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100+ Free Maharashtra HSC Electronics (Bifocal) Practice Questions

Maharashtra Higher Secondary Certificate (HSC / Class 12) Bifocal Electronics under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE), Pune — board subject code C2 practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Maharashtra HSC Electronics (Bifocal) Exam

Subject code C2

Official MSBSHSE bifocal subject code for Electronics

MSBSHSE HSC bifocal subject lists

2 theory papers

Paper I Applied Electronics + Paper II Digital Electronics

MSBSHSE/DVET Electronics (C2) course patterns

~50 marks each theory paper

Common written theory allocation per paper (about 3 hours each)

Public bifocal Electronics scheme summaries

~200 marks total

Historical subject total across theory + practical components

Junior-college bifocal Electronics mark-distribution reporting

35%

Common minimum passing floor per HSC subject under MSBSHSE

Maharashtra HSC pass-mark practice (~35% each subject)

English MCQ adaptation

This free local bank is not the official Higher Secondary paper format

OpenExamPrep practice policy

Maharashtra MSBSHSE HSC Bifocal Electronics (code C2) is a Class 12 two-paper technical elective (Applied + Digital, ~50 theory marks each) with heavy practical weight toward ~200 marks total and ~35% pass floor — not a pure MCQ board paper. This free 2026 bank is an English MCQ study adaptation for circuits, devices, digital logic, and calculation fluency.

Sample Maharashtra HSC Electronics (Bifocal) Practice Questions

Try these sample questions to test your Maharashtra HSC Electronics (Bifocal) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A 12 V battery is connected across a 4 Ω resistor. What is the current in the circuit?
A.48 A
B.3 A
C.0.33 A
D.16 A
Explanation: By Ohm’s law, I = V/R = 12 V / 4 Ω = 3 A. Current equals voltage divided by resistance for a simple resistive circuit.
2Two resistors of 6 Ω and 3 Ω are connected in series across a 18 V supply. What is the total circuit current?
A.3 A
B.2 A
C.6 A
D.9 A
Explanation: Series resistance R_total = 6 + 3 = 9 Ω. I = V/R_total = 18/9 = 2 A. In series, the same current flows through every resistor.
3Two resistors of 12 Ω and 6 Ω are connected in parallel. What is the equivalent resistance?
A.18 Ω
B.9 Ω
C.2 Ω
D.4 Ω
Explanation: For two parallel resistors, 1/R_eq = 1/12 + 1/6 = 1/12 + 2/12 = 3/12, so R_eq = 4 Ω. Parallel equivalent is always less than the smaller branch resistance.
4A 10 Ω resistor carries 2 A. What power is dissipated in the resistor?
A.40 W
B.20 W
C.5 W
D.12 W
Explanation: Power P = I²R = (2)² × 10 = 4 × 10 = 40 W. Equivalently P = VI with V = IR = 20 V gives P = 20 × 2 = 40 W.
5Three 6 Ω resistors are connected in parallel. What is the equivalent resistance?
A.18 Ω
B.6 Ω
C.2 Ω
D.3 Ω
Explanation: Equal resistors in parallel: R_eq = R/n = 6/3 = 2 Ω. Alternatively 1/R_eq = 3×(1/6) = 1/2.
6A series circuit has R1 = 4 Ω and R2 = 8 Ω across 24 V. What voltage appears across R2?
A.8 V
B.12 V
C.16 V
D.24 V
Explanation: Voltage divider: V2 = V × (R2/(R1+R2)) = 24 × (8/12) = 16 V. Current I = 24/12 = 2 A, so V2 = IR2 = 16 V.
7How much energy (in joules) does a 100 W lamp use in 5 minutes?
A.30 000 J
B.500 J
C.20 J
D.100 J
Explanation: Energy E = P × t. Convert 5 minutes to 300 s. E = 100 W × 300 s = 30 000 J.
8Four resistors of 8 Ω each are connected as two series pairs placed in parallel with each other. What is R_eq?
A.4 Ω
B.16 Ω
C.8 Ω
D.32 Ω
Explanation: Each series pair is 8+8 = 16 Ω. Two 16 Ω branches in parallel give R_eq = 8 Ω. Series-then-parallel reduction is the standard approach.
9A wire has resistance 10 Ω. If both its length and cross-sectional area are doubled, the new resistance is approximately:
A.10 Ω
B.5 Ω
C.20 Ω
D.40 Ω
Explanation: R = ρL/A. Doubling L multiplies R by 2; doubling A divides R by 2. Net factor is 1, so resistance stays 10 Ω.
10In a pure resistive AC circuit, the average power is maximum when the phase angle between voltage and current is:
A.90°
B.45°
C.180°
D.0°
Explanation: Average power P = VI cos φ. For a pure resistor, φ = 0°, cos 0° = 1, so power factor is unity and average power is maximum for given V and I rms.

About the Maharashtra HSC Electronics (Bifocal) Practice Questions

Verified exam format metadata for Maharashtra Higher Secondary Certificate (HSC / Class 12) Bifocal Electronics under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE), Pune — board subject code C2 is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.