Formula
Inductive reactance
X_L = ω L. Current lags voltage by π/2 in pure inductor. Reactance has units of ohm; like resistance but no power dissipation.
-- NCERT Class 12 Physics, Ch. 7, p. 182Reactance Impedance
8 MCQs3 revision cards9-step worked example
Source: NCERT Electromagnetic Induction and Alternating CurrentsPYQ coverage: NEET 2023, 2024Official key: NTA-verifiedLast updated: 26 Sep 2026
Reactance Impedance, explained for NEET
The distractor that earns marks off here is the swapped formula: writing X_C = ωC for a capacitor because ωL worked for the inductor. The corpus records it as the standard wrong option for this pattern — it is tempting precisely because both symbols sit in the same equation sheet and differ only by where ω lands.
Fix it with the physical reading rather than the algebra. An inductor opposes changes in current. Faster changes mean stronger opposition, so its reactance must rise with frequency: X_L = ωL = 2πfL. A capacitor is the opposite — it opposes steady current completely and passes rapid alternation easily, so its reactance must fall with frequency: X_C = 1/(ωC) = 1/(2πfC). If your answer has an inductor getting easier to drive at high frequency, you have swapped them.
NCERT Class 12 Physics Chapter 7 gives these on page 182 (inductive) and page 184 (capacitive). Both carry the ohm, and both play the role resistance plays in Ohm's law — the peak current through a pure element is V₀ divided by its reactance.
The phase behaviour differs too, and NEET tests it as often as the magnitude. In a pure inductor, voltage leads current by 90°. In a pure capacitor, current leads voltage by 90°. Neither element dissipates energy over a full cycle; both store and return it. That is what separates a reactance from a resistance, and it is why the word impedance exists — the combined opposition when resistive and reactive elements sit together, treated in the LCR-series lesson of this unit.
Two limiting cases are worth holding in memory, because they convert several exam stems into one-line answers. At DC (f = 0), an inductor behaves as a plain wire and a capacitor as an open break. At very high frequency the roles invert.
Watch out for the units of C. Microfarads and picofarads dominate question stems, and an unconverted µF puts your X_C off by 10⁶.
Can you answer these Reactance Impedance MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The SI unit of both inductive reactance and capacitive reactance is the
Show answer and why every option is right or wrong
Answer: C. Reactance is the voltage-to-current ratio for a reactive element, so it carries the ohm exactly as resistance does. NCERT Class 12 Physics Chapter 7, page 182, introduces X_L in ohms.
Why A is wrong: A is wrong because the henry is the unit of inductance L itself, not of the reactance ωL that the inductance produces at a given frequency.
Why B is wrong: B is wrong because the farad is the unit of capacitance C, not of the reactance 1/(ωC).
Why D is wrong: D is wrong because the weber is the unit of magnetic flux, which belongs to the induction formulas and not to AC reactance.
MCQ 2Easy RecallPractice
In a circuit containing a pure inductor driven by a sinusoidal source, the voltage across the inductor
Show answer and why every option is right or wrong
Answer: B. For a pure inductor the voltage reaches its maximum a quarter cycle ahead of the current, so voltage leads current by 90°. NCERT Class 12 Physics Chapter 7, page 182, states this alongside the X_L result.
Why A is wrong: A is wrong because it states the capacitor's relationship: it is in a pure capacitor that voltage lags current by 90°.
Why C is wrong: C is wrong because zero phase difference is the signature of a pure resistor, where energy is dissipated rather than stored and returned.
Why D is wrong: D is wrong because a 180° relationship would mean the voltage is simply inverted relative to the current; the quarter-cycle 90° offset is what makes the element reactive rather than dissipative.
MCQ 3Easy RecallPractice
As the frequency of the applied AC source is increased, the capacitive reactance of a fixed capacitor
Show answer and why every option is right or wrong
Answer: B. X_C = 1/(2πfC), so reactance falls as frequency rises — a capacitor passes rapid alternation easily. NCERT Class 12 Physics Chapter 7, page 186.
Why A is wrong: A is wrong because rising in direct proportion to frequency is the behaviour of X_L = 2πfL; applying it to the capacitor is the swapped-formula error.
Why C is wrong: C is wrong because only a resistance is frequency-independent; both reactances vary with f.
Why D is wrong: D is wrong because no reactance formula carries f²; the frequency appears to the first power in both X_L and X_C.
MCQ 4Direct ApplicationPractice
An inductor of 0.20 H is connected to a source of angular frequency 5.0 × 10² rad s⁻¹. Its inductive reactance is
Show answer and why every option is right or wrong
Answer: C. X_L = ωL = (5.0 × 10²)(0.20) = 1.0 × 10² Ω. NCERT Class 12 Physics Chapter 7, page 182.
