38 NEET previous-year questions on Current Electricity, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.
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19.2 W
19.2 kW
19.2 J
12.2 kW
Solution
The total resistance . The total potential drop . Using , . Therefore, option (b) is correct.
10 Ω
15 Ω
20 Ω
25 Ω
Solution
For the meter bridge, the balance condition is . When is shunted with an equal resistance, the new resistance is , and the new balance condition is . Simplifying, . From the first condition, . Substituting, . Solving, , so option (b) is correct.
0.25 Ω
0.95 Ω
0.5 Ω
0.75 Ω
Solution
The internal resistance can be found using the formula , where and are the balancing lengths for and , respectively. Substituting the values, , so option (c) is correct.
LE0r l r1
E0r (r+r1). l L
E0l L
LE0r (r+r1)l
Solution
K= potential gradient =( E0r r+r1 ) 1 L So E=Kℓ= E0rℓ (r+r1)L
2 σ1 σ2 σ1+σ2
σ1+σ2 2 σ1 σ2 www.vedantu.com 31
σ1+σ2 σ1 σ2
σ1 σ2 σ1+σ2
Solution
Rec= ℓ σ1A+ ℓ σ1A= ℓeq σeqAeq 𝟐ℓ σeqA=ℓ A(σ1+σ2 σ1σ2 ) σeq=2σ1σ2 σ1σ2
0.5 A
0.25 A
2 A
1 A
Solution
Resistance of ammeter = 480×20 480+20 =19.2Ω i= 30 40.8+19.2=0.5 A
a3R 6b
a3R 3b
a3R 2b
a3R b
Solution
Q =at− bt2 ∴ t ϵ[0,a b] i =dq dt = a − 2bt Note: i is +ve t ∈ (0, a 2b) And i is – ve t ∈ ( a 2b, a b) Positive current means current one direction and negative current means current in opposite direction. ∴ dH= i2Rdt = (a − 2bt)2R dt H = ∫(a − 2bt)2 R dt a b 0 = (a − 2bt)3R 3(−2b) | 0 a b = 1 −b[(a −2ba b) 3 − (a)3]R www.vedantu.com 11 = − 1 6b[(−a)3 − a3]R H =a3R 3b
5 : 1
5 : 4
3 : 4
3 : 2
Solution
E1 + E2 = λ50 E1 − E2 = λ10 E1 + E2 = 5E1 − 5E2 6E2 = 4E1 3 2 = E1 E2 www.vedantu.com 21
nR
R n
n2R
2 R n
Solution
2 22 21 1 Rl R l = 22 1 2 1 nl l = 22 1 R nR = R2 = n2R1 www.vedantu.com 12
Yellow – Green – Violet – Gold
Yellow – Violet – Orange – Silver
Violet – Yellow – Orange – Silver
Green – Orange – Violet – Gold
Solution
The colour code for a resistor is determined by the values and tolerances. For a resistor of , the first band is yellow (4), the second band is violet (7), the third band is orange (1000 or ), and the fourth band is gold (±5%). Therefore, the correct sequence is Yellow – Violet – Orange – Gold, so option (a) is correct.
20
11
10
9
Solution
For resistors in series, the total resistance is . The current . For resistors in parallel, the total resistance is . The current . Equating, , so option (c) is correct.
Rainbow is combined effect of dispersion, refraction and reflection of sunlight.
When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed.
The order of colours is reversed in the secondary rainbow.
An observer can see a rainbow when his front is towards the sun.
Solution
Conceptual
2 protons only
2 protons and 2 neutrons only
2 electrons, 2 protons and 2 neutrons
2 electrons and 4 protons only
Solution
Conceptual
2.25 × 10 15
2.5 × 10 6
2.5 × 10 − 6
2.25 × 10 − 15
Solution
Mobility is given by , where is the drift velocity and is the electric field. Substituting the values, , so option (b) is correct.
Solution
For copper, resistivity increases linearly with temperature, as described by the relation , where is the temperature coefficient of resistivity. Option (a) correctly shows this linear increase, so it is the correct choice.
470 k Ω , 5%
47 k Ω , 10%
4.7 k Ω , 5%
470 Ω , 5%
Solution
The color code for the resistance is given as yellow, violet, orange, and gold, corresponding to 4, 7, 000, and 5% tolerance. Thus, the resistance value is with a 5% tolerance, so option (a) is correct.
1.0 × 10 − 2 m
1.0 × 10 − 1 m
1.5 × 10 − 1 m
1.5 × 10 − 2 m
Solution
The length of 1 Ω of the resistance wire is calculated as . Therefore, option (b) is correct.
-(R),
-(Q),
-(S),
-(P)
Solution
- Drift Velocity (A) is given by , matching with (R).
- Electrical Resistivity (B) is given by , which is not directly listed but can be rearranged to match (S) as .
