JEE/NEET Physics · Alternating Current series · Part 3 of 5 · All parts →
- Series LCR: one current, three voltages with different phases
- Phasors: rotating arrows representing the waves — add them as vectors
- Impedance Z = √(R² + (X_L − X_C)²) — resistance and net reactance at right angles
- Voltage leads current by phase φ: tanφ = (X_L − X_C)/R
- VL and VC can EACH exceed the source voltage — they cancel each other
Put a resistor, coil and capacitor in series across AC and the coil’s and capacitor’s voltages fight each other — each individually dwarfing the supply, yet nearly cancelling. The phasor diagram turns this three-way tug into simple geometry. Part 3 of the Alternating Current series.
- One current, three voltages
- Phasors: waves as arrows
- The impedance triangle
- The phase angle
- Voltage magnification
- Solved examples
- Common mistakes
- This physics in your daily life
- Practice set
- Recap
One Current, Three Voltages
Series means ONE current everywhere. But the element voltages wear different phases: V_R in step, V_L leading the current by 90°, V_C lagging it by 90°. Adding them is vector addition, not arithmetic.
Phasors: Waves as Arrows
Represent each sinusoid as an arrow rotating at ω; its vertical projection is the instantaneous value. Now V_R, V_L, V_C are three fixed arrows (90° apart) — and their vector sum is the source voltage. Waves became geometry.
The Impedance Triangle
| Letter | What it means (plain words) | Value / unit |
|---|---|---|
| Z | impedance — total AC opposition | Ω |
| φ | phase angle between source V and current I | tanφ = (X_L−X_C)/R |
| V_R, V_L, V_C | element voltages (RMS or peak consistently) | V |
Voltage Magnification
Since V_L and V_C point oppositely, each can be enormous while their sum stays small: a series LCR at near-resonance may show 100 V across the capacitor from a 10 V source — voltage magnification = Q factor. Harmless in circuits, spectacular (and dangerous) on the grid.
Solved Examples
Z = √(30² + 40²) = 50 Ω — the 3-4-5 triangle again.
✔
Answer: 50 Ω
X = 30; Z = √(100+900) ≈ 31.6 Ω; I = 6.3 A.
tanφ = 30/10 = 3 → φ ≈ 71.6°, circuit inductive (current lags).
✔
Answer: 6.3 A, lagging 71.6°
I = 10/5 = 2 A; V_L = IX_L = 400 V — forty times the source, on each reactive element!
The two 400s cancel between L and C while R carries the modest 10 V: phasor magic.
✔
Answer: 400 V each
- Adding voltages arithmetically. V ≠ V_R + V_L + V_C in series AC — the phases forbid it; phasor-add.
- Z as plain sum. Z = R + X_L − X_C is WRONG: Pythagoras with the NET reactance is right.
- Sign-blind phase. X_L > X_C → inductive (lags); X_C > X_L → capacitive (leads): the net reactance’s sign sets the story.
- Panic at big V_L, V_C. They’re expected and cancel internally — check the phasor diagram before declaring an error.
This Physics in Your Daily Life
- Radio and TV tuners — LCR circuits where you vary C to tune: impedance minimum at your station’s frequency: channel selection by phasor cancellation.
- Induction heaters and wireless power — resonance magnifying voltages for efficient transfer: matched L and C doing more with less.
- Grid fault analysis — utilities model lines as series R-L (plus C) networks: protection relays compute phasors at microseconds’ notice.
- Audio equalizers — banks of LCR circuits boosting/cutting frequency bands: your bass and treble shaped by impedance triangles.
- Metal detectors — shifted resonance (target alters L) unbalances the phasors: treasure and security by geometry.
Two people on opposite swings of a merry-go-round: their height oscillations are sines, 90° apart. Freeze the carousel and each rider is an arrow — lengths are amplitudes, angles are phases. All AC algebra is frozen merry-go-rounds: add arrows, read heights later.
R = 30, X = 40: Z = 50 — a 3-4-5 triangle. Raise X to 300 with same R: Z ≈ 300.5 — reactance dominates, current shrinks to near-zero… unless X_L and X_C nearly cancel (resonance): then Z collapses back to R and current surges. The triangle’s shape IS the circuit’s character.
Draw the right triangle: base R (in phase), height X_L − X_C (perpendicular), hypotenuse Z. Beside it, the voltage triangle (V_R, V_L − V_C, V_source) — identical shape, scaled by I. One drawing solves both the current and the phase.
Practice set (answers hidden — try first)
(NEET-level) R=40, X=30: Z =
(JEE Main-level) R=8, X_L=6, X_C=12: Z =
(NEET-level) X_L = X_C: the circuit is
(Concept) Phasors are
(JEE Main-level) tanφ = 1 with R = 10: net X =
- series: one current, phased voltages
- phasor addition replaces arithmetic
- Z = √(R² + (X_L−X_C)²)
- tanφ = (X_L−X_C)/R
- V_L, V_C may each exceed the source
- 🔁 phasor concept
- 🔁 impedance triangle
- 🔁 phase formula
- 🧠 Chant: ‘R horizontal, net-X vertical, Z the hypotenuse’.
- 🧠 3-4-5 rule: ‘R=30, X=40 → Z=50’.
- 🏠 Daily: radio tuning = phasor cancellation at one f.
- 🏠 Daily: equalizers sculpt sound by impedance triangles.
Quick revision
- Series LCR: one current, three voltages with different phases
- Phasors: rotating arrows representing the waves — add them as vectors
- Impedance Z = √(R² + (X_L − X_C)²) — resistance and net reactance at right angles
- Voltage leads current by phase φ: tanφ = (X_L − X_C)/R
- VL and VC can EACH exceed the source voltage — they cancel each other
- One current, three voltages
Have a doubt on this topic?
