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CBSE Class 12 · Physics

Alternating Current

Official NCERT chapter from Physics Part I–II (book code leph1). ExamMaster notes are original teaching at CBSE Class 12 depth.

This lesson follows the official chapter “Alternating Current” in Physics Part I–II. The words below are ExamMaster’s teaching, not a paste from the book. Use the NCERT chapter for the classroom sequence; use these notes to hold the idea without copying exercises or figures.

  • CBSE Class 12
  • Medium level
  • 8 concepts

1AC Voltage Applied to a Resistor

AC on a resistor: I and V in phase as taught; I = V/R still. A resistor does not shift the sine. Power is VI average as a later heading. Resistor is the in-phase case.

Drawing a 90° shift on a lone R is a miss.

Figure. On a resistor, V and I rise and fall together. They share the mid-axis zeros; only the height differs.

How it works

  1. Name R and the AC VThe given.
  2. Write I=V/R, same phaseThe look.
  3. Keep no extra phase hereR only.

2Representation of AC Current and Voltage by

Phasor representation: a rotating arrow whose height is the instant value as taught. Phasors add like vectors for LCR. Representation is a picture of sines, not a new device.

A phasor as a battery-arrow is a miss.

Figure. A phasor is a rotating arrow frozen at one instant. For a resistor, V and I point the same way — they are in phase.

How it works

  1. Draw the taught rotating arrowThe phasor.
  2. Read the height as the instantThe use.
  3. Add phasors when several elements sitThe gift.

3AC Voltage Applied to an Inductor

AC on an inductor: V leads I by 90° as taught; XL = ωL. Inductor is a lag-of-I story. A larger ω or L raises XL.

Using R as XL is a miss.

Figure. On an inductor, I lags V by a quarter cycle. V hits its peak while I is still at zero.

How it works

  1. Write XL=ωLThe reactance.
  2. Say V leads I by π/2The phase.
  3. Keep ω=2πf if askedThe link.

4AC Voltage Applied to a Capacitor

AC on a capacitor: I leads V by 90° as taught; XC = 1/(ωC). Capacitor is a lead-of-I story. A larger C or ω lowers XC.

Using XL as XC is a swap.

Figure. On a capacitor, I leads V by a quarter cycle. I is already at its peak when V is still at zero.

How it works

  1. Write XC=1/(ωC)The reactance.
  2. Say I leads V by π/2The phase.
  3. Keep the inverse-ω lookThe trend.

5AC Voltage Applied to a Series LCR Circuit

Series LCR: Z = sqrt(R^2+(XL−XC)^2) as taught; tanφ from (XL−XC)/R. Resonance when XL=XC, Z=R, current largest. LCR is a triangle-of-phasors.

Adding R+XL+XC as Z is a miss.

Figure. Series LCR phasors: I and VR share one arm. VL is up, VC is down. V is the diagonal of VR and VL minus VC.

How it works

  1. Form XL, XC, then ZThe impedance.
  2. Read resonance as XL=XCThe special.
  3. Keep φ from the taught tanThe phase.

Resonance hint

XL=30 Ω, XC=30 Ω, R=10 Ω. Find Z.

  • XL−XC0
  • Z10 Ω
  • Readresonance — Z=R

Pro tip. When XL=XC, only R remains.

6Power in AC Circuit: The Power Factor

Power in AC: average P = Vrms Irms cosφ as taught. Power factor is cosφ. A pure L or C has cosφ=0 as framed — no average power. Power is the average, not the peak product.

Using Vpeak Ipeak as the household watt is a miss.

Figure. Average power is VI cos phi. At 30 degrees, cos 30 is 0.866, so most of the apparent power is real.

How it works

  1. Form Vrms Irms cosφAverage power.
  2. Keep cosφ as the factorThe extra.
  3. Refuse peak×peak as the billrms and φ.

7Transformers

Transformer: Vp/Vs = Np/Ns as taught (ideal); power in ≈ power out. A transformer is a mutual-induction pair for AC, not a DC trick. Step-up raises V and lowers I as framed.

A DC cell on the primary as a working transformer is a miss.

Figure. A step-up: twice the turns, twice the voltage. 80 to 160 turns, 110 V to 220 V. Flux crosses the core; the windings are boxes, not coils.

How it works

  1. Write the turn-ratio as the V-ratioThe ideal.
  2. Keep AC as the needChanging Φ.
  3. Read step-up as more V, less IPower-ish.

8A definition is a test you can run

AC-word is a test: phase on R/L/C, Z, power factor, or a transformer. If you only say “mains”, you have a heading.

A socket-sticker is not the test.

Figure. RMS is the DC value that heats the same resistor. For a sine, Vrms is V0 over root 2: 10 V peak is about 7.1 V rms.

How it works

  1. Name phase, Z, cosφ, or the ratioThe object.
  2. Give the school sentenceThe test.
  3. Then the word has contentThe definition ran.
At series resonance Z is
  1. R — XL cancels XC
  2. XL+XC
  3. 0 always

Only R remains.

Notes

  • Mapped to the official NCERT chapter “Alternating Current”. Original teaching only — no textbook sentences.
  • Science here is Physics, Chemistry and Biology ideas at this class, never a language or social-science chapter.

Formulas

  • XL=ωL
  • XC=1/(ωC)
  • Z=sqrt(R^2+(XL−XC)^2)
  • P=Vrms Irms cosφ
  • Vp/Vs=Np/Ns

Recap

Hold these pegs from the official chapter “Alternating Current”. The wording is ExamMaster’s teaching, not a textbook recap.

AC Voltage Applied to a Resistor
AC on a resistor: I and V in phase as taught; I = V/R still.
Representation of AC Current and Voltage by
Phasor representation: a rotating arrow whose height is the instant value as taught.
AC Voltage Applied to an Inductor
AC on an inductor: V leads I by 90° as taught; XL = ωL.
AC Voltage Applied to a Capacitor
AC on a capacitor: I leads V by 90° as taught; XC = 1/(ωC).
AC Voltage Applied to a Series LCR Circuit
Series LCR: Z = sqrt(R^2+(XL−XC)^2) as taught; tanφ from (XL−XC)/R.
Power in AC Circuit: The Power Factor
Power in AC: average P = Vrms Irms cosφ as taught.

Practise Alternating Current

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