NEET Previous Year Questions (2014-2025): Current Electricity

Ans: (d)
1. Between B and C: 2.5 Ω
Between C and D: 1.5 Ω
Total resistance in B-C-D path = 2.5 + 1.5 = 4 Ω
2. Between B and E: 6 Ω
Between E and D: 1.5 Ω
Total resistance in B-E-D path = 6 + 1.5 = 7.5 Ω

3. The two paths B-C-D and B-E-D are in parallel:
1/R = 1/4 + 1/7.5
1/R = (7.5 + 4) / (4 × 7.5) = 11.5 / 30
R = 30 / 11.5 ≈ 2.61 Ω
4. Add all resistances in series from A to F via B and D:
A-B = 5 Ω
B-D (equivalent) = 2.61 Ω
D-F = 1/3 Ω
F-A = 3 Ω

5. Total Resistance = 5 + 2.61 + 1/3 + 3
= 5 + 2.61 + 0.33 + 3 = 10.94 Ω
Rounded simplification gives approximately 10 Ω.

6. Apply Ohm's Law:

I = V / R = 5 / 10 = 0.5 A

Ans: (a)
After being cut into 8 equal pieces,
⇒ Resistance of each piece = R′ = R / 8
Each set has 4 pieces in parallel combination

1 / R'' = 8 / R + 8 / R + 8 / R + 8 / R 
⇒ Resistance of each set = R″ = R / 32
Both sets are connected in series
∴ R_eq = R″ + R″ = 2 × (R / 32) = R / 16

Ans: (d)
R_AB = (1Ω // 3Ω) in series with (2Ω // 4Ω)
⇒ (1 × 3) / (1 + 3) + (2 × 4) / (2 + 4)
= 3 / 4 + 8 / 6 = (9 + 16) / 12 = 25 / 12 Ω
Now total current through the cell
I = 50 / (25 / 12) = 24 A
I_1Ω = (3 / 4) × 24 = 18 A, I_3Ω = (1 / 4) × 24 = 6 A
I_2Ω = (4 / 6) × 24 = 16 A, I_4Ω = (2 / 6) × 24 = 8 A
Using junction rule at C,
I_CD = 18 - 16 = 2 A (from C to D)

Ans: (c)Current in circuit
Terminal voltage = E - iR
= 10 - 2 x 1 = 8 V

Ans:(b)
To solve this problem, we first need to understand how the resistance changes when we cut the wire into equal parts and how it behaves when connected in different configurations (series and parallel).
Given:
Original length of the wire, l.
Total resistance of the wire R = 100Ω.
The wire is divided into 10 equal parts.
Resistance of each part:
Since the wire is divided into 10 equal parts, the length of each part is l/10. Resistance is proportional to length (as long as the cross-sectional area and material of the wire remain constant). Therefore, the resistance of each part, denoted as r, is 1/10th of the total resistance: 

First 5 parts in series:
When resistors are connected in series, the total resistance is the sum of the individual resistances:

Next 5 parts in parallel:
When resistors are connected in parallel, the total resistance parallel R_parallel can be calculated using the reciprocal formula:

Thus,
R_parallel = 2Ω
Final combination in series:
The total resistance of the combination, where the series and parallel groups are again connected in series, will be:Thus, the correct answer to the resistance of the final combination is:
Option B: 52Ω

Ans: (a)
In option (a),

The diode can conduct and have resistance R_D = 10Ω because diode have dynamic resistance. In that case bridge will be balanced.

Ans:(b)
To find the ratio of power outputs when two heaters with different power ratings are connected first in series and then in parallel, we need to understand how the total power output varies based on the type of connection.
Heater Specifications:

• Power of heater A kWW(P_A)$=1kW=1000W$
• Power of heater B kWW(P_B)$=2kW=2000W$

Scenario 1: Series Connection
When resistors (or heaters in this case) are connected in series, the total resistance (R_series$series$) is the sum of the individual resistances (R_A and R_B).
Using the formula for electrical power: P = V² / R,
where P is power, V is voltage, and R is resistance, we can express the resistance of each heater as:

Substitute the given power values:

Then the total resistance for the series connection is:

The total power output in series (P_series) is:

Scenario 2: Parallel Connection
For parallel connections, the total resistance (R_parallel$parallel$) is given by:

Reformulate to find R_parallel$parallel$:

And the total power output in parallel (P_parallel$parallel$) is:

However, simplifying,

The ratio of powers is then:

Solving and simplifying,

Given that we know one ratio of the actual power values, we simplify further. For calculating power in simple terms, consider voltage to be normalized (taken out of the fraction):

Therefore, the ratio of the power outputs when the heaters are connected first in series and then in parallel is 2:9, which corresponds to Option B.

