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CraftExam
medium difficulty10 min

AC Through Pure Inductor and Pure Capacitor

In a pure inductor, current lags voltage by π/2 and opposition to AC is inductive reactance XL = ω L. In a pure capacitor, current leads voltage by π/2 and opposition to AC is capacitive reactance XC = 1/(ω C).

In one line: ELI the ICE: in an inductor, E leads I; in a capacitor, I leads E.

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Understand it

  • An inductor opposes change in current through induced emf, so current builds up after the applied voltage changes.
  • This produces a phase lag of π/2 for current.
  • A capacitor charges and discharges continuously in AC; current is maximum when the rate of change of voltage is maximum, so current leads voltage by π/2.
  • Reactance is measured in ohm, like resistance, but it depends on frequency.
  • XL increases with frequency, while XC decreases with frequency.

Example

At higher frequency, an inductor offers more opposition because XL = ω L increases. At higher frequency, a capacitor offers less opposition because XC = 1/(ω C) decreases.

Formula and rule

Inductor: XL = ω L, where XL is in ohm, ω is angular frequency in rad s−1, and L is inductance in henry.

Capacitor: XC = 1/(ω C), where XC is in ohm and C is capacitance in farad.

Current in pure L lags voltage by π/2; current in pure C leads voltage by π/2.

These ideal relations assume negligible resistance.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

An inductor of 0.5 H is connected to AC of angular frequency 200 rad s−1.

  1. 1XL = ω L = 200 × 0.5 = 100 ohm.
  2. 2If Vrms = 50 V, Irms = Vrms/XL = 50/100 = 0.5 A.
  3. 3The current lags voltage by π/2.
  4. 4For a capacitor of 100 microfarad at the same ω, XC = 1/(200 × 100 × 10−6) = 50 ohm, so with 50 V RMS, Irms = 1 A and current leads voltage by π/2.

Method you can reuse

  1. 1Identify whether the element is L or C.
  2. 2Use XL = ω L for an inductor or XC = 1/(ω C) for a capacitor.
  3. 3Use Irms = Vrms/X for current.
  4. 4Then add the correct phase statement: L means current lags; C means current leads.

Sketch it in your answer

Draw separate AC source-L and AC source-C circuits. In the phasor diagram for L, V is ahead of I by 90 degrees. In the phasor diagram for C, I is ahead of V by 90 degrees. Graphs should show one sine wave shifted by a quarter cycle.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

How does increasing frequency affect XL and XC?

Think first, then show answer
Increasing frequency increases inductive reactance XL because XL = ω L, but decreases capacitive reactance XC because XC = 1/(ω C). Thus an inductor blocks high-frequency AC more, while a capacitor allows it more easily.

Common wrong answer

Using XL = 1/(ω L) or XC = ω C is incorrect; it reverses the frequency behaviour and leads to wrong units.

Don't mix this up

A common error is to interchange the phase rule: students write that current leads in an inductor or lags in a capacitor. The correct rule is L: current lags, C: current leads.

Exam tip

For phase questions, mention the leading quantity clearly. For an inductor, voltage leads current by π/2. For a capacitor, current leads voltage by π/2.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Exams ask students to calculate reactance, compare frequency dependence, draw phasors, identify leading or lagging current, or reason why ideal L and C consume no average power.
  1. 1

    Enough for 1 mark

    In a pure inductor, current lags voltage by π/2 and opposition to AC is inductive reactance XL = ω L. In a pure capacitor, current leads voltage by π/2 and opposition to AC is capacitive reactance XC = 1/(ω C).

  2. 2

    Add for 2 marks

    Example: At higher frequency, an inductor offers more opposition because XL = ω L increases. At higher frequency, a capacitor offers less opposition because XC = 1/(ω C) decreases.

  3. 3

    Add for 3 marks

    An inductor opposes change in current through induced emf, so current builds up after the applied voltage changes.

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