of AC systems; it is not a claim that modern DC transmission is impossible.
Let’s look a little closer.
Go deeper: Zero mean does not mean zero energy
A sinusoidal current has zero signed mean over a complete cycle. The positive and negative charge transport cancel, but a resistor heats during both directions because i² is positive.
The carrier sketch shows local drift only. Electrons also have random microscopic motion, and electrical signals do not propagate at the drift speed.
Take a moment. What do you think?
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Question
A sine wave repeats every 0.025 s. Find f and ω.
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AC Voltage Applied to a Resistor
Compare the current’s sign with the resistor’s heating.Heat does not reverse when current does
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For R = 6 Ω and v = 12 sin θ V, i = 2 sin θ A. Both cross zero and reach their peaks together.
Why this works and model limits
The RMS value is the square root of the average of the squared waveform. The √2 conversion applies to a sinusoid, not every possible alternating waveform. The equal areas under p(t) and the horizontal 12 W line give the same energy per cycle; the signed mean current is zero.
- Root mean square value
- The square root of the time average of a waveform’s square. For a sinusoid, I_rms = i_peak/√2 and V_rms = v_peak/√2.
For an ideal resistor, v = iR at every instant. A sine-wave voltage produces a current with the same phase: both reach zero and their positive and negative peaks together.
Instantaneous power is p = vi = i²R ≥ 0. Over one sinusoidal cycle, mean power P = V_rms I_rms = I_rms²R = V_rms²/R. RMS gives the DC value with the same average heating.
Go deeper: Why the factor is √2
Since sin²θ = (1−cos2θ)/2, its full-cycle mean is 1/2. Therefore ⟨i²⟩ = i_peak²/2 and I_rms = i_peak/√2.
The √2 conversion is specific to sinusoidal waves. A quoted AC rating normally gives an RMS value, but always check whether a question specifies peak or RMS. The signed mean of i is different from its RMS.
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Question
A 30 Ω resistor receives a sinusoidal 18 V rms supply. Find RMS current, peak current and mean power.
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Representation by Phasors
Use the arrows to compare timing, not physical directions.A rotating arrow draws a sine wave
Slow teaching playback · actual frequency is independent of playback speed.
In phase. The vertical projection of each rotating arrow gives its instantaneous value; the arrow length gives its peak.
Why this works and model limits
These are mathematical arrows, not spatial vectors for voltage or current. Each waveform is normalised by its own peak; the two traces do not have the same units. Phasors describe same-frequency sinusoidal steady state, not arbitrary switching transients.
- Phasor
- A mathematical rotating arrow whose length and phase represent a sinusoid’s amplitude and phase. Its projection gives an instantaneous value.
Each phasor rotates at the common angular frequency. Its vertical projection follows a sine wave. Arrows in the same direction describe signals in phase; a fixed angular separation describes a fixed phase difference.
For a pure resistor, current and voltage are in phase. For a pure ideal inductor, current lags voltage by 90°; for a pure ideal capacitor, current leads by 90°. Mixed circuits can have other phase angles.
Go deeper: What a phasor does not mean
Voltage and current are scalars, not spatial vectors. Their sinusoidal amplitudes and phases combine like mathematical vectors, which is why the picture is useful.
This method describes same-frequency sinusoidal steady state. It does not by itself describe switching transients or an arbitrary waveform. Keep peak phasors together or RMS phasors together when adding.
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Question
At θ = 0° the voltage sine wave crosses zero upward. What is the normalised current of a pure inductor and of a pure capacitor?
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AC Voltage Applied to an Inductor
Connect the changing current to a changing magnetic field.Current follows voltage a quarter-cycle later
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Ideal L = 0.10 H: voltage follows di/dt, not i. Magnetic energy U = ½Li² grows and then returns to the source; average power is zero.
Why this works and model limits
This is the sinusoidal steady-state solution with no DC current offset. A perfectly ideal inductor can retain a current offset set by its initial condition; it does not automatically decay without resistance. Practical coils also dissipate energy in winding resistance and core losses.
- Inductive reactance
- X_L = ωL, measured in ohms. In sinusoidal steady state a pure ideal inductor has I_rms = V_rms/X_L, with current 90° behind voltage.
Using the passive terminal convention, the inductor’s voltage drop is v_L = L di/dt. A sinusoidal voltage gives i = i_peak sin(ωt−π/2) for the sinusoidal solution with no DC offset. Current responds to accumulated voltage, so it lags by one quarter-cycle.
The ideal coil stores U = ½Li². Energy is taken in while |i| grows and returned while |i| falls, twice per cycle. Its average net power is zero; a real coil’s winding resistance and core losses make its net power nonzero.
Go deeper: Frequency and initial conditions
X_L grows with f. At fixed sinusoidal RMS voltage, increasing f reduces current. This opposition is reactive rather than irreversible heat production.
