NEET PhysicsUpdated 19 Oct 2026

Alternating Current: NEET PYQs and weightage

Alternating Current carries 3.8 of 100 Physics marks in NEET by past-paper share: 29 of the 758 tagged NEET Physics PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 9 of 30 Physics chapters.

Take the 25-question Alternating Current test Formula sheet ↓All NEET Physics chapters

3.8/100 Weight in Physics

Exam importance (3/5, by past-paper share)

29 NEET PYQs

311 Practice questions

289 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Physics PYQs in the bank (29 of 758), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Alternating Current: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

RMS values
$I = \frac{i _{m}}{2} \approx 0.707 i_{m}$
$V = \frac{v _{m}}{2} \approx 0.707 v_{m}$
Use it when: a question gives "220 V mains" (that is rms) and you need the peak ( $2202 \approx 311$ V), or the reverse
Inductive reactance
$X_{L} = ω L = 2 π ν L (ohm)$
Use it when: finding how strongly a coil opposes ac; it grows with frequency
Capacitive reactance
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π ν C} (ohm)$
Use it when: finding how strongly a capacitor opposes ac; it falls as frequency rises (double $ν$ $\to$ half $X_{C}$ )
Impedance of series LCR
$Z = R^{2} + (X_{C} - X_{L})^{2}$
Use it when: any series circuit with $R$ , $L$ and/or $C$ ; drop the missing reactance (RC: $Z = R^{2} + X_{C}^{2}$ )
Current in series LCR
$i_{m} = \frac{v _{m}}{Z}$
$I = \frac{V}{Z}$
Use it when: you know the source voltage and $Z$ ; keep peak with peak, rms with rms
Phase angle
$tan ϕ = \frac{X _{C} - X _{L}}{R}$
$i = i_{m} sin (ω t + ϕ)$
Use it when: asked lead or lag; $X_{C} > X_{L} \to$ current leads, $X_{L} > X_{C} \to$ current lags
Phase in single elements
$R \to v and i in phase$
$pure L \to i lags v by \frac{π}{2}$
$pure C \to i leads v by \frac{π}{2}$
Use it when: drawing phasors or reading graphs; the quarter-cycle shift is $T /4$
Average power
$P = V I cos ϕ = I^{2} Z cos ϕ = I^{2} R (V, I rms)$
Use it when: power drawn by any ac circuit; only $R$ dissipates energy
Power factor
$cos ϕ = \frac{R}{Z}$
Use it when: asked for the power factor; pure $R \to 1$ , pure $L$ or pure $C \to 0$ (wattless current, $P = 0$ )
Resistor power
$P = \frac{1}{2} i_{m}^{2} R = I^{2} R = \frac{V ^{2}}{R} = V I$
Use it when: bulb or heater ratings (e.g. 100 W, 220 V $\to R = V^{2} / P = 484 Ω$ )
Resonance
$X_{L} = X_{C}$
$ω_{0} = \frac{1}{L C}$
$ν_{0} = \frac{1}{2 π L C}$
Use it when: "maximum current" or "tuning"; at resonance $Z = R$ , $ϕ = 0$ , $i_{m} = v_{m} / R$ , $P$ is maximum
Voltage addition in series
$V = V_{R}^{2} + (V_{L} - V_{C})^{2}$
Use it when: given separate element voltages and asked the source voltage (RC: $V = V_{R}^{2} + V_{C}^{2}$ )
Quality factor
$Q = \frac{ω _{0} L}{R} = \frac{1}{ω _{0} C R}$
Use it when: asked for the quality factor of a series LCR circuit (dimensionless); halving $R$ doubles the peak current at resonance
Ideal transformer
$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$
$\frac{I _{s}}{I _{p}} = \frac{N _{p}}{N _{s}}$
$so V_{p} I_{p} = V_{s} I_{s}$
Use it when: step-up ( $N_{s} > N_{p}$ : voltage up, current down) or step-down problems

Trap

Adding element voltages as plain numbers. V_R, V_L and V_C are out of phase, so V_R + V_C can exceed the source voltage. Always combine them with √(V_R² + (V_L − V_C)²).

How does NEET ask Alternating Current?

8 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1PhysicsNEET UG PYQAC Circuits

Try right here · No sign-in

An ac source is connected to a capacitor C. Due to decrease in its operating frequency:

