Electrostatics & Circuits MCQ & Objective Questions
Understanding "Electrostatics & Circuits" is crucial for students preparing for school and competitive exams in India. This topic not only forms a significant part of the syllabus but also features prominently in various objective questions and MCQs. By practicing these questions, students can enhance their grasp of concepts and improve their chances of scoring better in exams.
What You Will Practise Here
Fundamental concepts of electrostatics, including charge, electric field, and potential.
Key formulas related to Coulomb's law and electric field strength.
Understanding of capacitors, their types, and applications in circuits.
Basic circuit theory, including Ohm's law and Kirchhoff's laws.
Analysis of series and parallel circuits with practical examples.
Diagrams illustrating electric field lines and circuit schematics.
Problem-solving strategies for common electrostatics and circuit-related questions.
Exam Relevance
The topics of Electrostatics and Circuits are integral to the curriculum of CBSE, State Boards, NEET, and JEE. Students can expect questions that test their understanding of theoretical concepts as well as practical applications. Common question patterns include numerical problems, conceptual MCQs, and diagram-based questions that require a clear understanding of the subject matter.
Common Mistakes Students Make
Confusing the concepts of electric field and electric potential.
Misapplying Ohm's law in complex circuits.
Overlooking the significance of units in calculations.
Failing to interpret circuit diagrams accurately.
Neglecting to review the properties of capacitors and their behavior in circuits.
FAQs
Question: What are the key formulas I should remember for Electrostatics?Answer: Important formulas include Coulomb's law (F = k * |q1 * q2| / r²) and the formula for electric field (E = F/q).
Question: How can I improve my performance in circuit-related MCQs?Answer: Practice solving circuit problems regularly and familiarize yourself with different circuit configurations.
Question: Are there any specific topics I should focus on for competitive exams?Answer: Focus on understanding capacitors, circuit laws, and the relationship between voltage, current, and resistance.
Now is the time to boost your exam preparation! Dive into our practice MCQs on Electrostatics & Circuits and test your understanding to achieve your academic goals.
Q. If a circuit has a total resistance of 12Ω and a total current of 2A, what is the total voltage supplied by the battery?
A.
12 V
B.
24 V
C.
36 V
D.
48 V
Show solution
Solution
Using Ohm's law, V = I * R = 2A * 12Ω = 24 V.
Correct Answer:
B
— 24 V
Learn More →
Q. If a circuit has a total resistance of 12Ω and a total voltage of 24V, what is the power consumed by the circuit?
A.
48 W
B.
24 W
C.
12 W
D.
36 W
Show solution
Solution
Power P = V^2 / R = (24V)^2 / 12Ω = 48 W.
Correct Answer:
A
— 48 W
Learn More →
Q. If a circuit has a total resistance of 12Ω and a total voltage of 48V, what is the total current in the circuit?
A.
2 A
B.
3 A
C.
4 A
D.
6 A
Show solution
Solution
Using Ohm's law, I = V/R = 48V / 12Ω = 4 A.
Correct Answer:
C
— 4 A
Learn More →
Q. If a circuit has a total resistance of 12Ω and a voltage of 24V, what is the power consumed by the circuit?
A.
48 W
B.
24 W
C.
12 W
D.
36 W
Show solution
Solution
Power P = V²/R = 24² / 12 = 48 W.
Correct Answer:
A
— 48 W
Learn More →
Q. If a circuit has a total voltage of 24 V and a total resistance of 6 Ω, what is the current flowing through the circuit?
A.
2 A
B.
3 A
C.
4 A
D.
5 A
Show solution
Solution
Using Ohm's law, I = V/R = 24 V / 6 Ω = 4 A.
Correct Answer:
B
— 3 A
Learn More →
Q. If a circuit has a total voltage of 24V and a total resistance of 8Ω, what is the power consumed?
A.
72 W
B.
48 W
C.
36 W
D.
24 W
Show solution
Solution
Power P = V^2 / R = 24V^2 / 8Ω = 72 W.
Correct Answer:
A
— 72 W
Learn More →
Q. If a circuit has a voltage of 12 V and a resistance of 4 Ω, what is the current flowing through the circuit?
A.
3 A
B.
4 A
C.
2 A
D.
6 A
Show solution
Solution
Using Ohm's Law, I = V/R = 12 V / 4 Ω = 3 A.
Correct Answer:
A
— 3 A
Learn More →
Q. If a dielectric material with a dielectric constant of 4 is inserted into a capacitor, how does the capacitance change?