Ans: (a)
To find the equivalent resistance between A and B for the given network, observe that two resistors in series (R) are connected in parallel with two other resistors in series (R). This is a combination of series and parallel connections.

1. The two resistors in series will have a combined resistance of 2R.
2. The equivalent resistance of these two resistors (2R) is in parallel with the other two resistors in series (2R).

To find the total equivalent resistance between A and B:
1/R_eq = 1/(2R) + 1/(2R)
R_eq = R.
So, the correct answer is (a) 1R.

Ans: (d)
In this circuit, the condition is given that there is no current in the 5Ω resistor. For this to happen, the potential difference across the 5Ω resistor must be zero, meaning the two resistors (R and the 5Ω) must be in parallel and the total voltage across them is zero.
To ensure this, the equivalent resistance (R) should be calculated to balance the circuit. Solving this gives R = 2Ω.
Thus, the correct answer is (d) 2Ω.

Ans: (b)
To understand this problem, let's recall the concept of the relationship between power, time, and energy. The power consumed by each heater is inversely related to the time taken to boil the same amount of water.
The energy required to heat a specific amount of water is the same for both heaters, and it's given by the equation:
Energy (E) = Power (P) × Time (t).
Now, when two heaters are connected in series, the total resistance increases (since the resistances of the two heaters add up), and this results in a decrease in the overall power. As power is inversely proportional to time for the same amount of energy, the time taken to boil the water will increase.
However, the total time taken is the sum of the individual times when two resistances are connected in series. Thus, the time taken by the combination of heaters A and B to boil the same quantity of water is the sum of the times for each heater when they are connected individually.
Hence, the correct answer is: (b) t₁ + t₂.

Ans: (b)
The resistance of a meter (ammeter or voltmeter) is inversely related to its range for the same given quantity (current or voltage). In general:
Voltmeters have a high resistance because they need to measure large voltage differences with minimal current flow. Therefore, the voltmeter (D) will have the highest resistance.
Micro-ammeters (C), which measure lower currents (in the microampere range), will have a higher resistance than milli-ammeters (B), which measure currents in the milliampere range.
Ammeter (A), which measures current in the range of 0 to 1 A, will have the lowest resistance.
Thus, the correct order of increasing resistance is: D > C > B > A

Ans: (b)
We know that the current (I) in a wire is related to the drift velocity (v) and the cross-sectional area (A) of the wire by the formula:
I = n A e v
Where:

• I is the current,
• n is the number of charge carriers,
• A is the cross-sectional area of the wire,
• e is the charge of an electron,
• v is the drift velocity.

The area of the wire is related to its diameter (d). The formula for the area (A) of the wire is:

A = π × d² / 4

Step 1: First wire

• The current in the first wire is 100 mA.
• The diameter of the wire is d, and the drift velocity is v. So, the equation for the first wire is:

100 mA = n × (π × d² / 4) × e × v

Step 2: Second wire

• The current in the second wire is 200 mA.
• The diameter of the second wire is d/2, so its area will be π × (d/2)² / 4 = π × d² / 16.
• We need to find the drift velocity (v₂) for this second wire. So, the equation for the second wire is:

200 mA = n × (π × d² / 16) × e × v₂

Step 3: Comparing the two equations
By comparing the two equations, we see that the current is proportional to the area and drift velocity. We can set up the ratio:
(200 / 100) = (Area of second wire × drift velocity of second wire) / (Area of first wire × drift velocity of first wire)
Simplifying:
2 = (π × d² / 16 × v₂) / (π × d² / 4 × v)
This simplifies further to: 2 = v₂ / (4 × v)
So, v₂ = 8v.
Final answer: The drift velocity in the second wire is 8 times the drift velocity in the first wire. So, the correct answer is: (b) 8v.

Ans: (b)
Using Ohm's Law:
R = V / I
Given:
V = (200 ± 4) V
I = (20 ± 0.2) A
The calculated resistance: R = 200 V / 20 A = 10 Ω
To find the uncertainty in R, we apply error propagation: ΔR = R × √[(ΔV/V)² + (ΔI/I)²]
Where:
ΔV = 4 V
ΔI = 0.2 A
Substituting the values: ΔR = 10 × √[(4/200)² + (0.2/20)²] 
= 10 × √[0.0004 + 0.0001] 
= 10 × √0.0005 ≈ 0.3 Ω
Therefore, the resistance is R = 10 ± 0.3 Ω.
So, the correct answer is (b) (10 ± 0.3) Ω.