- Relaxation Period (C) is given by , matching with (P).
- Current Density (D) is given by , matching with (Q).
Thus, the correct matches are (A)-(R), (B)-(S), (C)-(P), (D)-(Q), corresponding to option (1).
60 cm
21.6 cm
64 cm
62 cm
Solution
The balance point in a potentiometer is given by the ratio of the EMFs of the cells. Using the formula , where , , and , we get . Therefore, option (a) is correct.
0.25 Ω
0.5 Ω
1 Ω
4 Ω
Solution
For four resistors of equal resistance in parallel, the effective resistance . Given , . For the same resistors in series, , so option (d) is correct.
1 2 3 r r r +
2 2 3 r r r +
1 1 2 r r r +
2 1 3 r r r +
Solution
In the given circuit, the current splits between and in parallel. The ratio of currents to is given by the inverse ratio of the resistances in parallel: . Therefore, the correct option is (b).
Increases for both conductors and semiconductors
Decreases for both conductors and semiconductors
Increases for conductors but decreases for semiconductors
Decreases for conductors but increases for semiconductors
Solution
As the temperature increases the resistivity of the conductor increases hence the electrical resistance increases. However for semiconductor the resistivity decreases with the temperature. Hence electrical resistance of semiconductor decreases.
1 : 2
2 : 1
1 : 4
4 : 1
Solution
For parallel combination 2 = VP R 12 21 =PR PR ⇒ 1 2 200 2 100 1= =P P
104 A/m2
106 A/m2
10–5 A/m2
105 A/m2
Solution
Resistance, = ρ= σ LLR AA ⇒ σ= L RA Also current density = σ= LEjE RA 2 42 10 10 100 1010 1010 −− ×= = ×π× ×π π π j = 105 A/m2
Should be approximately equal to 2X
Should be approximately equal and are small
Should be very large and unequal
Do not play any significant role
Solution
We know, a wheatstone bridge is said to be most precise when it is most sensitive. This can be done by making ratio arms equal. Thus (2) is correct option.
400 Ω
200 Ω
50 Ω
100 Ω
Solution
For the galvanometer to show no deflection, the bridge must be balanced, so . Solving, , but the closest option is 100 Ω, so option (d) is correct.
Yellow
Red
Green
Orange
Solution
The third band in a resistor color code indicates the multiplier. For a resistance of , the multiplier is , which corresponds to the color orange. The tolerance band (±5%) is typically gold, but this is not asked. Thus, the correct option is (d).
1.5 A from B to A through E
0.2 A from B to A through E
0.5 A from A to B through E
5/9 A from A to B through E
Solution
Using Ohm's law, . For the circuit, and , so . The current flows from the positive terminal (A) to the negative terminal (B) through E, making option (c) correct.
1000
10
100
1
Solution
In series, the total resistance is , and the current is . In parallel, the total resistance is , and the current is . The ratio of the currents is , so option (c) is correct.
3×10⁻¹ °C⁻¹
3×10⁻⁴ °C⁻¹
3×10⁻³ °C⁻¹
3×10⁻² °C⁻¹
Solution
The temperature coefficient of resistance is given by . Substituting the values, , so option (d) is correct.
4 V
6 V
8 V
10 V
Solution
Current in circuit i = 10 2A41 =+ Terminal voltage = E – iR = 10 – 2 × 1 = 8 V
26 Ω
52 Ω
55 Ω
60 Ω
Solution
Divided into 10 parts ρ= lR A 10 10 ρ′ == lRR A 5 10=×S RR [series] RS = 50 50=P RR [parallel] Req = RS + RP = 52 Ω
(Image option — will be added soon) (1)
(Image option — will be added soon) (2)
(Image option — will be added soon) (3)
(Image option — will be added soon) (4)
Solution
In option (1), 10 10 15 5 DR= + The diode can conduct and have resistance RD = 10 Ω because diode have dynamic resistance. In that case bridge will be balanced.
1 : 1
2 : 9
1 : 2
2 : 3
Solution
Power Consumed = 2VP R= AB BA PR PR= RA = 2RB For Series Combination 2 3S B VP R= For Parallel Combination 23 2P B VP R= 2 9 S P P P = - 20 - NEET (UG)-2024 (Code-Q1)
(1) 2.0 A
(2) 0.5 A
(3) 2.5 A
(4) 1.5 A
Solution
Sol.
Balanced wheatstone bridge
its equivalent
Circuit can be redrawn as
5 volt
6 volt
9 volt
10 volt
Solution
Given, and
Solution
Sol. After being cut into 8 equal pieces,
⇒ Resistance of each piece =
Each set has 4 pieces in parallel combination
⇒ Resistance of each set =
Both sets are connected in series
∴
1.5 A
2.0 A
2.5 A
3.0 A
Solution
Total current through the cell:
Using the junction rule at C: (from C to D).
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