Answer key (B) & worked solution

Correct answer: Option B

i_{C} = i_{D} = \frac{V _{O}}{X _{C}} sin ω t

i_{C} = i_{D} = (V_{O} ω C) sin ω t

On decreasing frequency

i_{C} ↓

Q2PhysicsNEET UG PYQAC Circuits

A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when

AFrequency of the AC source is decreased

BNumber of turns in the coil is reduced

CA capacitance of reactance X

_{C}

= X

_{L}

is included in the same circuit

DAn iron rod is inserted in the coil

Answer key (D) & worked solution

Correct answer: Option D

Impedance is

Z = R^{2} + X^{2}_{L} = R^{2} + (L \times 2 π f)^{2}

I = \frac{E}{Z}

I = \frac{E}{R ^{2} + ( L \times 2 π f ) ^{2}}

Brightness of bulb decreases if

I ↓

I ↓

L ↑, f ↑

and we know

L \propto M

If an iron rod is inserted in the coil then

M ↑, L ↑, I ↓

Q3PhysicsNEET UG PYQAC Circuits

The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is

A0.5

B0.8

C1.0

D0.4

Answer key (B) & worked solution

Correct answer: Option B

tan ϕ = \frac{V _{L} - V _{C}}{V _{R}} = \frac{100 - 40}{80} = \frac{3}{4}

ϕ = 3 7^{\circ}

Power factor

= cos ϕ = cos 3 7^{\circ} = \frac{4}{5}

or 0.8

Q4PhysicsNEET UG PYQAC Circuits

100 Ω

resistance and a capacitor of

100 Ω

reactance are connected in series across a 220 volt source. When the capacitor is 50% charged, the peak value of the displacement current is

A11 A

B4.4 A

112 A

D2.2 A

Answer key (D) & worked solution

Correct answer: Option D

(i_{d})_{m a x} = (i_{c})_{m a x} = \frac{E _{0}}{Z} = \frac{220 2}{100 ^{2} + 100 ^{2}}

\Rightarrow (i_{d})_{m a x} = \frac{220 2}{100 2} = 2.2 A

Q5PhysicsNEET UG PYQAC Circuits

An inductor 20 mH, a capacitor 100 μ F and a resistor

50Ω

re connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is

A2.74 W

B0.43 W

C0.79 W

D1.13 W

Answer key (C) & worked solution

Correct answer: Option C

P_{a v} = (\frac{V _{R M S}}{Z})^{2} R

Z = R^{2} + (ω L - \frac{1}{ω C})^{2} = 56 Ω

∴ P_{a v} = (\frac{10}{( 2 ) 56})^{2} \times 50 = 0.79 W

Q6PhysicsNEET UG PYQTransformer

In an ideal transformer, the turns ratio is

\frac{N _{p}}{N _{s}} = \frac{1}{2}

. The ratio

V_{s} : V_{p}

is equal to (the symbols carry their usual meaning):

A2 : 1

B1 : 1

C1 : 4

D1 : 2

Answer key (A) & worked solution

Correct answer: Option A

\frac{e _{s}}{e _{p}} = \frac{N _{s}}{N _{p}} = \frac{2}{1}

Q7PhysicsNEET UG PYQAC Circuits

An ac voltage

V = 220 sin (2 \times 1 0^{3} t)

Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given :

L = 10 mH, C = 25 μ F, R = 100Ω

)

A5.5 A

B11.0 A

C22.0 A

D2.2 A

Answer key (D) & worked solution

Correct answer: Option D

Angular frequency of AC source,

ω = 2 \times 1 0^{3} rad / s

Inductive reactance,

X_{L} = ω L = (2 \times 1 0^{3}) \times (10 \times 1 0^{- 3}) = 20Ω

Capacitive reactance,

X_{C} = \frac{1}{ω C} = \frac{1}{( 2 \times 1 0 ^{3} ) \times ( 25 \times 1 0 ^{- 6} )} = 20Ω

Impedance

Z

for circuit

= R (∵ X_{L} = X_{C}

and circuit is at resonance

)

∴ current amplitude,

i_{0} = \frac{V _{0}}{Z} = \frac{V _{0}}{R} = \frac{220}{100} = 2.2 A

Q8PhysicsNEET UG PYQAC Circuits

In a series LCR circuit, the inductance

L

is 10 mH, capacitance

C

is 1

μ F

and resistance

R

100 Ω

. The frequency at which resonance occurs is:

A15.9 kH

B1.59 rad/s

C1.59 kHz

D15.9 rad/s

Answer key (C) & worked solution

Correct answer: Option C

L = 10 \times 1 0^{- 3} H

C = 1 \times 1 0^{- 6} F

R = 100 Ω

At resonance

X_{L} = X_{C}

ω L = \frac{1}{ω C}

f = \frac{1}{2 π L C} = \frac{1}{2 π 10 \times 10 ^{- 3} \times 10 ^{- 6}} = 1.59

KHz

Which topics in Alternating Current carry the most questions?

All 5 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
AC Circuits	19	183
Transformer	5	45
Power in AC Circuit	3	43
Average, Peak and RMS Value of Alternating Current/Voltage	2	38
Miscellaneous Examples	0	2

Why does Alternating Current show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Alternating Current, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Alternating Current?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Alternating Current test

Chapters either side of Alternating Current by weight

Electrostatic Potential and Capacitance Weight 4.2/100 Laws of Motion Weight 4.0/100 Gravitation Weight 3.8/100 Kinetic Theory of Gases Weight 3.6/100

Alternating Current: NEET PYQs and weightage

Alternating Current: formula sheet for NEET

RMS values

Inductive reactance

Capacitive reactance

Impedance of series LCR

Current in series LCR

Phase angle

Phase in single elements

Average power

Power factor

Resistor power

Resonance

Voltage addition in series

Quality factor

Ideal transformer

How does NEET ask Alternating Current?

Which topics in Alternating Current carry the most questions?

Why does Alternating Current show "–" until you answer 3 questions?

What happens when you test Alternating Current?

Chapters either side of Alternating Current by weight