WBJEE 2025 Current Electricity Practice Questions with Solutions

WBJEE 2025 Current Electricity Practice Questions with Solutions: Current Electricity is a fundamental chapter in intermediate Physics syllabus. The chapter is also a part of WBJEE syllabus. The WBJEE 2025 exam paper is expected to have a total of 2-3 questions (approximately 6% of the total syllabus weightage) from this chapter. Therefore, to score full marks in this section, you must be adept with the WBJEE Current Electricity sample questions. Current Electricity being an important chapter in the WBJEE 2025 Physics syllabus PDF covers various topics, such as Ohm's Law, Electrical Resistance, Ohmic and Non Ohmic Conductors, Electrical Energy & Power etc. Since these topics are majorly based on numerical solutions and application of various laws and theories of Physics, you must solve practice questions for WBJEE 2025 Current Electricity while studying these. Over the years, there has not been any major shift in the question pattern, and so, based on the past few years' question paper anlaysis, we have provided a series of WBJEE 2025 Current Electricity practice questions with solutions in this article.

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List of WBJEE 2025 Current Electricity Sub Topics for Exam Preparation

The chapter on Current Electricity in WBJEE is important and to score well in the exam, you must be well aware of all the topics covered in the WBJEE 2025 syllabus PDF . The table below features the complete list of topics from which WBJEE Current Electricity sample questions will be asked.

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WBJEE 2025 Current Electricity Expected Weightage

The marks weightage makes it easier to comprehend how many questions from the Current Electricity might be asked and how many marks you can score by solving the amximum practice questions for WBJEE 2025 Current Electricity. You can get an idea about the WBJEE Current Electricity marks weightage and expected number of questions from the table below.

WBJEE 2025 Current Electricity Practice Questions Practice Questions with Solutions

The WBJEE 2025 Current Electricity practice questions with solutions serve as an excellent preparation tool for exam preparations. These questions can be used to evaluate knowledge and exam readiness as well as to revise the core concepts. View the following crucial WBJEE 2025 Current Electricity practice questions hand-picked by our experts based on the previous years' paper analysis.

Q.1 A battery of e.m.f. E and internal resistance r is connected to an external resistance R the condition for maximum power transfer is:

a) r < R

b) r > R

c) r = 1/R

d) r  = R

Correct Answer: d) r  = R

Solution:

Current in the circuit I = E/R+f

Power delivered to the resistance R is

P = I ² R = E ² R/(R+r) ²

It is maximum when dP/dR = 100

dP/dR = E ² [(r+R) ² -2R(r+R)/(r+R) ⁴ ] = 0

or (r+R) ² = 2R(r+R)

or r = R

Q.2 A thin wire of resistance 4 Ω is bent to form a circle. The resistance across any diameter is:

a) 4 Ω

b) 2 Ω

c) 1 Ω

d) 8 Ω

Correct Answer: c) 1Ω

Solution:

The resistance of the total wire is 4Ω

The two semicircles are connected in parallel between the two ends of the diameter.

For two resistors in parallel, the effective resistance Reff is given by:

1/R _eff = 1/R ₁ + 1/R ₂

Here, R ₁ = R2 = 2Ω

Substituting:

1/R _ef = 1/2 + 1/2 = 2/2 = 1

Therefore:

R _eff = 1Ω

Q.3 Six wires, each of resistance r, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is:

a) r

b) 2r

c) r/3

d) r/2

Correct Answer: d) r/2

Solution:

Six wires of resistance r each form a tetrahedron. For current entering at vertex A and leaving at B, symmetry ensures equal current distribution. Using Kirchhoff’s laws and symmetry, the equivalent resistance between A and B is:

R _eq = r/2

Q.4 Two equal resistances, 400 Ω each, are connected in series with a 8 V battery. If the resistance of first one increases by 0.5%, the change required in the resistance of the second one in order to keep the potential difference across it unaltered is to:

a) Increase it by 1 Ω

b) Increase it by 2 Ω

c) Increase it by 4 Ω

d) Decrease it by 4 Ω

Correct Answer: b) Increase it by 2 Ω

Solution:

In series combination, R = R ₁ + R ₂

The same current will pass in R ₁ and R ₂ . According to the question, R ₁ is increased by 0.5%.

