Current Electricity 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 Current Electricity 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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Current Electricity previous year questions
39 questions. Tap an option to check your answer.
Q1NEET 2021
Column - I gives certain physical terms associated with flow of current through a metallic conductor. Column - II gives some mathematical relations involving electrical quantities. Match Column - I and Column - II with appropriate relations. Column - I | Column - II (a) Drift Velocity | (P) m/(ne²ρ) (b) Electrical Resistivity | (Q) nev_d (c) Relaxation Period | (R) (eE/m)τ (d) Current Density | (S) E/J
Not quite — the answer is A.
Drift velocity v_d = (eE/m)τ; resistivity ρ = E/J; relaxation time τ = m/(ne²ρ); current density J = nev_d. So drift velocity–(eE/m)τ, resistivity–E/J, relaxation period–m/(ne²ρ), current density–nev_d.
Q2NEET 2021
In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs ?
Not quite — the answer is A.
With the same driver circuit, the balancing length is proportional to the EMF: l₂ = 36 × (2.5/1.5) = 60 cm. 21.6 cm results from inverting the ratio.
Q3NEET 2021
The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series ?
Not quite — the answer is D.
Four identical wires of resistance R in parallel give R/4 = 0.25 Ω, so R = 1 Ω. In series the resistance is 4R = 4 Ω (16 times the parallel value).
Q4NEET 2020 · Q144
The solids which have the negative temperature coefficient of resistance are :
Not quite — the answer is D.
In semiconductors and insulators, heating frees more charge carriers across the band gap, so resistance falls with temperature (negative coefficient). In metals the carrier number is fixed and increased lattice scattering raises resistance, giving a positive coefficient.
Q5NEET 2020 · Q164
The color code of a resistance is given below : bands Yellow, Violet, Brown, Gold. The values of resistance and tolerance, respectively, are :
Not quite — the answer is D.
Yellow = 4 and violet = 7 give the digits 47, brown is the multiplier 10¹, so R = 470 Ω; a gold band means ±5% tolerance. The kΩ values use the wrong multiplier, and 10% tolerance would need a silver band.
Q6NEET 2020 · Q173
A charged particle having drift velocity of 7.5×10⁻⁴ m s⁻¹ in an electric field of 3×10⁻¹⁰ V m⁻¹, has a mobility in m² V⁻¹ s⁻¹ of :
Not quite — the answer is B.
Mobility µ = v_d/E = 7.5×10⁻⁴ / 3×10⁻¹⁰ = 2.5×10⁶ m² V⁻¹ s⁻¹. The 2.25 values come from multiplying the two quantities, which is dimensionally wrong.
Q7NEET 2020 · Q179
A resistance wire connected in the left gap of a metre bridge balances a 10 Ω resistance in the right gap at a point which divides the bridge wire in the ratio 3 : 2. If the length of the resistance wire is 1.5 m, then the length of 1 Ω of the resistance wire is :
Not quite — the answer is B.
At balance R/10 = 3/2, so R = 15 Ω. The wire is 1.5 m long, so 1 Ω corresponds to 1.5/15 = 0.1 m = 1.0×10⁻¹ m. Taking R as 10 Ω (forgetting the balance ratio) gives 1.5×10⁻¹ m.
Q8NEET 2019 · Q23
Which of the following acts as a circuit protection device?
Not quite — the answer is D.
A fuse melts and breaks the circuit when the current exceeds a safe value, protecting the circuit. A switch only opens or closes the circuit manually and does not respond to overload.
Q9NEET 2018 · Q5
A carbon resistor of (47 ± 4.7) kΩ is to be marked with rings of different colours for its identification. The colour code sequence will be
Not quite — the answer is B.
47 kΩ = 47 × 10³ Ω: first digit 4 is yellow, second digit 7 is violet, multiplier 10³ is orange. The tolerance 4.7/47 = 10% is silver (gold would be 5%).
Q10NEET 2018 · Q6
A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is
Not quite — the answer is C.
Series: I = E/(nR + R). Parallel: 10I = E/(R/n + R) = nE/((n + 1)R). Dividing, 10 = n(n + 1)R/((n + 1)R) = n, so n = 10.
Q11NEET 2017 · Q142
The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be:
Not quite — the answer is C.