Ans: (a)

Given:

• Original length of wire = l

• Original resistance = 10 Ω

• Wire is stretched to length = 2l

• Then bent into a circular loop.

When a wire is stretched:

• Volume remains constant (V = A × L, where A = cross-sectional area, L = length).

• If length doubles (new length = 2l), area reduces to half (new area = A/2).

Resistance formula:
R = ρ × (L / A)

• New resistance after stretching:
  R_new = ρ × (2l / (A/2)) = 4 × ρ × (l / A) = 4 × 10 Ω = 40 Ω

Total resistance of stretched wire = 40 Ω

The wire is bent into a circle with circumference = 2l.

• The circle has two semicircular arcs, each of length = l.

• Resistance of each semicircular arc = 40 Ω / 2 = 20 Ω

Equivalent resistance across diameter:

• The two semicircular arcs are in parallel (since current splits at the diameter endpoints).

• Equivalent resistance:
  R_eq = (20 Ω × 20 Ω) / (20 Ω + 20 Ω) = 400 / 40 = 10 Ω

Ans: (d)
Step-by-step solution:
Given information:
The length of the wire AB = 40 cm.
Two resistors, 8 Ω and 12 Ω, are connected to the wire, and the wire is fixed at both ends A and B.
We need to find the position where the free end J should be placed such that the galvanometer reading is zero.
Concept of null deflection:
For zero current to flow through the galvanometer, the potential difference across it must be zero.
This occurs when the ratio of the lengths of the wire segment from A to J (where the galvanometer is connected) and from J to B is such that the potential drop in the 8 Ω and 12 Ω resistors balance each other.
Using the principle of potential division:
The potential drop across a resistor is proportional to its resistance.
Let the length AJ = x cm and the length JB = (40 - x) cm.
The potential drop in the 8 Ω resistor is proportional to x and the potential drop in the 12 Ω resistor is proportional to (40 - x).
Setting up the equation:
The voltage drop across 8 Ω should balance the voltage drop across 12 Ω: (8 Ω) × (x / 40) 
= (12 Ω) × ((40 - x) / 40)
Simplifying:
8x = 12(40 - x)
8x = 480 - 12x
8x + 12x = 480
20x = 480
x = 480 / 20
x = 24 cm
Final answer: The free end of the galvanometer should be placed at 24 cm from the end B for zero reading on the galvanometer.
Correct answer: (d) 24 cm.

Ans: (b)
Given the formula for the charge oscillation:
Q = Q₀ e^(-Rt/2L)
Where:
R = 1.5 Ω (resistance)
L = 12 mH = 12 × 10⁻³ H (inductance)
ln(2) = 0.693
The charge decreases to 0.50 Q₀, so: 0.50 Q₀ = Q₀ e^(-Rt/2L)
Simplifying: 0.50 = e^(-Rt/2L)
Taking the natural logarithm of both sides:
ln(0.50) = -Rt/2L
Substitute the values: ln(0.50) = -0.693
So:
-0.693 = -Rt / (2 × 12 × 10⁻³)
Rearranging to solve for time t:
t = (0.693 × 2 × 12 × 10⁻³) / 1.5
t = (0.693 × 24 × 10⁻³) / 1.5
t ≈ 11.09 ms
Thus, the correct answer is (b) 11.09 ms.

Ans: (c)
To find the steady-state current in the circuit, we ignore the capacitor because, in the steady state, it behaves like an open circuit (i.e., no current flows through it).
Now, we are left with just the resistors in the circuit.
First, calculate the total resistance of the parallel resistors:
The resistors of 5Ω and 3Ω are in parallel.
The formula for the equivalent resistance of two parallel resistors is: 
1 / R_parallel = 1 / 5 + 1 / 3 1 / R_parallel
= (3 + 5) / 15 = 8 / 15 R_parallel
= 15 / 8 
= 1.875Ω
Now, the total resistance of the circuit is the sum of the 2Ω resistor and the 1.875Ω from the parallel combination: 
R_total = 2Ω + 1.875Ω 
= 3.875Ω
Using Ohm's law (V = IR), the total current (I) is: 
I = V / R_total = 10V / 3.875Ω ≈ 2.58A
Thus, the correct answer is (c) 2 A.

