12 min readConcept sheet1-mark MCQ, 2-mark phase questions and 3-mark derivations of reactance with phasor diagrams; also a NEET favourite

AC through a resistor, an inductor and a capacitor

In a resistor current and voltage are in phase. In an inductor current lags by π/2; in a capacitor current leads by π/2. Reactances are ωL and 1/ωC.

PhasorA rotating arrow whose length is the peak value; it turns anticlockwise at ω, and its vertical projection gives the instantaneous value

Inductive reactanceX_L = ωL = 2πνL; unit ohm (Ω); rises with frequency

Capacitive reactanceX_C = 1/(ωC) = 1/(2πνC); unit ohm (Ω); falls as frequency rises

Average powerZero over a full cycle in a pure inductor or a pure capacitor

Phasors

A phasor is a rotating arrow that stands for an alternating quantity. Its length equals the peak value. It turns anticlockwise about the origin with angular speed ω. Its projection on the vertical axis gives the instantaneous value, such as vₘ sin ωt. When two quantities differ in phase, their phasors are drawn at an angle equal to the phase difference. The phasor ahead in the anticlockwise sense leads; the one behind lags. Phasors let us add alternating voltages as we add vectors.

Pure inductor: the current lags the voltage by π/2

1
Apply v = vₘ sin ωt across an inductor L with negligible resistance.
2
The self-induced emf is −L di/dt. By Kirchhoff's loop rule, v − L di/dt = 0, so di/dt = (vₘ/L) sin ωt.
3
Integrate: i = −(vₘ/ωL) cos ωt = (vₘ/ωL) sin(ωt − π/2).
4
Peak current iₘ = vₘ/(ωL) = vₘ/X_L, where X_L = ωL is the inductive reactance.
5★
So, the current lags the voltage by π/2 (a quarter cycle), and X_L = ωL = 2πνL.

★ key step

Pure capacitor: the current leads the voltage by π/2

1
Apply v = vₘ sin ωt across a capacitor C. The charge on it is q = Cv = Cvₘ sin ωt.
2
Current i = dq/dt = ωCvₘ cos ωt.
3
Write it as i = iₘ sin(ωt + π/2), with iₘ = ωCvₘ = vₘ/X_C.
4
Here X_C = 1/(ωC) is the capacitive reactance.
5★
So, the current leads the voltage by π/2, and X_C = 1/(ωC) = 1/(2πνC).

★ key step

R, L and C compared

Feature	Resistor R	Inductor L	Capacitor C
Opposition to current	Resistance R	Reactance X_L = ωL	Reactance X_C = 1/(ωC)
Phase of current	In phase with v	Lags v by π/2	Leads v by π/2
When frequency rises	No change	X_L rises (X_L ∝ ν)	X_C falls (X_C ∝ 1/ν)
For DC (ν = 0)	R	Zero: acts as a plain wire	Infinite: blocks DC
Average power over a cycle	VI (not zero)	Zero	Zero

Worked example: reactance and frequency

Problem

A (1/π) H inductor and, separately, a (100/π) μF capacitor are each connected to a 220 V, 50 Hz supply. Find the reactance and the rms current for each. What happens to each current if the frequency becomes 100 Hz?

Working

Inductor: X_L = 2πνL = 2π × 50 × (1/π) = 100 Ω.
I = V/X_L = 220 V ÷ 100 Ω = 2.2 A.
Capacitor: X_C = 1/(2πνC) = 1/(2π × 50 × (100/π) × 10⁻⁶) = 1/(10⁻²) = 100 Ω.
I = V/X_C = 220 V ÷ 100 Ω = 2.2 A.
At 100 Hz: X_L doubles to 200 Ω, so I = 1.1 A; X_C halves to 50 Ω, so I = 4.4 A.

Answer

At 50 Hz, both reactances are 100 Ω and both currents are 2.2 A. At 100 Hz, the inductor current falls to 1.1 A and the capacitor current rises to 4.4 A.

How to draw the phasor diagrams

For each element, draw a horizontal arrow for the voltage phasor and label it V.
Resistor: draw the current phasor I along V, in the same direction. Write 'in phase'.
Inductor: draw I a quarter turn clockwise from V (behind it). Mark the angle π/2 and write 'I lags V'.
Capacitor: draw I a quarter turn anticlockwise from V (ahead of it). Mark π/2 and write 'I leads V'.
Draw a curved arrow to show anticlockwise rotation at ω.
Below each diagram you may sketch the v and i sine curves, shifted by a quarter cycle for L and C.

In real life

In a home theatre speaker, a capacitor sends the high notes to the small tweeter (X_C is low at high frequency), and an inductor sends the bass to the big woofer (X_L is low at low frequency).

Exam angle

1 marks: “What is the phase difference between the current and the voltage in a pure capacitor?” π/2; the current leads the voltage.

2 marks: “Why does a capacitor allow AC to pass but block DC?” X_C = 1/(2πνC). For DC, ν = 0, so X_C is infinite and no steady current flows; for AC, X_C is finite.

3 marks: “An AC voltage v = vₘ sin ωt is applied to a pure inductor. Show that the current lags the voltage by π/2, define inductive reactance and draw the phasor diagram.” v = L di/dt; integrate to get i = iₘ sin(ωt − π/2), iₘ = vₘ/ωL; X_L = ωL; phasor I drawn π/2 behind V.

Remember it

ELI the ICE man: in L, the voltage E comes before I; in C, I comes before E.