Why A is wrong: A is wrong because 4.0 × 10⁻³ Ω does not follow even from inverting the relation: L/ω = 0.20/5.0 × 10² = 4.0 × 10⁻⁴ Ω. X_L = ωL = 5.0 × 10² × 0.20 = 1.0 × 10² Ω.
Why B is wrong: B is wrong because it divides ω by L, which is the swapped-formula error of applying the capacitive 1/(ωC) structure to an inductor.
Why D is wrong: D is wrong because it is the product ωL with the power of ten mishandled; 5.0 × 10² times 0.20 is 1.0 × 10², not 1.0 × 10⁻².
MCQ 5Direct ApplicationPractice
A capacitor of 5.0 µF is connected to an AC source of angular frequency 1.0 × 10³ rad s⁻¹. Its capacitive reactance is closest to
Show answer and why every option is right or wrong
Answer: D. X_C = 1/(ωC) = 1/[(1.0 × 10³)(5.0 × 10⁻⁶)] = 1/(5.0 × 10⁻³) = 2.0 × 10² Ω. NCERT Class 12 Physics Chapter 7, page 186.
Why A is wrong: A is wrong because it is the product ωC rather than its reciprocal — the swapped-formula error, applying the inductor's product structure to a capacitor.
Why B is wrong: B is wrong because it reads µF as mF, taking C = 5.0 × 10⁻³ F: 1/[(1.0 × 10³)(5.0 × 10⁻³)] = 2.0 × 10⁻¹ Ω, a factor of 10³ too small.
Why C is wrong: C is wrong because it uses the capacitance value directly with a stray power of ten; the reciprocal of 5.0 × 10⁻³ is 2.0 × 10², not 5.0 × 10³.
MCQ 6Direct ApplicationPractice
A pure inductor of reactance 5.0 × 10¹ Ω is connected across a source whose peak voltage is 1.0 × 10² V. The peak current in the inductor is
Show answer and why every option is right or wrong
Answer: A. Reactance plays the role of resistance for peak values: I₀ = V₀/X_L = (1.0 × 10²)/(5.0 × 10¹) = 2.0 A. NCERT Class 12 Physics Chapter 7, page 182.
Why B is wrong: B is wrong because it multiplies voltage by reactance instead of dividing; reactance sits in the denominator exactly as resistance does in Ohm's law.
Why C is wrong: C is wrong because it inverts the division, giving X_L/V₀ rather than V₀/X_L.
Why D is wrong: D is wrong because it applies a √2 factor that the question does not call for: both the given voltage and the required current are peak values, so no peak-to-RMS conversion enters.
MCQ 7Concept TrapPractice
A steady direct current is to be passed through a circuit containing an ideal inductor in one branch and an ideal capacitor in another. In the steady DC state, the inductor behaves as
Show answer and why every option is right or wrong
Answer: C. At f = 0, X_L = 2πfL = 0 so the inductor offers no opposition, while X_C = 1/(2πfC) grows without limit so the capacitor blocks completely. Both follow directly from the reactance formulas of NCERT Class 12 Physics Chapter 7, pages 182 and 184.
Why A is wrong: A is wrong because it is the exact reversal: it assigns the capacitor's blocking behaviour to the inductor and vice versa, the same swap that produces wrong reactance formulas.
Why B is wrong: B is wrong on the inductor: setting f = 0 in X_L = 2πfL gives zero reactance, not infinite.
Why D is wrong: D is wrong on the capacitor: setting f = 0 in X_C = 1/(2πfC) makes the reactance unbounded, so no steady current can be sustained through it.
MCQ 8CalculationPractice
An inductor of 0.10 H and a capacitor of 1.0 × 10⁻⁵ F are each connected, separately, to sources of the same angular frequency ω. The two reactances are found to be equal in magnitude. The value of ω is
Show answer and why every option is right or wrong
Answer: D. Setting ωL = 1/(ωC) gives ω² = 1/(LC) = 1/[(0.10)(1.0 × 10⁻⁵)] = 1.0 × 10⁶, so ω = 1.0 × 10³ rad s⁻¹. Both reactance definitions are from NCERT Class 12 Physics Chapter 7, pages 182 and 184.
Why A is wrong: A is wrong because it is √(LC) rather than 1/√(LC); the product LC sits in the denominator once the two reactances are equated.