Melting and redrawing keeps the volume constant, so if length becomes nl the area becomes A/n. R = ρl/A then becomes ρ(nl)/(A/n) = n²R. Taking only the length change gives nR.
Q12NEET 2017 · Q154
A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves:
Not quite — the answer is C.
At the balance point no current flows through the galvanometer, so no current is drawn from the cell being measured and its terminal voltage equals its emf. This null method is what makes the potentiometer accurate; a voltmeter always draws some current.
Q13NEET 2016 Phase 2 · Q26
A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is
Not quite — the answer is C.
Rated current I = P/V = 500/100 = 5 A and bulb resistance = 100²/500 = 20 Ω. To draw 5 A from 230 V the total resistance must be 230/5 = 46 Ω, so R = 46 − 20 = 26 Ω. 46 Ω is the total resistance, not the series resistor.
Q14NEET 2016 Phase 1 · Q14
The charge flowing through a resistance R varies with time t as Q = at − bt², where a and b are positive constants. The total heat produced in R is:
Not quite — the answer is A.
Current i = dQ/dt = a − 2bt, which falls to zero at t = a/2b. Heat H = ∫₀^(a/2b) (a − 2bt)²R dt = R·a³/(6b). Forgetting to square the time-varying current (using an average current) gives the larger wrong values.
Q15AIPMT 2015
A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is:
Not quite — the answer is B.
1 mV/cm = 0.1 V/m, so 0.4 V must drop across the 4 m, 8 Ω wire: I = 0.4/8 = 0.05 A. Total resistance = 2/0.05 = 40 Ω, so the series resistance is 40 − 8 = 32 Ω. 40 Ω is the total, not the added resistance.
Q16AIPMT 2015
Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is:
Not quite — the answer is C.
The conductor is a single series path, so charge conservation makes the current the same at every cross-section. Current density, field and drift velocity all vary inversely with the local area.
Q17AIPMT 2014
The resistances in the two arms of the meter bridge are 5 Ω and R Ω, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6 l₁. The resistance 'R', is: (l₁ is the initial balance length measured from end A, on the side of the 5 Ω resistance)
Not quite — the answer is A.
Initially 5/R = l₁/(100 − l₁). After shunting, the arm becomes R/2: 10/R = 1.6l₁/(100 − 1.6l₁). Dividing, 2 = 1.6(100 − l₁)/(100 − 1.6l₁) gives l₁ = 25 cm, so R = 5 × 75/25 = 15 Ω.
Q18AIPMT 2014
Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 Ω. The power loss in the wire is:
Not quite — the answer is A.
Over 150 km, the voltage drop is 150 × 8 = 1200 V and the resistance is 150 × 0.5 = 75 Ω. Power lost = V²/R = 1200²/75 = 19 200 W = 19.2 kW.
Q19AIPMT 2014
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of (i) infinity (ii) 9.5 Ω, the 'balancing lengths', on the potentiometer wire are found to be 3 m and 2.85 m, respectively. The value of internal resistance of the cell is:
Not quite — the answer is B.
With R = ∞ the balancing length gives the emf, and with R connected it gives the terminal voltage, so r = R(l₁/l₂ − 1) = 9.5 × (3/2.85 − 1) = 9.5 × 0.15/2.85 = 0.5 Ω.
Q20NEET 2013
A wire of resistance 4Ω is stretched to twice its original length. The resistance of stretched wire would be -
Not quite — the answer is D.
Stretching keeps the volume constant, so doubling the length halves the cross-section. R = ρl/A becomes ρ(2l)/(A/2) = 4R = 4 × 4 = 16 Ω. 8 Ω only accounts for the change in length.
Q21NEET 2013
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is
Not quite — the answer is B.
I = E/(R + r): 0.2 = 2.1/(10 + r), so 10 + r = 10.5 and r = 0.5 Ω.
Q22AIPMT 2012 Prelims
A ring is made of a wire having a resistance R₀ = 12 Ω. Points A and B on the ring divide it into two arcs of lengths l₁ and l₂. At which points A and B should a current carrying conductor be connected so that the resistance R of the sub circuit between these points is equal to 8/3 Ω?
Not quite — the answer is A.
