Alternating Current for NEET UG

The electricity in our homes reverses direction 50 times a second. Learn rms values, reactance, impedance, resonance and transformers.

PhysicsClass 1265 MCQs3 lessons13 flashcards3 memory tricks

Practise Alternating Current free

What you'll learn

  1. AC voltage and rms values
    Relate peak and rms values, and explain why rms is used.
  2. Reactance, impedance and phase
    Find the reactance of L and C, the impedance of an LCR circuit, and the phase difference.
  3. Resonance and transformers
    Find the resonant frequency of an LCR circuit and use the transformer equation.

Key formulas

AC voltagev = V₀ sin ωt
rms voltageV_rms = V₀/√2 ≈ 0.707V₀
rms currentI_rms = I₀/√2
Average power in a resistorP = V_rms I_rms
Impedance of series LCRZ = √(R² + (X_L − X_C)²)
Phase angletan φ = (X_L − X_C)/R
Power factorcos φ = R/Z
Average powerP = V_rms I_rms cos φ

Memory tricks

Phase in L and C circuitsELI the ICE man.

In an inductor (L), voltage E leads current I by 90°. In a capacitor (C), current I leads voltage E by 90°.

Resonance in an LCR circuitTuning the FM radio to your favourite station is LCR resonance.

At resonance X_L = X_C, impedance is at its minimum (just R) and current is at its maximum. f₀ = 1/(2π√LC).

Why power lines use high voltageStep up the voltage, step down the current, and less is lost on the way from the power station to your town.

Transmission power P = VI. Higher V means lower I, and the line loss I²R falls with the square of the current.

Try 3 NEET-style questions

Q1. A series circuit contains R = 30 Ω and a capacitor of reactance 40 Ω. Its impedance and power factor are

  1. 70 Ω and 0.43
  2. 10 Ω and 3
  3. 50 Ω and 0.8
  4. 50 Ω and 0.6
Show answer

Answer: (4) 50 Ω and 0.6

Z = √(R² + X_C²) = √(900 + 1600) = 50 Ω. Power factor cos φ = R/Z = 30/50 = 0.6.

Q2. The capacitive reactance of a 10 μF capacitor on a 50 Hz supply is about

  1. 3.18 Ω
  2. 318 Ω
  3. 3180 Ω
  4. 31.8 Ω
Show answer

Answer: (2) 318 Ω

X_C = 1/(2πfC) = 1/(2π × 50 × 10 × 10⁻⁶) = 1/(3.14 × 10⁻³) ≈ 318 Ω.

Q3. A choke coil is preferred over a resistor for controlling AC current because it

  1. stores charge
  2. dissipates negligible average power
  3. has zero reactance
  4. is cheaper
Show answer

Answer: (2) dissipates negligible average power

An ideal choke coil has reactance but no resistance, so it limits the current without dissipating average power as heat.

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