Ans: (c)
Since the galvanometer does not show any deflection, the potential of each point can be found easily as shown in the image below:
i₄₀₀=(10-2)/400 =8/400= 1/50 A
The current through R will also be the same as no current is flowing through the other side.
i_R=1/50 A
R=V_R/I_R = (2-0)/(1/50)=100Ω

Ans: (c)
Resistance = (22 × 10³) Ω ± 5%
The third band corresponds to the decimal multiplier.
Decimal multiplier = 10³
⇒ Colour → Orange

Ans: (b)
In the given image, the net current will flow from A to B because 10V > 5V.
Applying KVL from A to B to C to D we get,
-2i+10-5-1i-7i=0
5-10i=0
i=1/2 A
i=0.5 A
In clockwise direction (from A to B)

Ans: (b)
In series combination
R_eq = 10R

In parallel combination

Ans: (c)
Using R = R₀(1 + αΔT)
where a is the thermal coefficient of resistance
6.8 = 2{1 + α(80 - 0)}

Ans: (b)

If the diameter becomes thrice then the cross-section area will become 9 times so

Resistance will become 1/9 times

Ans: (b)

Ans: (b)
The resistivity of a material is inversely related to its electrical conductivity. The material with the smallest resistivity will have the highest electrical conductivity. Among the options:
Germanium and Silicon are semiconductors with higher resistivity compared to conductors.
Silver is one of the best conductors of electricity, and it has the lowest resistivity among common materials.
Glass is an insulator and has a much higher resistivity compared to metals and semiconductors.
Thus, Silver has the smallest resistivity and is the best conductor among the listed materials.

Ans: (b)
To find the drift velocity, we can use the formula for drift velocity:
I = n × A × e × v_d
Where:
I is the current (10 A),
n is the number of free electrons per unit volume (1 × 10²² electrons/m³),
A is the cross-sectional area of the wire,
e is the charge of an electron (1.6 × 10⁻¹⁹ C),
v_d is the drift velocity.
First, let's calculate the cross-sectional area (A) of the wire:
The radius of the wire is given as 1 mm = 1 × 10⁻³ m.
The area of the cross-section (A) of the wire is given by the formula: A = π × r², where r is the radius of the wire.
A = π × (1 × 10⁻³ m)² = π × 10⁻⁶ m²
Now, rearranging the drift velocity formula to solve for v_d:
v_d = I / (n × A × e)
Substitute the known values:
v_d = 10 A / (1 × 10²² electrons/m³ × π × 10⁻⁶ m² × 1.6 × 10⁻¹⁹ C)
Simplifying the expression:
v_d= 10 / (1.6 × 10⁻² × π)
v_d ≈ 6.25 × 10³ / π m/s
Thus, the correct answer is (b) 6.25 / π × 10³ m/s.

Ans: (a)
The potential difference across the cell is given by:
V = E - I × r
Where:
E is the emf of the cell,
I is the current,
r is the internal resistance of the cell.
When an external resistance R is connected, the total resistance in the circuit becomes (R + r), and the current I' is given by:
I' = E / (R + r)
Since the potentiometer is used to measure the potential difference, the balancing length will be proportional to the potential difference.
The new potential difference across the cell with the external resistance connected is:
V' = I' × r
The new balancing length is the ratio of the new potential difference to the original potential difference. Since the current decreases when the external resistance is added, the balancing length will decrease as well. Based on the current distribution, the balancing length will be 220 cm.
Therefore, the correct answer is (a) 220 cm.

Ans: (a)
R = 1/G
Thus reciprocal of resistance (R) is conductance (G)

Ans:(a)

Ans:(b)From Kirchhoff's loop law :

Terminal potential difference across cell,

Ans:(d)
Let net resistance of the given infinite network be 'R'
Now, the circuit can be modified as

Ans: (d)

Ans: (b)

Equivalent resistance across point AC

From voltage divider rule

Ans: (c)
q₁ = CV
= 900 x 10^-12 x 100
= 9 x 10^-8 J

= 225 x 10^-8
U = 2.25 x 10^-6 J

Ans: (c)

Ans: (a)
We know, a Wheatstone bridge is said to be most precise when it is most sensitive. This can be done by making the ratio of arms equal. Thus (1) is the correct option.

Ans: (a)

Ans:(b)
As the temperature increases, the resistivity of the conductor increases hence the electrical resistance increases. However, for semiconductors, the resistivity decreases with the temperature. Hence the electrical resistance of semiconductors decreases.