The two arcs have resistances R₁ and R₂ with R₁ + R₂ = 12 Ω, and they are in parallel between A and B: R₁R₂/12 = 8/3, so R₁R₂ = 32. Then R₁ = 4 Ω and R₂ = 8 Ω, and since resistance is proportional to length, l₁/l₂ = 1/2.
Q23AIPMT 2012 Prelims
If voltage across a bulb rated 220 V–100 W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is
Not quite — the answer is D.
For a fixed resistance, P = V²/R, so ΔP/P ≈ 2ΔV/V = 2 × 2.5% = 5%. (Exactly, (0.975)² = 0.951, a 4.9% drop.) 2.5% wrongly assumes power proportional to voltage.
Q24AIPMT 2011 Mains
A galvanometer of resistance G is shunted by a resistance S ohm. To keep the main current in the circuit unchanged, the resistance to be put in series with the galvanometer is
Not quite — the answer is A.
For the main current to stay unchanged, the shunted galvanometer plus the series resistance must still equal G. The shunted combination is GS/(G + S), so R = G − GS/(G + S) = G²/(G + S). The answer GS/(G + S) is only the shunted combination itself, not the extra resistance needed.
Q25AIPMT 2011 Mains
A thermocouple of negligible resistance produces an e.m.f. of 40 µV/°C in the linear range of temperature. A galvanometer of resistance 10 ohm whose sensitivity is 1 µA/div, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be
Not quite — the answer is B.
The smallest detectable current is 1 µA (one division), so the smallest detectable emf is 1 µA × 10 Ω = 10 µV. With 40 µV per °C, ΔT = 10/40 = 0.25°C. Answers like 0.1°C or 0.5°C come from mixing up the ratio.
Q26AIPMT 2011 Prelims
A current of 2 A flows through a 2 Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω resistor. The internal resistance of the battery is
Not quite — the answer is C.
The emf is the same in both cases: ε = 2(2 + r) = 0.5(9 + r). So 4 + 2r = 4.5 + 0.5r, giving 1.5r = 0.5 and r = 1/3 Ω (ε = 4.67 V). Checking 1/3 Ω: 2 × 2.333 = 4.67 and 0.5 × 9.333 = 4.67, consistent.
Q27AIPMT 2010 Prelims · Q30
Consider the following two statements: (A) Kirchhoff's junction law follows from the conservation of charge. (B) Kirchhoff's loop law follows from the conservation of energy. Which of the following is correct?
Not quite — the answer is A.
The junction rule says charge does not pile up at a node: conservation of charge. The loop rule says the net potential change round a closed loop is zero, i.e. work per unit charge round a loop is zero: conservation of energy. Both statements are correct.
Q28AIPMT 2010 Prelims · Q31
In producing chlorine by electrolysis 100 kW power at 125 V is being consumed. How much chlorine per minute is liberated? (E.C.E. of chlorine is 0.367 × 10⁻⁶ kg/C)
Not quite — the answer is D.
I = P/V = 100000/125 = 800 A. In one minute m = ZIt = 0.367 × 10⁻⁶ × 800 × 60 = 17.6 × 10⁻³ kg. 3.67 × 10⁻³ kg would need a far smaller charge.
Q29AIPMT 2009 Prelims · Q33
A student measures the terminal potential difference (V) of a cell (of emf ε and internal resistance r) as a function of the current (I) flowing through it. The slope, and intercept, of the graph between V and I, then, respectively, equal:
Not quite — the answer is A.
Terminal voltage V = ε − Ir, a straight line in I. Its slope is −r and its intercept on the V axis (I = 0) is ε. A positive slope r would mean the voltage rises as more current is drawn.
Q30AIPMT 2009 Prelims · Q46
The mean free path of electrons in a metal is 4 × 10⁻⁸ m. The electric field which can give on an average 2 eV energy to an electron in the metal will be in units of V/m:
Not quite — the answer is C.
An electron gains energy eEλ while travelling one mean free path. Setting eEλ = 2 eV gives E = 2 V/(4 × 10⁻⁸ m) = 5 × 10⁷ V/m. The very small values (10⁻¹¹) would come from multiplying by λ instead of dividing.
Q31AIPMT 2008 Prelims · Q29
A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively
Not quite — the answer is C.
