Alternating Current NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Alternating Current question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Alternating Current previous year questions
27 questions. Tap an option to check your answer.
Q1NEET 2021
A series LCR circuit containing 5.0 H inductor, 80 μF capacitor and 40 Ω resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be :
Not quite — the answer is C.
ω₀ = 1/√(LC) = 1/√(5 × 80 × 10⁻⁶) = 1/0.02 = 50 rad/s. Half-power frequencies are ω₀ ± R/(2L) = 50 ± 40/10 = 50 ± 4, i.e. 46 rad/s and 54 rad/s.
Q2NEET 2021
A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit ?
Not quite — the answer is A.
For an ideal transformer input power equals output power, 44 W. Primary current = P/V = 44/220 = 0.2 A. 4 A is the secondary (lamp) current, 44/11.
Q3NEET 2020 · Q154
A 40 µF capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly :
Not quite — the answer is C.
X_C = 1/(2πfC) = 1/(2π × 50 × 40×10⁻⁶) ≈ 79.6 Ω, so I_rms = V_rms/X_C = 200/79.6 ≈ 2.5 A. Taking 200 V as a peak value and dividing by √2 gives about 1.8 A, which is not what the rms supply rating means.
Q4NEET 2020 · Q169
A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is π/3. If instead C is removed from the circuit, the phase difference is again π/3 between current and voltage. The power factor of the circuit is :
Not quite — the answer is C.
With L removed, tan(π/3) = X_C/R; with C removed, tan(π/3) = X_L/R. So X_L = X_C, the full LCR circuit is at resonance, Z = R and the power factor cos φ = R/Z = 1. 0.5 is cos(π/3), the power factor of the reduced circuits, not of the full one.
Q5NEET 2018 · Q28
An inductor 20 mH, a capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is
Not quite — the answer is C.
X_L = 314 × 0.02 = 6.28 Ω, X_C = 1/(314 × 10⁻⁴) = 31.85 Ω, Z = √(50² + 25.57²) ≈ 56.2 Ω. I_rms = (10/√2)/56.2 ≈ 0.126 A, so P = I_rms²R ≈ 0.79 W. Using peak instead of rms current doubles the power.
Q6NEET 2017 (Gujarati paper) · Q157
A bulb connected in series with an air-cored solenoid is lit by an a.c. source. If a soft iron core is introduced in the solenoid:
Not quite — the answer is C.
Soft iron raises the permeability and hence the inductance L, so the inductive reactance ωL and the circuit impedance increase. The rms current falls and the bulb glows dimmer. It does not go out completely, because the impedance is still finite.
Q7NEET 2016 Phase 2 · Q30
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication?
Not quite — the answer is C.
Better tuning means a sharper resonance, i.e. a larger quality factor Q = (1/R)√(L/C). The values are about 10.4 for 20 Ω/1.5 H/35 μF, 9.4 for 25 Ω/2.5 H/45 μF, 22.8 for 15 Ω/3.5 H/30 μF and 7.3 for 25 Ω/1.5 H/45 μF. The 15 Ω, 3.5 H, 30 μF combination has the highest Q.
Q8NEET 2016 Phase 2 · Q32
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
Not quite — the answer is C.
Net voltage V = √(V_R² + (V_L − V_C)²) = √(80² + 60²) = 100 V. Power factor cosφ = V_R/V = 80/100 = 0.8. Taking the ratio of V_C to V_L or assuming resonance gives the other values.
Q9NEET 2016 Phase 1 · Q43
A small signal voltage V(t) = V₀ sin ωt is applied across an ideal capacitor C:
Not quite — the answer is B.
In an ideal capacitor the current leads the voltage by 90°, so the power factor cos90° = 0 and the average power over a full cycle is zero: energy stored in one quarter-cycle is returned in the next. Current lagging by 90° describes an ideal inductor, not a capacitor.
Q10AIPMT 2015
A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be:
Not quite — the answer is B.