Therefore, increment in the resistance:-

0.5/100 x 400 =  0.2Ω

So to keep the potential unchanged in the second resistance the change required will 2.0 Ω increment

Q.5 If all the cells have negligible internal resistance, what will be the current through the 2 Ω resistor when steady state is reached?

a) 0.66 A

b) 0.29 A

c) 0 A

d) 0.14 A

Correct Answer: c) 0 A

Solution:

Identify the configuration of the 2Ω resistor (series or parallel) and the total circuit resistance.

Using Ohm’s law ( I = V / R ), calculate the current based on the voltage of the power supply and the equivalent resistance.

If the 2 Ω resistor is in a branch with a capacitor, no current flows through it in steady state because the capacitor blocks the current. Hence, the current through the

2Ω resistor is: 0A

Q.6 Two wires of same radius having lengths l1 and l2 and resistivities and p1 and p2 are connected in series. The equivalent resistivity will be:

a) p ₁ l ₂ +p ₂ l ₁ /p ₁ +p ₂

b) p ₁ l ₁ +p ₂ l ₂ /l ₁ +l ₂

c) p ₁ l ₁ - p ₂ l ₂ /l ₁ - l ₂

d) p ₁ l ₂ +p ₂ l ₁ /l ₁ +l ₂

Correct Answer: b) ₁ l ₁ +p ₂ l ₂ /l ₁ +l ₂

Solution:

R=pl/A

where, p = resistivity

l = length of the resistance wire

a = area of cross section

When the wires are connected in series, then

R = R ₁ + R ₂

R = p ₁ l ₁ /A ₁ + p ₁ l ₂ /A ₂

= p ₁ l ₁ /A + p ₂ l ₂ /A

Therefore, (A ₁ = A ₂ )

p _eq = resistivity of combination

l = l ₁ + l ₂

A = A

Thus, p _eq l ₁ +l ₂ /A = p ₁ l ₁ + p ₂ l ₂ /A

p _eq = p ₁ l ₁ }p ₂ l ₂ /l ₁ +l ₂

Q.7 Four resistors, 100 Ω, 200 Ω, 300 Ω, and 400 Ω are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?

a) 210 Ω

b) 240 Ω

c) 300 Ω

d) 250 Ω

Correct Answer: a) 210 Ω

Solution:

R _series1 = R ₃ + R ₄ = 300  + 400 = 700Ω

The equivalent resistance across the diagonal is given by:

1/R _eq = 1/R _series1 +1/R _series2

Substitute the values:

1/R _eq = 1/700+1/300

Simplify:

1/R _eq = 300+700/700x300 = 1000/21000

R _eq = 21000/1000 = 210Ω

Q.8 What will be current through the 200 Ω resistor in the given circuit, a long time after the switch K is made on?

a) Zero

b) 100 mA

c) 10 mA

d) 1 mA

Correct Answer: a) Zero

Solution:

A long time after the switch K is closed, the capacitor in the circuit becomes fully charged and behaves like an open circuit in a steady state. If the 200Ω resistor is in series with the capacitor, no current flows through it since the capacitor blocks the current. Thus, the current through the 200Ω resistor is zero in a steady state.

Q.9 An electric bulb, a capacitor, a battery and a switch are all in series in a circuit. How does the intensity of light vary when the switch is turn on?

a) continues to increase gradually

b) gradually increases for sometime and then becomes steady

c) sharply rises initially and then gradually decreases

d) gradually decreases for sometime and then gradually decreases

Correct Answer: c) sharply rises initially and then gradually decreases

Solution:

Initially, there will be no voltage drop across the capacitor, so the intensity of the bulb will rise sharply and gradually voltage drop across the capacitor will increase as a result voltage drop across the bulb decreases, so the intensity of the bulb will decrease.

Q. 10 Consider the circuit shown in the figure where all the resistances are of magnitude 1 k Ω. If the current in the extreme right resistance X is 1 mA, the potential difference between A and B is:

a) 34 V

b) 21 V

c) 68 V

d) 55V

Correct Answer: a) 34 V

Solution:

All resistances are 1kΩ

Current through the extreme right resistance X is 1 mA

We need to find the potential difference between points A and B.

The extreme right resistor X has a current of 1mA, so the voltage drop across it is:

V _x = I ⋅ R = (1mA) (1kΩ)=1V

​Calculate the Voltage Across A and B:

V _AB ​=34V

The potential difference between A and B is: 34V

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While WBJEE 2025 Current Electricity practice questions with solutions are a good way to identify the most expected/ sample questions, taking WBJEE 2025 mock test from time to time is also highly recommended to evaluate performance after each attempt and improve all shortcomings before the final examination.

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