Integrating v_L = L di/dt allows a constant current offset fixed by initial conditions. A perfectly ideal inductor has no resistance to remove that offset. The standard 90° relation refers to the sinusoidal steady-state component.
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Question
An ideal 0.080 H coil is driven at 25 Hz by 10 V rms. Find X_L and I. What happens at 50 Hz?
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AC Voltage Applied to a Capacitor
Follow charge onto the plates without sending electrons across the gap.Current leads while the plates charge
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C = 100 µF, v = 12 sin θ V. At this phase, the left plate carries 0.00 mC; i = C dv/dt is largest when the plate charge crosses zero.
Why this works and model limits
Electrons accumulate on one plate and leave the other through the external wires. They do not cross the ideal insulating gap. The electric field stores U = ½Cv²; when |v| falls, stored energy returns. A steady DC source gives no continuing current after charging in a real circuit.
- Capacitive reactance
- X_C = 1/(ωC), measured in ohms. For a pure ideal capacitor in sinusoidal steady state, current leads voltage by 90°.
| Property | Resistor (R) | Inductor (L) | Capacitor (C) |
|---|---|---|---|
| Phase Relation | Current in phase | Current lags by π/2 | Current leads by π/2 |
| Opposition | Resistance (R) | Reactance Xₗ = ωL | Reactance Xc = 1/ωC |
| Avg. Power | IV | Zero | Zero |
With q = Cv, the external-wire current is i = dq/dt = C dv/dt. For v = v_peak sinωt this becomes i = ωC v_peak cosωt: a sine wave 90° ahead of voltage. Current is largest as voltage and plate charge cross zero.
X_C falls as frequency or capacitance rises. The capacitor stores electric energy U = ½Cv², and returns it when |v| falls. A pure ideal capacitor dissipates zero average power over a cycle.
Go deeper: AC current without a conducting bridge
Charge accumulates on one plate while it leaves the other through the external circuit. The changing field in the dielectric does not mean electrons traverse the ideal insulating gap.
After charging from a steady DC source, a real circuit’s current stops. There can be a transient during charging. Under AC the charge and field keep changing, so the external wires continue carrying alternating current.
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Question
A 50 µF ideal capacitor receives 8 V rms at 40 Hz. Find X_C and I, then double the frequency.
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Series LCR Circuit and Impedance
Use one current, three voltage phases and one resultant.Add voltages with their phases
At 50.0 Hz: Z = 20.00 Ω, I = 0.600 A rms. Current leads the source by 1.2°.
Why this works and model limits
The current is identical in all three series elements. This diagram uses current as the horizontal phase reference: V_R is parallel, V_L points +90°, V_C points −90°. δ = atan((X_L−X_C)/R) is the source voltage angle relative to current, so i(t) = i_peak sin(ωt−δ). RMS and peak phasors must not be mixed.
- Impedance
- For series R, L and C at angular frequency ω, Z = √[R² + (ωL−1/ωC)²]. RMS current is V_rms/Z.
All series elements carry the same instantaneous current. With current as the horizontal phasor reference, V_R lies along it, V_L is +90° and V_C is −90°. The source phasor is their vector sum, so V² = V_R² + (V_L−V_C)².
Define δ as the source-voltage angle relative to current: tanδ = (X_L−X_C)/R. For source v = v_peak sinωt, current is i = i_peak sin(ωt−δ). Positive δ means current lags; negative δ means it leads.
Go deeper: Why ordinary addition fails
The terminal loop equation v = v_R+v_L+v_C holds for instantaneous signed voltages. Their RMS magnitudes do not generally add arithmetically because their peaks occur at different times.
The sinusoidal solution is the steady-state response. Switching can add a transient that decays when resistance is present; a phasor triangle alone does not determine the complete initial-condition response.
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Question
A series circuit has R = 6 Ω, X_L = 12 Ω and X_C = 4 Ω, with a 10 V rms source. Find current and phase.
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Resonance in AC Circuits
Tune the frequency and compare the sharpness of the peak.Find the peak — then see why it happens
L = 0.10 H, C = 100 µF: f₀ = 50.33 Hz (ω₀ = 316.23 rad/s). At resonance X_L = X_C = 31.62 Ω, Z = R; with R = 20 Ω the 12 V rms source drives 0.600 A.
Why this works and model limits
At resonance, the non-zero inductor and capacitor voltages cancel as phasors. Their individual RMS values are 18.97 V in this model, each greater than the 12 V source. Lower R raises the current peak and narrows the response; Q = ω₀L/R is 1.58 for 20 Ω and 0.79 for 40 Ω. A lone ideal RL or RC circuit has no finite-frequency cancellation of these two reactances.
- Series resonance
- At ω₀ = 1/√(LC), X_L = X_C. With R > 0, series impedance is R and current is maximal for a fixed sinusoidal source voltage.