The volume stays constant, so R = ρl/A = ρl²/V is proportional to l². A 10% stretch gives R' = (1.1)²R = 1.21R. Specific resistance (resistivity) depends only on the material and temperature, so it stays the same.
Q32AIPMT 2008 Prelims · Q31
An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20°C? The temperature of boiling water is 100°C.
Not quite — the answer is B.
Heat needed = mcΔθ = 1 × 4200 × 80 = 3.36 × 10⁵ J; power = VI = 220 × 4 = 880 W. t = 3.36 × 10⁵/880 ≈ 382 s ≈ 6.4 min, closest to 6.3 min. Heating from 0°C to 100°C instead of 20°C would give about 8 min.
Q33AIPMT 2008 Prelims · Q36
A galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
Not quite — the answer is A.
Full-scale current = 3/(50 + 2950) = 1 mA for 30 divisions. For 20 divisions I = (2/3) mA, so total resistance = 3/(2/3 × 10⁻³) = 4500 Ω and the series resistance = 4500 − 50 = 4450 Ω. 4500 Ω would forget to subtract the galvanometer's own resistance.
Q34AIPMT 2007 Prelims · Q20
Three resistances P, Q, R each of 2Ω and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of 6Ω is connected in parallel to S the bridge gets balanced. What is the value of S?
Not quite — the answer is A.
Balance requires P/Q = R/S', so the effective arm S' = 2 Ω. S in parallel with 6 Ω: 6S/(S + 6) = 2 gives 4S = 12, so S = 3 Ω. Answering 2 Ω forgets that S is in parallel with the 6 Ω.
Q35AIPMT 2007 Prelims · Q41
The resistance of an ammeter is 13 Ω and its scale is graduated for a current upto 100 amps. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto 750 amperes by this meter. The value of shunt-resistance is
Not quite — the answer is A.
The meter still carries 100 A, so the shunt carries 650 A at the same voltage: S = 13 × 100/650 = 2 Ω. 0.2 Ω would be a power-of-ten slip.
Q36AIPMT 2006 Prelims · Q6
Two cells, having the same emf, are connected in series through an external resistance R. Cells have internal resistances r₁ and r₂ (r₁ > r₂) respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of R is :
Not quite — the answer is A.
The current is I = 2E/(R + r₁ + r₂). The p.d. across the first cell is E − Ir₁ = 0, so I = E/r₁. Equating: 2r₁ = R + r₁ + r₂, giving R = r₁ − r₂. Setting the second cell's p.d. to zero would give r₂ − r₁, which is negative here.
Q37AIPMT 2006 Prelims · Q37
Kirchhoff's first and second laws for electrical circuits are consequences of :
Not quite — the answer is B.
Kirchhoff's junction law (sum of currents at a node is zero) expresses conservation of charge. The loop law (sum of potential changes round a loop is zero) expresses conservation of energy. The reverse pairing swaps the two laws.
Q38AIPMT 2005 Prelims · Q37
When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between two of diametrically opposite points will be :
Not quite — the answer is A.
Each semicircle has resistance R/2 and the two halves are in parallel between diametrically opposite points: (R/2 × R/2)/(R/2 + R/2) = R/4. R/2 is the resistance of only one half.
Q39AIPMT 2005 Prelims · Q41
A 5-A fuse wire can withstand a maximum power of 1 W in circuit. The resistance of the fuse wire is :
Not quite — the answer is D.
P = I²R, so R = P/I² = 1/25 = 0.04 Ω. Using P = IR gives 0.2 Ω.
How often Current Electricity came up
Questions from this chapter in each paper on this site.
- NEET 2021: 3
- NEET 2020: 4
- NEET 2019: 1
- NEET 2018: 2
- NEET 2017: 2
- NEET 2016 Phase 2: 1
- NEET 2016 Phase 1: 1
- AIPMT 2015: 2
- AIPMT 2014: 3
- NEET 2013: 2
- AIPMT 2012 Prelims: 2
- AIPMT 2011 Mains: 2
- AIPMT 2011 Prelims: 1
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 2
- AIPMT 2008 Prelims: 3
- AIPMT 2007 Prelims: 2
- AIPMT 2006 Prelims: 2
- AIPMT 2005 Prelims: 2
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