12 min readConcept sheetSection B short answer (3 marks): show that the current lags (or leads) the voltage by 90°; very short answer (2 marks): rms numericals; Section A MCQ (1 mark) on reactance and frequency

Alternating current: peak and rms values, phasors, and AC through R, L and C

AC varies as V = V_m cos ωt and V_rms = V_m/√2. A resistor keeps I in phase; an inductor makes it lag by 90°; a capacitor makes it lead.

rms valuesV_rms = V_m/√2 ≈ 0.707 V_m and I_rms = I_m/√2; AC meters read rms values

Inductive reactanceX_L = ωL = 2πνL; the current lags the voltage by π/2 (90°)

Capacitive reactanceX_C = 1/ωC = 1/2πνC; the current leads the voltage by π/2 (90°)

Indian household supply230 V rms at 50 Hz, so the peak is about 325 V (older books, including your NIOS lesson, give 220 V and 310 V)

AC, its peak value and its rms value

A **direct current** (DC), as from a cell, flows one way at a steady value. An **alternating current** (AC) changes its size all the time and reverses its direction regularly.

1
An AC voltage is written V = V_m cos ωt, and its current I = I_m cos ωt. V_m and I_m are the peak values; ω = 2πν is the angular frequency, and ν is the frequency.
2
Over a whole cycle, the positive and negative halves cancel, so the average current is zero.
3
But the heat I²R is never negative. Over a cycle the average of I² is I_m²/2, so the average power is P = I_m²R/2.
4
The rms (root mean square) value is the steady current that gives the same heat: I_rms²R = I_m²R/2.
5★
So, I_rms = I_m/√2 ≈ 0.707 I_m and V_rms = V_m/√2. AC ammeters and voltmeters show rms values.
6
The standard Indian household supply is 230 V rms at 50 Hz. Its peak is √2 × 230 ≈ 325 V, which is why a shock from the mains is so dangerous. Your NIOS lesson, like many older books, gives 220 V (peak about 310 V); the standard in India is now 230 V.

★ key step

Phasors, and how to draw them

A phasor is an arrow that stands for an alternating quantity. Its length is the peak value, and it turns anticlockwise at an angular speed ω.
Two quantities that are out of step are drawn as two arrows with the phase difference between them. The arrow ahead in the anticlockwise sense leads; the one behind lags.
Resistor: draw the current arrow I along the voltage arrow V. Write 'in phase'.
Inductor: draw I a quarter turn (90°) behind V, that is, clockwise from it. Write 'I lags V by 90°'.
Capacitor: draw I a quarter turn (90°) ahead of V, that is, anticlockwise from it. Write 'I leads V by 90°'.
Phasors let us add alternating voltages like vectors, as in the series LCR circuit.

Pure inductor: the current lags the voltage by 90°

1
Apply V = V_m cos ωt across an inductor L of negligible resistance.
2
The applied voltage balances the back emf: L dI/dt = V_m cos ωt.
3
Integrate: I = (V_m/ωL) sin ωt = (V_m/ωL) cos(ωt − π/2).
4★
Peak current I_m = V_m/ωL = V_m/X_L, where X_L = ωL = 2πνL is the inductive reactance, in ohm.
5
So, the current lags the voltage by π/2. X_L rises with frequency; for DC (ν = 0), X_L = 0.
6
The average power over a cycle is zero: energy is stored in the magnetic field and then given back.

★ key step

Pure capacitor: the current leads the voltage by 90°

1
Apply V = V_m cos ωt across a capacitor C. Its charge is q = CV = CV_m cos ωt.
2
Current I = dq/dt = −ωCV_m sin ωt = ωCV_m cos(ωt + π/2).
3★
Peak current I_m = ωCV_m = V_m/X_C, where X_C = 1/ωC = 1/2πνC is the capacitive reactance, in ohm.
4
So, the current leads the voltage by π/2. X_C falls as the frequency rises; for DC, X_C is infinite, so a capacitor blocks DC.
5
The average power over a cycle is zero: energy is stored in the electric field and then given back.

★ key step

R, L and C in an AC circuit compared

Feature	Resistor R	Inductor L	Capacitor C
Opposition to current	Resistance R	Reactance X_L = ωL	Reactance X_C = 1/ωC
Phase of I compared with V	In phase	Lags by 90°	Leads by 90°
At low frequency	R	Small (zero for DC)	Large (infinite for DC)
At high frequency	R	Large	Small
Average power	V_rms I_rms	Zero	Zero
rms current	V_rms/R	V_rms/X_L	V_rms/X_C

Worked example: a kettle, an inductor and a capacitor on the mains

Problem

Each of these is joined in turn to the 230 V, 50 Hz mains: (a) a kettle of resistance 46 Ω, (b) an inductor of 0.50 H with negligible resistance, (c) a 10 μF capacitor. Find the rms current in each. For (a), also find the peak current and the power. (π = 3.14, √2 = 1.414)

Working

(a) I_rms = V_rms/R = 230 V ÷ 46 Ω = 5.0 A. Peak I_m = √2 × 5.0 A = 7.07 A. Power P = V_rms I_rms = 230 × 5.0 W = 1150 W.
(b) X_L = 2πνL = 2 × 3.14 × 50 × 0.50 Ω = 157 Ω.
I_rms = 230 V ÷ 157 Ω ≈ 1.5 A, lagging the voltage by 90°.
(c) X_C = 1/(2πνC) = 1/(2 × 3.14 × 50 × 10 × 10⁻⁶) Ω = 1/(3.14 × 10⁻³) Ω ≈ 318 Ω.
I_rms = 230 V ÷ 318 Ω ≈ 0.72 A, leading the voltage by 90°.
If the frequency were doubled, X_L would double and X_C would halve.

Answer

(a) 5.0 A rms, about 7.1 A peak, 1150 W; (b) about 1.5 A; (c) about 0.72 A

In real life

Your home's 230 V, 50 Hz supply is AC because transformers can raise and lower AC voltages easily. The capacitor in a ceiling fan helps start its motor. It makes the current in one winding lead the current in the other, which sets up a turning magnetic field.

Exam angle

3 marks: “Show that in a pure inductor joined to an AC source, the current lags the voltage by π/2. How does the inductive reactance change with frequency?” L dI/dt = V_m cos ωt; I = (V_m/ωL) cos(ωt − π/2), so a lag of π/2; X_L = ωL = 2πνL rises in proportion to the frequency and is zero for DC.

2 marks: “An AC voltmeter reads 230 V across the mains. What is the peak voltage? Why do we use rms values?” V_m = √2 × 230 ≈ 325 V. The average of AC over a cycle is zero, but the rms value is the steady value that gives the same heat, so DC formulas can be used.

1 marks: “What is the average power drawn by a pure capacitor in an AC circuit?” Zero.

Remember it

ELI the ICE man: in an inductor (L), the voltage E comes before the current I; in a capacitor (C), the current I comes before the voltage E.