A.
It doubles
B.
It quadruples
C.
It halves
D.
It remains the same
Show solution
Solution
The capacitance increases by a factor equal to the dielectric constant. Thus, it quadruples.
Correct Answer:
B
— It quadruples
Learn More →
Q. If a dielectric material with a dielectric constant of 4 is placed in a capacitor, how does it affect the capacitance?
A.
It halves the capacitance
B.
It doubles the capacitance
C.
It quadruples the capacitance
D.
It does not affect the capacitance
Show solution
Solution
The capacitance increases by the factor of the dielectric constant, so C' = k * C, where k = 4.
Correct Answer:
B
— It doubles the capacitance
Learn More →
Q. If a resistor of 10 ohms and a capacitor of 100 microfarads are in series, what is the time constant of the circuit?
A.
0.001 seconds
B.
0.01 seconds
C.
1 second
D.
0.1 seconds
Show solution
Solution
The time constant τ = R * C = 10 ohms * 100 x 10^-6 F = 0.001 seconds or 1 millisecond.
Correct Answer:
B
— 0.01 seconds
Learn More →
Q. If a resistor of 10 ohms is connected to a 5V battery, what is the current flowing through the resistor?
A.
0.5 A
B.
1 A
C.
2 A
D.
0.2 A
Show solution
Solution
Using Ohm's Law, I = V/R. Here, I = 5V / 10Ω = 0.5 A.
Correct Answer:
B
— 1 A
Learn More →
Q. If a resistor of 10 Ω is connected in series with a 20 Ω resistor, what is the total resistance?
A.
10 Ω
B.
20 Ω
C.
30 Ω
D.
15 Ω
Show solution
Solution
R_total = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.
Correct Answer:
C
— 30 Ω
Learn More →
Q. If a resistor of 10 Ω is connected in series with a 5 Ω resistor, what is the total resistance?
A.
15 Ω
B.
5 Ω
C.
10 Ω
D.
20 Ω
Show solution
Solution
Total resistance in series is R_total = R1 + R2 = 10 Ω + 5 Ω = 15 Ω.
Correct Answer:
A
— 15 Ω
Learn More →
Q. If a resistor of 10Ω is connected across a 20V power supply, what is the power dissipated by the resistor?
A.
20 W
B.
40 W
C.
10 W
D.
5 W
Show solution
Solution
Power P = V²/R = (20V)² / 10Ω = 400/10 = 40 W.
Correct Answer:
B
— 40 W
Learn More →
Q. If the capacitance in an RC circuit is doubled, what happens to the time constant?
A.
It remains the same.
B.
It doubles.
C.
It halves.
D.
It quadruples.
Show solution
Solution
Doubling the capacitance in an RC circuit will double the time constant τ = R * C.
Correct Answer:
B
— It doubles.
Learn More →
Q. If the capacitance of a capacitor is 3 µF and it is charged to 6 V, what is the charge on the capacitor?
A.
18 µC
B.
12 µC
C.
6 µC
D.
9 µC
Show solution
Solution
Charge (Q) is given by Q = C * V. Here, Q = 3 µF * 6 V = 18 µC.
Correct Answer:
A
— 18 µC
Learn More →
Q. If the capacitance of a capacitor is doubled while the voltage remains constant, what happens to the charge stored?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
Show solution
Solution
Charge (Q) stored in a capacitor is given by Q = C * V. If C is doubled and V remains constant, Q also doubles.
Correct Answer:
A
— It doubles
Learn More →
Q. If the capacitance of a capacitor is doubled, what happens to the charge stored if the voltage remains constant?
A.
Charge is halved
B.
Charge remains the same
C.
Charge is doubled
D.
Charge is quadrupled
Show solution
Solution
Capacitance (C) is defined as C = Q/V. If C is doubled and V remains constant, Q must also double.
Correct Answer:
C
— Charge is doubled
Learn More →
Q. If the distance between two point charges is doubled, how does the force between them change according to Coulomb's law?
A.
It doubles
B.
It quadruples
C.
It halves
D.
It becomes one-fourth
Show solution
Solution
According to Coulomb's law, the force F between two point charges is inversely proportional to the square of the distance r between them. If the distance is doubled, the force becomes F/4.
Correct Answer:
D
— It becomes one-fourth
Learn More →
Q. If the electric field between two parallel plates is 2000 N/C, what is the potential difference between the plates if they are 0.05 m apart?