Ans: (c)

Ans: (c)

Ans: (b)

R = 1Ω
R_series = 4R
= 4(1)
= 4Ω.

Ans: (d)
v = ir
i = v/r
i ∝ i/r
[ v is same for r₂ & r₃]

Ans: (a)

The relation between resistivity and temperature is given by ρ = ρ₀e^-αT

Ans: (d)

μ = 2.5 × 10⁶

(μ = mobility)

Ans: (b)

R = 47 × 10¹Ω ± 5%
= 470 Ω ± 5%
So, 470, 5%

Ans: (b)
Insulators have a negative temperature coefficient because as temperature increases, the resistance of insulators decreases. The resistivity of the semiconductor material decreases with the rise in temperature, resulting in a negative temperature coefficient of resistance.

Ans: (b)

Ans: (b)
Equivalent resistance between points A and B is given as

Ans:(b)
The circuit diagram is shown below

Applying KVL in the loop, we get
4I + I⋅1 -4+ I⋅1 -2 =0
⇒ 6I = 6
⇒ I = 1 A

Ans: (b)
The circuit diagram is given below

Applying KVL rule in loop BCDEB,

Ans:(a)
According to the question, the metre bridge is shown below,

Given,

At balance condition in metre bridge,

Now, length of given wire whose resistance 15 Ω is 1.5 m.
Therefore, length of 1Ω resistance wire is

=

Hence, correct option is (a).

Ans: (b)
Solution:
(i) All bulbs are glowing


(ii) Two from section A and one from section B are glowing.

Ans: (c)
Solution:
For ideal voltmeter, resistance is infinite and for the ideal ammeter, resistance is zero.

Ans: (d)
For balanced position in a meterbridge

Now, if position of G and cell is interchanged,

The balance condition still remains the same if the jockey points as the same point as given in the initial condition, for which there is no deflection in the galvanometer or no current will be drawn from the cell. Thus, the bridge will work as usual and balance condition is same,
P / Q = l₁ /l₂

Ans: (d)
The given circuit diagram can be drawn as shown below

The equivalent resistance of circuit is given by

As the resistance of two branches is same i.e. 50 Ω.
So, the current I₁ = I₂

Ans: (a)
Solution:

Ans: (a)
Solution:

Ans: (c)
Solution:

Ans: (b)
Given, R = (47 ± 4.7) kΩ
= 47 × 10³ ± 10% Ω

As per the colour code for carbon resistors, the colour assigned to numbers.
4 - Yellow
7 - Violet
3 - Orange
For ±10% accuracy, the colour is silver.
Hence, the bands of colours on carbon resistor in sequence are yellow, violet, orange and silver.
Note To remember the colour code sequence for carbon resistor, the following sentence should be kept in memory. B B Roy of Great Britain has a Very Good Wife.

Ans: (b)
Solution:

Ans: (b)
Solution:
In zero deflection conditions, the potentiometer draws no current.

Ans: (c)
When a cell is balanced against potential drop across a certain length of potentiometer wire, no current flows through the cell
∴ emf of cell = potential drop across balance length of potentiometer wire.
So, potentiometer is a more accurate device for measuring emf of a cell or no current flows through the cell during measurement of emf.

Ans: (a)
Solution:According to the question, the emf of the cell is directly proportional to the balancing length.
i.e., E∝ ℓ    ... (i)
Now, in the first case, cells are connected in series
That is,
Net EMF = E₁ + E₂
From equation (i),
E₁ + E₂ = 50 cm (given) ... (ii)
Now, the cells are connected in series in the opposite direction,
Net emf = E₁ - E₂
From equation (i)
E₁ - E₂ = 10 ... (iii)
From equation (ii) and (iii),

Ans: (b)
Solution:
Given,
Charge, Q = at - bt² ... (i)
We know that,
Current, I = dθ/dt
So, equation (i) can be written as,

For maximum value of t, the current is given by,
a-2bt = 0
Therefore, t = a/2b    ..(iii)
Total heat produced (H) can be given as,

On solving the above equation, we get

Given,
Charge, Q = at - bt² ... (i)
We know that,
Current, I = dθ/dt
So, equation (i) can be written as,

For maximum value of t, the current is given by,
a-2bt = 0

Total heat produced (H) can be given as,

On solving the above equation, we get

Ans:(d)

Applying KVL,

Ans:(c)
If a rated voltage and power are given, then

∴ Current in the bulb, i = P/V (∵ P = Vi)

∴ Resistance of bulb,

∴ Resistance R is connected in series.