Without L: P = V²/R. With L, I = V/Z and power = I²R = V²R/Z² = (V²/R)(R/Z)² = P(R/Z)². P(R/Z) misses that both current and power factor fall by R/Z.
Q11AIPMT 2014
A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:
Not quite — the answer is A.
Primary current = 3000/200 = 15 A. Output power = 0.9 × 3000 = 2700 W, so secondary voltage = 2700/6 = 450 V; ignoring efficiency gives 500 V, which is not offered.
Q12NEET 2013
A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when -
Not quite — the answer is D.
An iron rod increases the inductance L, raising XL = ωL and the impedance, so the current and the bulb's brightness drop. Lowering the frequency, reducing the turns or adding a capacitor with XC = XL all reduce the net reactance and make the bulb brighter.
Q13AIPMT 2012 Mains
The instantaneous values of alternating current and voltage in a circuit are given as i = (1/√2) sin(100πt) ampere e = (1/√2) sin(100πt + π/3) volt The average power in watts consumed in the circuit is
Not quite — the answer is D.
Peak values are 1/√2, so Irms = Vrms = 1/2. The phase difference is π/3, so P = Vrms·Irms·cos φ = ½ × ½ × ½ = 1/8 W. Using peak values instead of rms values gives a larger, wrong answer.
Q14AIPMT 2012 Prelims
In an electrical circuit R, L, C and an a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is π/3. If instead, C is removed from the circuit, the phase difference is again π/3. The power factor of the circuit is
Not quite — the answer is D.
Removing L: tan(π/3) = X_C/R; removing C: tan(π/3) = X_L/R. So X_L = X_C = √3R and with all three present the circuit is at resonance: Z = R, power factor cos φ = 1. 1/2 is the power factor of each reduced circuit, not of the full RLC circuit.
Q15AIPMT 2011 Mains
A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source of 200 volt, 100 Hz is connected across the coil, the current in the coil will be
Not quite — the answer is C.
Inductive reactance doubles with frequency: at 100 Hz XL = 40 Ω. Z = √(30² + 40²) = 50 Ω, so I = 200/50 = 4 A. Keeping XL = 20 Ω gives 20/√13 A, which is the trap.
Q16AIPMT 2011 Prelims
In an ac circuit an alternating voltage e = 200√2 sin 100t volts is connected to a capacitor of capacity 1 µF. The r.m.s. value of the current in the circuit is
Not quite — the answer is A.
V(rms) = 200√2/√2 = 200 V and ω = 100 rad/s, so X_C = 1/(ωC) = 1/(100 × 10⁻⁶) = 10⁴ Ω. I(rms) = 200/10⁴ = 0.02 A = 20 mA. Using the peak voltage 200√2 V would give about 28 mA, which is not the rms value.
Q17AIPMT 2011 Prelims
An ac voltage is applied to a resistance R and inductor L in series. If R and the inductive reactance are both equal to 3 Ω, the phase difference between the applied voltage and the current in the circuit is
Not quite — the answer is C.
For a series LR circuit tan φ = X_L/R = 3/3 = 1, so φ = π/4 with voltage leading the current. π/2 would apply only for a pure inductor.
Q18AIPMT 2010 Mains · Q81
A condenser of capacity C is charged to a potential difference of V₁. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V₂ is
Not quite — the answer is D.
In the LC circuit energy is conserved: ½CV₁² = ½CV₂² + ½LI². So I = [C(V₁² − V₂²)/L]^(1/2). Using (V₁ − V₂)² instead of V₁² − V₂² treats the voltage drop wrongly, and omitting the square root gives I² rather than I.
Q19AIPMT 2010 Prelims · Q38
A 220 volt input is supplied to a transformer. The output circuit draws a current of 2.0 ampere at 440 volts. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is
Not quite — the answer is A.
Output power = 440 × 2 = 880 W. Input = 880/0.8 = 1100 W, so primary current = 1100/220 = 5.0 A. Multiplying by 0.8 instead of dividing gives 3.2 A; ignoring efficiency gives 4 A.