Why B is wrong: B is wrong because it reports ω² = 1/(LC) without taking the square root — the value 1.0 × 10⁶ is the square of the answer, not the answer.
Why C is wrong: C is wrong because it divides 1/(LC) by 10⁴ through a slip in handling the powers of ten; 1/(1.0 × 10⁻⁶) is 1.0 × 10⁶, whose root is 1.0 × 10³.
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Reactance Impedance: quick recall before you leave
How do you solve a Reactance Impedance question? A worked example
- 1
Given
• Capacitance C = 2.0 µF = 2.0 × 10⁻⁶ F• Source frequency f = 50 Hz• Source RMS voltage V_rms = 2.2 × 10² V• Pure capacitor (no resistance in the branch)
- 2
Required
The capacitive reactance, and the RMS current drawn.
- 3
Concept
A capacitor in an AC circuit opposes current through its reactance, which falls as frequency rises. Reactance occupies the same slot as resistance in the current-voltage relation, so once X_C is known the current follows by division. The current in a pure capacitor leads the source voltage by 90°, but that phase information does not alter the magnitude of the RMS current.
- 4
Formula
X_C = 1/(2πfC), then I_rms = V_rms/X_C.
- 5
Substitution
X_C = 1 / [2π × 50 × (2.0 × 10⁻⁶)]
- 6
Calculation
Denominator: 2π × 50 = 3.1416 × 10²; multiplied by 2.0 × 10⁻⁶ gives 6.283 × 10⁻⁴.
X_C = 1/(6.283 × 10⁻⁴) = 1.59 × 10³ Ω.
I_rms = (2.2 × 10²)/(1.59 × 10³) = 0.138 A.
The constant 2π here is a mathematical constant and the 50 Hz is an exactly stated line frequency; neither limits the significant figures. The precision is set by the two-significant-figure data (2.0 µF and 2.2 × 10² V). - 7
Final answer
X_C ≈ 1.6 × 10³ Ω; I_rms ≈ 0.14 A.
- 8
Common trap
The swapped-formula distractor: computing 2πfC instead of its reciprocal, which yields 6.3 × 10⁻⁴ and then an absurd current of 3.5 × 10⁵ A. A sanity check on magnitude catches it — a 2 µF capacitor on mains is a near-block, so a current of a fraction of an ampere is expected and hundreds of kiloamperes are not. The second trap is leaving C in microfarads, which shifts X_C by 10⁶.
- 9
Similar NEET-style question
An inductor of 0.50 H is connected to a 50 Hz supply of RMS voltage 1.1 × 10² V. Find its inductive reactance and the RMS current drawn. *(Expected: X_L = 2π × 50 × 0.50 ≈ 1.6 × 10² Ω; I_rms ≈ 0.70 A.)*
What to remember before solving Reactance Impedance questions
Formula
Capacitive reactance
X_C = 1/(ω C). Current leads voltage by π/2 in pure capacitor. Decreases with frequency (capacitor passes high-frequency AC).
-- NCERT Class 12 Physics, Ch. 7, p. 186Which Reactance Impedance formulas do you need for NEET?
2 formulas — click to collapse
Capacitive reactance
Resistance-equivalent of capacitor in AC. Current leads voltage by 90°.
| Symbol | Quantity | SI Unit |
|---|---|---|
| X_C | capacitive reactance | Ω |
| omega | angular freq | rad/s |
| C | capacitance | F |
Valid when
- Pure capacitor
- AC steady state
Inductive reactance
Resistance-equivalent of inductor in AC circuit. Voltage leads current by 90°.
| Symbol | Quantity | SI Unit |
|---|---|---|
| X_L | inductive reactance | Ω |
| omega | angular freq | rad/s |
| L | inductance | H |
Valid when
- Pure inductor
- AC steady state
More in Electromagnetic Induction and Alternating Currents: 4 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
Reactance Impedance questions from past NEET papers
2 questions from NEET 2023, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
10.58 A
20.93 A
31.20 A
40.35 A
NTA Answer: Option 2(revised_final)
NEET 2023
An ac source is connected to a capacitor C. Due to decrease in its operating frequency
1Displacement current increases
2Displacement current decreases
3Capacitive reactance remains constant
4Capacitive reactance decreases
NTA Answer: Option 2(final)
All 20 past-paper questions from Electromagnetic Induction and Alternating Currents →
How does NEET ask about Reactance Impedance?
1 recurring pattern from past papers — click to collapse
Compute X_L = ωL or X_C = 1/(ωC) at given frequency. Compare phase relationships.
Direct ApplicationEasy
Common distractors
swaps XL XC formulas
Uses XL formula for capacitor
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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