Alternating current

NUST NET (Engineering) · Physics · Magnetism and alternating current. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

An alternating current changes size and direction all the time, so we quote its rms value: the steady current that would give the same heating. Mains voltages and meter readings are rms values.

Inductors and capacitors oppose AC without wasting energy. This opposition is called reactance and it depends on frequency: an inductor opposes high frequencies more, a capacitor opposes low frequencies more. In an inductor the current lags the voltage by 90°; in a capacitor it leads by 90°.

Resistance and reactance are out of phase, so they are combined like the sides of a right-angled triangle to give the impedance. At resonance the two reactances cancel and the current is greatest.

Rules to remember

Common mistake

Adding resistance and reactance directly. A 30 Ω resistor in series with a 40 Ω reactance gives 50 Ω, not 70 Ω, because they combine as √(R2 + X2).

Worked example

A series circuit has R = 6.0 Ω, XL = 12 Ω and XC = 4.0 Ω and is connected to a 50 V rms supply. What is the rms current?

  1. Net reactance: XL − XC = 12 − 4.0 = 8.0 Ω.
  2. Impedance: Z = √(6.02 + 8.02) = √(36 + 64) = √100 = 10 Ω.
  3. Current: Irms = Vrms/Z = 50/10 = 5.0 A.

Answer: 5.0 A

Practice questions

Try each one, then open the answer.

1. The frequency of the AC supply to a circuit is doubled. The reactance XL of an inductor and XC of a capacitor in it

  1. A
    XL halves and XC doubles
  2. B
    both double
  3. C
    XL doubles and XC halves
  4. D
    both are unchanged
Show answer

Answer: C. XL = 2πfL rises with frequency, while XC = 1/(2πfC) falls with it. So XL doubles and XC halves.

2. A series circuit contains an inductor of 0.10 H and a capacitor of 10 μF. Its resonant angular frequency ω0 is

  1. A
    100 rad s−1
  2. B
    159 rad s−1
  3. C
    1000 rad s−1
  4. D
    106 rad s−1
Show answer

Answer: C. ω0 = 1/√(LC) = 1/√(0.10 × 10 × 10−6) = 1/√(10−6) = 1000 rad s−1. 106 forgets the square root; 159 is f0 = ω0/2π, which is in hertz, not rad s−1.

3. A 30 Ω resistor and an inductor of reactance 40 Ω are connected in series to an AC supply. The impedance of the circuit is

  1. A
    70 Ω
  2. B
    10 Ω
  3. C
    35 Ω
  4. D
    50 Ω
Show answer

Answer: D. The voltages across R and L are 90° out of phase, so Z = √(R2 + XL2) = √(900 + 1600) = 50 Ω. Simply adding them (70 Ω) ignores the phase difference.

More questions on this topic

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