Q20AIPMT 2009 Prelims · Q25
Power dissipated in an LCR series circuit connected to an a.c. source of emf ε is:
Not quite — the answer is D.
Average power P = εrms Irms cos φ with I = ε/Z and cos φ = R/Z, so P = ε²R/Z² = ε²R/[R² + (Lω − 1/Cω)²]. Using only one factor of Z (the square-root form) gives the wrong power.
Q21AIPMT 2008 Prelims · Q39
In an a.c. circuit the e.m.f. (e) and the current (i) at any instant are given respectively by e = E₀ sin ωt i = I₀ sin(ωt − φ) The average power in the circuit over one cycle of a.c. is
Not quite — the answer is D.
Average power = Erms × Irms × cos φ = (E₀/√2)(I₀/√2) cos φ = (E₀I₀/2) cos φ. E₀I₀/2 holds only when current and emf are in phase (φ = 0), and the sin φ form gives the wattless (reactive) component.
Q22AIPMT 2007 Prelims · Q23
A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 amp, the efficiency of the transformer is approximately
Not quite — the answer is B.
Efficiency = output/input = 100/(220 × 0.5) = 100/110 ≈ 0.91, i.e. about 90%. 50% comes from comparing the voltages instead of the powers.
Q23AIPMT 2007 Prelims · Q44
What is the value of inductance L for which the current is maximum in a series LCR circuit with C = 10 μF and ω = 1000 s⁻¹ ?
Not quite — the answer is D.
Current is maximum at resonance, ω²LC = 1: L = 1/(ω²C) = 1/(10⁶ × 10 × 10⁻⁶) = 0.1 H = 100 mH. R only sets the height of the peak, not where it occurs.
Q24AIPMT 2006 Prelims · Q9
A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be :
Not quite — the answer is C.
The resonant frequency is f = 1/(2π√(LC)). Replacing L by 2L and C by 4C multiplies LC by 8, so the frequency becomes f/√8 = f/(2√2). The frequency decreases; it does not increase to 8f.
Q25AIPMT 2006 Prelims · Q23
The core of a transformer is laminated because :
Not quite — the answer is A.
A solid iron core lets large eddy currents circulate, wasting energy as heat. Thin insulated laminations break up these current paths and minimise eddy-current loss. Lamination does not change the turns ratio, and its purpose is not to reduce weight or prevent rusting.
Q26AIPMT 2006 Prelims · Q28
A coil of inductive reactance 31 Ω has a resistance of 8 Ω. It is placed in series with a condenser of capacitative reactance 25 Ω. The combination is connected to an a.c. source of 110 V. The power factor of the circuit is :
Not quite — the answer is C.
The net reactance is X_L − X_C = 31 − 25 = 6 Ω, so Z = √(8² + 6²) = 10 Ω. The power factor is cos φ = R/Z = 8/10 = 0.80. Using X_L + X_C would give a much smaller value; the supply voltage is not needed.
Q27AIPMT 2005 Prelims · Q34
In a circuit, L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of C is :
Not quite — the answer is A.
Current leading the voltage means the circuit is net capacitive: tan 45° = (X_C − X_L)/R = 1, so X_C = X_L + R. 1/(2πfC) = 2πfL + R gives C = 1/[2πf(2πfL + R)]. The form with (2πfL − R) would come from X_C = X_L − R, which makes the circuit inductive with current lagging.
How often Alternating Current came up
Questions from this chapter in each paper on this site.
- NEET 2021: 2
- NEET 2020: 2
- NEET 2018: 1
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 2: 2
- NEET 2016 Phase 1: 1
- AIPMT 2015: 1
- AIPMT 2014: 1
- NEET 2013: 1
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 1
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 2
- AIPMT 2010 Mains: 1
- AIPMT 2010 Prelims: 1
- AIPMT 2009 Prelims: 1
- AIPMT 2008 Prelims: 1
- AIPMT 2007 Prelims: 2
- AIPMT 2006 Prelims: 3
- AIPMT 2005 Prelims: 1
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