Alternating Current

Alternating CurrentNEET Physics · Class 12 · NCERT Chapter 7

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7 interactive concept widgets for Alternating Current. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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RMS, peak and average AC values

AC through a resistor

AC through an inductor

AC through a capacitor

Series LCR impedance

Series LCR resonance

Transformer turns and current ratios

AC basics: RMS and elements

RMS / peak / average values, plus how V and I behave through R, L and C.

RMS / Peak

RMS, peak and average AC values

The three different averages used in AC circuits.

For a sinusoidal AC, peak V_0 is the maximum value. Mean over a full cycle is zero. RMS = V_0 / sqrt(2). Mean over a half-cycle = 2 V_0 / pi.

Peak V_0: 311 V

Indian household mains: V_rms = 220 V, so V_0 ≈ 311 V.

RMS

219.91 V

V_{rms} = V_{0} / 2

Avg (full cycle)

0 V

Avg (half cycle)

197.99 V

Try this

RMS / peak = 1/sqrt(2) ≈ 0.707 (only for sinusoidal). For square wave, RMS = peak.
When NEET says "220 V mains", that's RMS, not peak.
Heating effect of AC equals heating effect of DC at the same RMS value.

AC + R

AC through a resistor

V and I are in phase. Same shape, peak at the same time.

AC through a pure resistor: voltage and current are exactly in phase. No phase shift.

Peak V₀: 10 V

R: 5 Ω

Peak I₀ = V₀/R = 2.00 A

Phase: 0 (in phase)

V = V_{0} sin ω t, I = (V_{0} / R) sin ω t

● V (solid)

● I (dashed)

Try this

V and I peak at the same instant.
Power dissipated in R: P_avg = V_rms² / R = I_rms² R.
No reactance for a resistor; only resistance.

AC + L

AC through an inductor

V leads I by 90°. Reactance grows with frequency.

AC through an inductor: V leads I by π/2 (90°). Reactance X_L = ω L grows with frequency.

L: 50 mH

V₀: 10 V

Frequency f: 50 Hz

X_L = ωL = 15.71 Ω

I₀ = V₀/X_L = 0.637 A

X_{L} = ω L = 2 π f L

● V (leads)

● I (lags by π/2)

Try this

V peaks first; I peaks a quarter-cycle later.
At very high f: X_L is huge, very little I flows. Inductor blocks high frequencies.
At very low f: X_L is small. At DC (f = 0), inductor is just a wire.
No real average power dissipated by an ideal inductor.

AC + C

AC through a capacitor

I leads V by 90°. Reactance falls with frequency.

AC through a capacitor: I LEADS V by π/2 (90°). Reactance X_C = 1/(ω C) DROPS with frequency.

C: 10.00 µF

V₀: 10 V

Frequency f: 50 Hz

X_C = 1/(ωC) = 318.31 Ω

I₀ = V₀/X_C = 0.031 A

X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}

● V (lags)

● I (leads by π/2)

Try this

I peaks first; V peaks a quarter-cycle later.
At very high f: X_C is small. Capacitor passes high frequencies.
At DC (f = 0): X_C is infinite. Capacitor blocks DC.
Memory aid: ELI the ICE man. E (V) leads I in L; I leads E (V) in C.

LCR circuits and resonance

Combine R, L and C: impedance, phase angle and the special frequency where it all balances.

LCR

Series LCR impedance

Z combines R, X_L and X_C using a Pythagorean-like formula.

Series LCR: total opposition is the impedance Z. Phase angle phi between V and I depends on which reactance wins.

R: 10 Ω

L: 50 mH

C: 40.00 µF

f: 50 Hz

Impedance Z

64.65 Ω

Z = R^{2} + (X_{L} - X_{C})^{2}

X_L

15.71

X_C

79.58

cos φ

0.155

Phase φ = -81.1°. Capacitive (I leads V).

Try this

When X_L = X_C: Z = R (minimum), φ = 0. This is RESONANCE.
cos φ is the power factor; only this fraction of apparent power is real power dissipated.
High Q (low R relative to X_L = X_C) means a sharp resonance peak.

Resonance

Series LCR resonance

The frequency at which X_L = X_C; impedance is minimum.

At resonance, X_L = X_C. The reactive parts cancel; impedance equals R alone. Current is maximum at the resonance frequency.

L: 10.00 mH

C: 10.00 µF

R: 2 Ω

Resonance frequency

503.29 Hz

f_{0} = \frac{1}{2 π L C}

Quality factor Q

15.81

Q = \frac{1}{R} \frac{L}{C}

Try this

Smaller R: higher Q, sharper resonance peak. Larger R: broader peak.
Used in radio tuning: adjust C to match the broadcast frequency.
At resonance: current is V_rms / R, phase angle is zero, power factor is 1.

Transformer

How a transformer changes voltage and current using turns ratio.

Transformer

Transformer turns and current ratios

Voltage scales with turns; current scales with the inverse of turns (ideal).

Ideal transformer: turns ratio sets the voltage ratio. Power in = power out (no losses).

Primary V_p: 220 V

Primary turns N_p: 1000

Secondary turns N_s: 50

Primary I_p: 2 A

Secondary V_s

11.0 V

\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}

Secondary I_s (ideal)

40.00 A

Power: P_p = 440 W, P_s = 440 W

Step-DOWN transformer

Try this

Step-up: more secondary turns. Voltage rises, current drops.
Power transmission uses high-voltage AC because P loss = I² R; lower current means smaller losses.
Real transformers are 95-99% efficient; they lose energy to copper resistance, hysteresis and eddy currents.
A transformer cannot work on DC. It needs a CHANGING flux, which only AC provides.

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