MHT-CET Physics · AC Circuits

Reactance and Single-Element AC Circuits

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

All AC Circuits notes MHT-CET Physics strategy

Why this matters

23 PYQs, none HARD — reliable marks. Three shapes: how a reactance changes when the frequency, L or C changes, the current and its phase in a circuit of one element, and a bulb in series with a coil or capacitor that brightens or dims.

Concept 1 of 3: How Reactance Depends on Frequency, L and C

A coil fights a changing current, so the faster the change the more it resists: X_L = ωL rises with frequency and is zero for steady d.c. A capacitor passes change easily, so X_C = 1/ωC falls with frequency and is infinite for d.c.

Definition

$X_L = \omega L = 2\pi fL$ : directly proportional to $f$ and $L$ . $L \times 3$ , $f \times 2$ ⇒ $X_L \times 6$ .
$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi fC}$ : $f$ and $C$ both doubled ⇒ $\dfrac{X_C}{4}$ .
For d.c. ( $f = 0$ ): $X_L = 0$ , so $\dfrac{X_{\text{ac}}}{X_{\text{dc}}} = \infty$ .
$X_L = X_C$ at $\omega$ ; at $2\omega$ , $X_C : X_L = 1 : 4$ .
Inductors combine like resistors: series add, parallel as reciprocals.

Reactances

X_L = \omega L, \qquad X_C = \frac{1}{\omega C}

Worked example

Reactance of a 0.1 H inductor and of a

100\,\mu

F capacitor at 50 Hz?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 1 · Q12Easy

Example 1 · AC Circuits · Reactance — Inductive, Capacitive, and Single-Element Circuits

The inductive reactance of a coil is '

X_L

'. If inductance is tripled and frequency is doubled, the new inductive reactance will be

Treating X_C like X_L

X_C

goes DOWN as frequency or capacitance goes up. Doubling both divides it by 4; the options offer

4X

for the student who multiplied.

Drill 9 more on how reactance depends on frequency, l and c

Concept 2 of 3: Current and Phase With a Single L or C

In a pure inductor the current lags the voltage by 90°; in a pure capacitor it leads by 90° — remember CIVIL: in a C, I leads V; V leads I in an L. The size of the current is just V over the reactance.

Definition

Pure $L$ : $I = \dfrac{V}{X_L}$ , current LAGS the voltage by $\dfrac{\pi}{2}$ .
Pure $C$ : $I = \dfrac{V}{X_C} = V\omega C$ , current LEADS by $\dfrac{\pi}{2}$ : $E = E_0\sin\omega t \Rightarrow I = E_0\omega C\sin\left(\omega t + \dfrac{\pi}{2}\right)$ .
Pure $R$ : in phase. So in a series LCR, current and voltage are out of phase in $L$ and in $C$ , in phase in $R$ .
An a.c. ammeter reads r.m.s.: $V_0 = 200\sqrt{2}$ , $\omega = 100$ , $C = 1\,\mu$ F ⇒ $X_C = 10^4\,\Omega$ , $I = 20$ mA.

CIVIL

\text{C: } I \text{ leads } V \text{ by } 90^\circ; \qquad \text{L: } V \text{ leads } I \text{ by } 90^\circ

Worked example

10\,\mu

F capacitor is connected to 220 V, 50 Hz. R.m.s. current?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q39Moderate

Example 2 · AC Circuits · Reactance — Inductive, Capacitive, and Single-Element Circuits

An alternating voltage

E = 100\sqrt{2}\sin(50t)

is connected to a

2\mu\text{ }F

capacitor through an a.c. ammeter. The ammeter reading will be

Swapping lead and lag

Every phase question offers both. CIVIL settles it: Capacitor — I before V; inducto-L — V before I.

Drill 8 more on current and phase with a single l or c

Concept 3 of 3: A Bulb in Series With a Coil or a Capacitor

The bulb glows with the current, and the current falls as the reactance rises. So ask one question of every change: does it raise or lower the reactance?

Definition

With a coil: an iron core raises $L$ , so $X_L$ rises and the bulb DIMS; fewer turns or lower frequency brighten it.
With a capacitor: a larger $C$ or higher frequency lowers $X_C$ and BRIGHTENS it; a smaller $C$ or lower frequency dims it.
Adding a capacitor to a coil circuit partly cancels $X_L$ and can raise the current.

Current falls as reactance rises

I = \frac{V}{\sqrt{R^2 + X^2}}

Worked example

A bulb in series with a coil glows on an a.c. supply. An iron rod is pushed into the coil. What happens?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q39Easy

Example 3 · AC Circuits · Reactance — Inductive, Capacitive, and Single-Element Circuits

An electric lamp connected in series with a capacitor and an a.c. source is glowing with certain brightness. On increasing the value of capacitance, the brightness of the lamp

Reversing the capacitor's rule

For a coil, more L or more frequency means dimmer; for a capacitor, more C or more frequency means BRIGHTER. The two rules run in opposite directions.

Drill 3 more on a bulb in series with a coil or a capacitor

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 (3)

How Reactance Depends on Frequency, L and C
Reactances
$X_L = \omega L, \qquad X_C = \frac{1}{\omega C}$
Current and Phase With a Single L or C
CIVIL
$\text{C: } I \text{ leads } V \text{ by } 90^\circ; \qquad \text{L: } V \text{ leads } I \text{ by } 90^\circ$
A Bulb in Series With a Coil or a Capacitor
Current falls as reactance rises
$I = \frac{V}{\sqrt{R^2 + X^2}}$

Watch out for (3)

Treating X_C like X_L
→ How Reactance Depends on Frequency, L and C
Swapping lead and lag
→ Current and Phase With a Single L or C
Reversing the capacitor's rule
→ A Bulb in Series With a Coil or a Capacitor

Test yourself on AC Circuits

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the AC Circuits chapter test Drill all 23 questions