At resonance the opposite reactive voltage phasors cancel. Z reaches its minimum R and I_rms = V_rms/R. The individual L and C voltages need not be zero and can each exceed the source voltage.
The angular resonant frequency is ω₀ in rad/s. The cyclic frequency is f₀ = ω₀/(2π) in hertz. Changing C changes the frequency selected by a tuning circuit. A pure series RL or RC circuit has no finite-frequency L–C cancellation.
Go deeper: Damping and selectivity
For fixed L and C, reducing R raises and narrows the current-response peak. The dimensionless series quality factor is Q = ω₀L/R = 1/(ω₀CR).
This model has a nonzero resistance; the ideal R = 0 resonance limit is not a finite-amplitude steady response to a maintained ideal voltage source. The radio example illustrates frequency selection without predicting a real receiver’s full circuitry.
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Question
For L = 0.20 H and C = 20 µF, find ω₀ and f₀. If R = 25 Ω and the source is 15 V rms, find current at resonance.
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Power in AC Circuits and Power Factor
Distinguish average delivered power from momentary energy return.Delivered energy and returned energy
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P = VI cos δ. For V = 12 V rms and I = 2 A rms, average power is 24 W. Negative instantaneous power means stored energy is returning to the source.
Why this works and model limits
The product vi can be negative for part of a cycle even though a passive load has non-negative average power. A pure ideal reactive element takes energy and returns it, with zero net energy per cycle. The capacitor correction diagram keeps the load and delivered real power unchanged; it cancels the lagging reactive current drawn from the supply. Choosing an excessive capacitance would produce a leading power factor.
- Power factor
- For sinusoidal voltage and current with phase difference δ, P_average = V_rms I_rms cosδ. The power factor is cosδ.
The instantaneous power is p = vi and can become negative when a reactive element returns stored energy. Over a full cycle a passive series RLC circuit dissipates P = I²R = VI cosδ. A pure ideal L or C returns everything it takes, giving zero average power.
A lower power factor requires more supply current for the same real power and voltage. That increases resistive line loss. A suitable capacitor in parallel with a lagging load supplies leading reactive current, reducing the reactive current required from the source.
Go deeper: Keep the load unchanged during correction
Resolve load current into an in-phase component and a quadrature component. The in-phase component supplies real power; the quadrature component exchanges energy without net delivery over a cycle.
The capacitor correction model cancels the lagging quadrature current at the design frequency. It leaves real load power unchanged. Too much correction gives a leading power factor; it does not produce free energy.
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Question
A 100 V rms supply delivers 300 W to a sinusoidal load at power factor 0.60. Find supply current. What would it be with full reactive-current compensation?
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Transformers
Follow magnetic coupling between two electrically separate coils.More voltage does not mean more power
Ideal model: N_p = 200, N_s = 400. A 24 V rms primary supplies 48 W to a resistive secondary load: V_s = 48.0 V, I_s = 1.00 A. Input current is 2 A. At phase 0°, flux / maximum = 1.00, instantaneous primary voltage = 0.00 V and secondary voltage = 0.00 V.
Why this works and model limits
The turns ratio assumes common changing flux, negligible winding resistance and negligible leakage. The current ratio shown assumes an ideal loaded transformer with magnetising current neglected and a resistive load. A real transformer needs excitation current even with its secondary open. The two circuits have no conducting connection; energy couples magnetically through the core. Coil counts are schematic, with their true turn numbers labelled.
- Ideal transformer ratio
- With the same changing flux linking both coils, V_s/V_p = N_s/N_p. For an ideal loaded transformer neglecting excitation current, I_s/I_p = N_p/N_s.
An alternating primary supply creates changing core flux that links the secondary. Each coil’s induced emf is proportional to its turn count, giving the ideal voltage ratio. More secondary turns steps voltage up; fewer steps it down.
For an ideal transformer and a resistive load, V_p I_p = V_s I_s. Raising secondary voltage reduces secondary current in the same proportion. After a steady-DC switching transient, unchanged flux gives no continuing secondary emf.
Go deeper: Where the ideal approximation ends
The ideal ratios assume negligible winding resistance and leakage, common flux and magnetising current neglected. A real transformer takes excitation current with its secondary open; loading changes primary current rather than enforcing one universal input current.
Flux leakage reduces coupling. Copper resistance produces I²R heat. Insulated laminations reduce eddy-current loops in the core, while low-hysteresis core material reduces energy lost during repeated magnetisation reversal. Thicker wire helps high-current windings.
For a given delivered power, stepping transmission voltage up reduces current and I²R line losses. Step-down stages then supply lower voltages near consumers. Do not connect an ordinary transformer primary to DC to test this sketch; overheating can result.
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Question
An ideal transformer has 300 primary turns and 75 secondary turns. With 40 V rms primary voltage, a resistive secondary load draws 2 A. Find secondary voltage and primary current.
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