JEE Mains Physics · Alternating Current

Reactance of a Resistor, Inductor and Capacitor

An inductor opposes a.c. with reactance ωL, which grows with frequency; a capacitor with 1/ωC, which falls with it; and in each the current is a quarter cycle out of step with the voltage.

All Alternating Current notes JEE Mains Physics strategy

Why this matters

Twenty-two PYQs, fifteen of them multiple choice, and eight from 2024. Twelve compute a reactance or the current through one inductor or capacitor. Five ask whether the current leads or lags the voltage. Five test a real coil on d.c. and then on a.c., or a network at a frequency so high that capacitors short and inductors open.

Concept 1 of 3: Reactance: ωL and 1/ωC

A coil fights changes in current, so the faster the current changes, the more it opposes: its reactance ωL grows with frequency and is zero for steady d.c. A capacitor passes changes easily, so its reactance 1/ωC falls as the frequency rises and is infinite for d.c. A resistor does not care about frequency at all.

Definition

$X_L = \omega L = 2\pi fL$ : a straight line through the origin on an $X$ –f graph.
$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi fC}$ : a falling curve, $X_C \propto \dfrac{1}{f}$ .
R does not change with f: a horizontal line.
Current through one element: $I = \dfrac{V}{X}$ , peak with peak and rms with rms. Pure capacitor: $I_0 = V_0\omega C$ . Pure inductor: $I_0 = \dfrac{V_0}{\omega L}$ .
Halve f: $X_L$ halves, so the inductor current doubles; $X_C$ doubles, so the capacitor current halves.
More capacitance (a dielectric slipped in) lowers $X_C$ , so more current flows through anything in series with it.
Between the plates the displacement current equals the conduction current in the wires: $I_d = V\omega C$ .
R, $X_L$ and $X_C$ are all in ohm, so a ratio of two of them has no unit, while $X_LX_C = \dfrac{L}{C}$ is in $\text{ohm}^{2}$ .

Reactances

X_L = \omega L = 2\pi fL \qquad X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}

Worked example

20\ \mu\text{F}

capacitor is connected to

e = 50\sin(500t)\ \text{V}

. Find its reactance, the peak current, and the reading of an a.c. ammeter in series.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q92Moderate

Example 1 · Alternating Current · Reactance of a Resistor, Inductor and Capacitor

A capacitor of capacitance

150.0\mu F

is connected to an alternating source of emf given by

e = 36

\sin(120\pi t)V

. The maximum value of current in the circuit is approximately equal to:

Making the capacitive reactance rise with frequency

Capacitive reactance goes DOWN as f or C goes up. Doubling both divides it by 4.

Multiplying a given ω by 2π

In sin(500t) the 500 is already ω in rad/s. Use 2πf only when the frequency is given in hertz.

Peak or rms

An emf written as E₀ sin ωt gives the PEAK current. An a.c. ammeter shows the rms value, √2 times smaller.

Drill 11 more on reactance: ωl and 1/ωc

Concept 2 of 3: Phase of the current in R, L and C

In a resistor the current follows the voltage step for step. In an inductor the voltage is L di/dt, so it peaks when the current changes fastest, a quarter cycle ahead of the current. In a capacitor the current is C dv/dt, so the current is the one a quarter cycle ahead. When the current is zero just as the voltage peaks, the two are 90° apart.

Definition

Pure L: if $i = I_0\sin\omega t$ , then $v = L\dfrac{di}{dt} = I_0\omega L\sin\left(\omega t + \dfrac{\pi}{2}\right)$ . The voltage leads.
Pure C: if $v = V_0\sin\omega t$ , then $i = C\dfrac{dv}{dt} = V_0\omega C\sin\left(\omega t + \dfrac{\pi}{2}\right)$ . The current leads.
Memory aid CIVIL: in C, I leads V; V leads I in L.
On a phasor diagram the arrows turn anticlockwise; the arrow further round in that direction leads.
To write the voltage across an inductor from its current: multiply the amplitude by $\omega L$ and add $\dfrac{\pi}{2}$ to the phase.

Element	Opposition	Change with frequency	Current compared with voltage	Average power
Pure resistor	R	None	In phase	$V_{rms}I_{rms}$
Pure inductor	$X_L = \omega L$	Grows in proportion to f	Lags by $\dfrac{\pi}{2}$	Zero
Pure capacitor	$X_C = \dfrac{1}{\omega C}$	Falls as $\dfrac{1}{f}$	Leads by $\dfrac{\pi}{2}$	Zero
Ideal L and C in series	$\|X_L - X_C\|$	Falls to zero at resonance	Lags by $\dfrac{\pi}{2}$ if $X_L > X_C$ , leads if $X_C > X_L$	Zero No resistance anywhere, so the gap is exactly 90° whichever reactance wins.
Series LCR	$\sqrt{R^{2} + (X_L - X_C)^{2}}$	Least at resonance	Angle $\phi$ with $\tan\phi = \dfrac{X_L - X_C}{R}$	$V_{rms}I_{rms}\cos\phi$

Current zero while the voltage is at its peak means a 90° gap: no resistance in the circuit.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q20Moderate

Example 2 · Alternating Current · Reactance of a Resistor, Inductor and Capacitor

In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to: (A) pure inductor. (B) pure capacitor. (C) pure resistor. (D) combination of an inductor and capacitor. Choose the correct answer from the options given below:

Which one leads in an inductor

In an inductor the VOLTAGE leads the current. Saying 'the current leads' is the capacitor's rule. CIVIL settles it.