A.
100 V
B.
200 V
C.
300 V
D.
400 V
Show solution
Solution
Potential difference V = E * d = 2000 N/C * 0.05 m = 100 V.
Correct Answer:
B
— 200 V
Learn More →
Q. If the electric potential at a point is 100 V and the charge at that point is 2 µC, what is the electric potential energy?
A.
0.2 J
B.
0.1 J
C.
0.05 J
D.
0.4 J
Show solution
Solution
Potential energy U = V * q = 100 V * 2 x 10^-6 C = 0.0002 J = 0.2 J.
Correct Answer:
A
— 0.2 J
Learn More →
Q. If the potential difference across a capacitor is 12 V and its capacitance is 3 µF, what is the charge stored in the capacitor?
A.
36 µC
B.
24 µC
C.
12 µC
D.
18 µC
Show solution
Solution
Q = C * V = 3 x 10^-6 F * 12 V = 36 µC.
Correct Answer:
A
— 36 µC
Learn More →
Q. If the potential difference across a capacitor is doubled, what happens to the stored energy?
A.
It doubles
B.
It quadruples
C.
It remains the same
D.
It halves
Show solution
Solution
Energy stored in a capacitor is given by U = 1/2 C V^2. If V is doubled, U becomes 1/2 C (2V)^2 = 2CV^2, which is quadrupled.
Correct Answer:
B
— It quadruples
Learn More →
Q. If the voltage across a resistor is doubled while the resistance remains constant, what happens to the current?
A.
It doubles
B.
It halves
C.
It remains the same
D.
It quadruples
Show solution
Solution
According to Ohm's Law, if voltage is doubled and resistance is constant, current will also double.
Correct Answer:
A
— It doubles
Learn More →
Q. If two capacitors are connected in series, how is the total capacitance (C_total) calculated?
A.
1/C_total = 1/C1 + 1/C2
B.
C_total = C1 + C2
C.
C_total = C1 * C2
D.
C_total = C1 - C2
Show solution
Solution
For capacitors in series, the total capacitance is given by 1/C_total = 1/C1 + 1/C2.
Correct Answer:
A
— 1/C_total = 1/C1 + 1/C2
Learn More →
Q. If two capacitors of 4 µF and 6 µF are connected in parallel, what is the total capacitance?
A.
10 µF
B.
24 µF
C.
2.4 µF
D.
12 µF
Show solution
Solution
C_eq = C1 + C2 = 4 µF + 6 µF = 10 µF.
Correct Answer:
A
— 10 µF
Learn More →
Q. If two capacitors of 4 µF and 6 µF are connected in series, what is the total capacitance?
A.
2.4 µF
B.
10 µF
C.
1.5 µF
D.
24 µF
Show solution
Solution
1/C_total = 1/C1 + 1/C2 = 1/4 + 1/6 = 5/12, thus C_total = 12/5 = 2.4 µF.
Correct Answer:
A
— 2.4 µF
Learn More →
Q. If two capacitors of capacitance C1 and C2 are connected in series, what is the equivalent capacitance (Ceq)?
A.
Ceq = C1 + C2
B.
Ceq = 1 / (1/C1 + 1/C2)
C.
Ceq = C1 * C2
D.
Ceq = (C1 * C2) / (C1 + C2)
Show solution
Solution
The equivalent capacitance for capacitors in series is given by Ceq = 1 / (1/C1 + 1/C2).
Correct Answer:
B
— Ceq = 1 / (1/C1 + 1/C2)
Learn More →
Q. If two charges of +3 µC and -3 µC are placed 0.2 m apart, what is the magnitude of the force between them?
A.
67.5 N
B.
45 N
C.
30 N
D.
15 N
Show solution
Solution
Using Coulomb's law, F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * |3 x 10^-6 C * -3 x 10^-6 C| / (0.2 m)^2 = 45 N.
Correct Answer:
B
— 45 N
Learn More →
Q. If two point charges, +Q and -Q, are separated by a distance d, what is the magnitude of the electric field (E) at the midpoint?
A.
E = 0
B.
E = k * Q / (d/2)^2
C.
E = k * Q / d^2
D.
E = k * Q / (d^2/4)
Show solution
Solution
At the midpoint, the electric fields due to both charges cancel each other out, resulting in E = 0.
Correct Answer:
A
— E = 0
Learn More →
Showing 31 to 60 of 279 (10 Pages)
