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AC Through Pure Inductor and Pure Capacitor

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

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Student-friendly explanation

An inductor opposes change in current through induced emf, so current builds up after the applied voltage changes. This produces a phase lag of pi/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 pi/2. Reactance is measured in ohm, like resistance, but it depends on frequency. XL increases with frequency, while XC decreases with frequency.

How to write this in exams

  1. 1

    Start with the exact idea

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

  2. 2

    Then show how to use it

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

  3. 3

    Add one concrete example

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

  4. 4

    Avoid this incomplete answer

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

Definition

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

Example

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

Rule to remember

Inductor: XL = omega L, where XL is in ohm, omega is angular frequency in rad s^-1, and L is inductance in henry. Capacitor: XC = 1/(omega C), where XC is in ohm and C is capacitance in farad. Current in pure L lags voltage by pi/2; current in pure C leads voltage by pi/2. These ideal relations assume negligible resistance.

Memory hook

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

Examples and method

Worked example

An inductor of 0.5 H is connected to AC of angular frequency 200 rad s^-1. XL = omega L = 200 x 0.5 = 100 ohm. If Vrms = 50 V, Irms = Vrms/XL = 50/100 = 0.5 A. The current lags voltage by pi/2. For a capacitor of 100 microfarad at the same omega, XC = 1/(200 x 100 x 10^-6) = 50 ohm, so with 50 V RMS, Irms = 1 A and current leads voltage by pi/2.

Method to apply

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

Diagram support

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.

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.

Avoid common mistakes

Common confusion

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.

Common wrong answer

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

Exam tip

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

Quick check

How does increasing frequency affect XL and XC?

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

Answer writing and exam use

1-mark answer

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

2-mark answer

In a pure inductor, current lags voltage by pi/2 and opposition to AC is inductive reactance XL = omega L. In a pure capacitor, current leads voltage by pi/2 and opposition to AC is capacitive reactance XC = 1/(omega C). Inductor: XL = omega L, where XL is in ohm, omega is angular frequency in rad s^-1, and L is inductance in henry. Capacitor: XC = 1/(omega C), where XC is in ohm and C is capacitance in farad. Current in pure L lags voltage by pi/2; current in pure C leads voltage by pi/2. These ideal relations assume negligible resistance. At higher frequency, an inductor offers more opposition because XL = omega L increases. At higher frequency, a capacitor offers less opposition because XC = 1/(omega C) decreases.

3-mark answer

An inductor opposes change in current through induced emf, so current builds up after the applied voltage changes. This produces a phase lag of pi/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 pi/2. Reactance is measured in ohm, like resistance, but it depends on frequency. XL increases with frequency, while XC decreases with frequency. Inductor: XL = omega L, where XL is in ohm, omega is angular frequency in rad s^-1, and L is inductance in henry. Capacitor: XC = 1/(omega C), where XC is in ohm and C is capacitance in farad. Current in pure L lags voltage by pi/2; current in pure C leads voltage by pi/2. These ideal relations assume negligible resistance. An inductor of 0.5 H is connected to AC of angular frequency 200 rad s^-1. XL = omega L = 200 x 0.5 = 100 ohm. If Vrms = 50 V, Irms = Vrms/XL = 50/100 = 0.5 A. The current lags voltage by pi/2. For a capacitor of 100 microfarad at the same omega, XC = 1/(200 x 100 x 10^-6) = 50 ohm, so with 50 V RMS, Irms = 1 A and current leads voltage by pi/2. 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. Using XL = 1/(omega L) or XC = omega C is incorrect; it reverses the frequency behaviour and leads to wrong units.
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