Adding 90° to the wrong quantity

Going from current to voltage in an inductor, add π/2. Going from voltage to current, subtract it. For a capacitor it is the other way round.

Drill 4 more on phase of the current in r, l and c

Concept 3 of 3: A real coil on d.c. and a.c., and the frequency limits

A real coil is a resistance and an inductance in series. On d.c. the inductance does nothing, so the current shows the resistance alone. On a.c. the current shows the impedance √(R² + X²). Only the resistance takes power. The same idea of extremes works for whole networks: at a very high frequency capacitors act like wires and inductors like breaks; on d.c. it is the reverse.

Definition

On d.c.: $R = \dfrac{V_{dc}}{I_{dc}}$ .
On a.c.: $Z = \dfrac{V_{rms}}{I_{rms}}$ , $X_L = \sqrt{Z^{2} - R^{2}}$ , $L = \dfrac{X_L}{2\pi f}$ .
Power taken by the coil: $P = I_{rms}^{2}R$ . The inductance takes none on average.
Magnetic energy: $\dfrac{1}{2}LI^{2}$ . Averaged over a cycle it is $\dfrac{1}{2}LI_{rms}^{2}$ ; its peak is $\dfrac{1}{2}LI_0^{2}$ . Read which one the question means.
Very high f: $X_C \to 0$ (a wire) and $X_L \to \infty$ (a break). Very low f or d.c.: $X_C \to \infty$ and $X_L \to 0$ .
Redraw the network with those wires and breaks, then combine the resistors that are left.

Coil on d.c. and on a.c.

R = \frac{V_{dc}}{I_{dc}} \qquad Z = \frac{V_{rms}}{I_{rms}} = \sqrt{R^{2} + X_L^{2}}

Worked example

A coil draws 2 A from a 24 V d.c. supply and 1.2 A from a 24 V (rms), 50 Hz supply. Find its resistance, reactance, inductance and the power it takes on a.c.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q28Moderate

Example 3 · Alternating Current · Reactance of a Resistor, Inductor and Capacitor

When a coil is connected across a

20\text{ }V

dc supply, it draws a current of

5\text{ }A

. When it is connected across

20\text{ }V,50\text{ }Hz

ac supply, it draws a current of

4\text{ }A

. The self inductance of the coil is ______

mH

. (Take

\pi= 3

)

Taking V/I on a.c. as the reactance

V/I on a.c. is the impedance, which still contains the coil's resistance. Find R on d.c. first, then

X_L = \sqrt{Z^{2} - R^{2}}

Power in the inductance

Only the resistance takes power: P = I²R with the rms current. I²Z overstates it.

Average or peak stored energy

½LI² with the rms current is the energy averaged over a cycle; with the peak current it is the maximum. The two differ by a factor of 2.

Drill 4 more on a real coil on d.c. and a.c., and the frequency limits

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Reactance: ωL and 1/ωC
Reactances
$X_L = \omega L = 2\pi fL \qquad X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}$
A real coil on d.c. and a.c., and the frequency limits
Coil on d.c. and on a.c.
$R = \frac{V_{dc}}{I_{dc}} \qquad Z = \frac{V_{rms}}{I_{rms}} = \sqrt{R^{2} + X_L^{2}}$

Reference tables (1)

Phase of the current in R, L and C5 rows

Element	Opposition	Change with frequency	Current compared with voltage	Average power
Pure resistor	R	None	In phase	$V_{rms}I_{rms}$
Pure inductor	$X_L = \omega L$	Grows in proportion to f	Lags by $\dfrac{\pi}{2}$	Zero
Pure capacitor	$X_C = \dfrac{1}{\omega C}$	Falls as $\dfrac{1}{f}$	Leads by $\dfrac{\pi}{2}$	Zero
Ideal L and C in series	$\|X_L - X_C\|$	Falls to zero at resonance	Lags by $\dfrac{\pi}{2}$ if $X_L > X_C$ , leads if $X_C > X_L$	Zero No resistance anywhere, so the gap is exactly 90° whichever reactance wins.
Series LCR	$\sqrt{R^{2} + (X_L - X_C)^{2}}$	Least at resonance	Angle $\phi$ with $\tan\phi = \dfrac{X_L - X_C}{R}$	$V_{rms}I_{rms}\cos\phi$

Current zero while the voltage is at its peak means a 90° gap: no resistance in the circuit.

Watch out for (8)

Making the capacitive reactance rise with frequency
→ Reactance: ωL and 1/ωC
Multiplying a given ω by 2π
→ Reactance: ωL and 1/ωC
Peak or rms
→ Reactance: ωL and 1/ωC
Which one leads in an inductor
→ Phase of the current in R, L and C
Adding 90° to the wrong quantity
→ Phase of the current in R, L and C
Taking V/I on a.c. as the reactance
→ A real coil on d.c. and a.c., and the frequency limits
Power in the inductance
→ A real coil on d.c. and a.c., and the frequency limits
Average or peak stored energy
→ A real coil on d.c. and a.c., and the frequency limits

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