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100+ Free Karnataka II PUC Electronics (KSEAB) Practice Questions

Karnataka School Examination and Assessment Board (KSEAB) II PUC Class 12 Electronics (Code 40) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Karnataka II PUC Electronics (KSEAB) Exam

~70 + 30

Common theory + practical mark split (total 100)

KSEAB II PUC science practical subject pattern

~3 h 15 m

Typical theory duration (often includes reading time)

KSEAB II PUC Electronics model paper logistics

Code 40

Official subject code for II PUC Electronics

KSEAB / DPUE Electronics blueprint

~35%

Commonly reported overall pass threshold (confirm circular)

KSEAB II PUC pass criteria reporting

Mixed + practical

Official format is mixed written theory plus practical, not pure MCQ

KSEAB II PUC Electronics assessment pattern

English MCQ adaptation

This free local bank is not the official mixed paper format

OpenExamPrep practice policy

KSEAB II PUC Electronics (code 40) is a Class 12 board subject with ~70 theory + ~30 practical and theory lasting about 3 h 15 m. Pass is commonly ~35% overall (~30–35% careful wording) as notified. Fees are paid through colleges. This free 2026 bank is an English MCQ study adaptation — not an official paper simulation.

Sample Karnataka II PUC Electronics (KSEAB) Practice Questions

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

1In a junction field-effect transistor (JFET), the current between drain and source is primarily controlled by:
A.The gate-to-source voltage
B.The base current
C.The collector current
D.The emitter resistance only
Explanation: A JFET is a voltage-controlled device. The reverse-biased gate–channel junction modulates channel width (and thus channel conductivity), so drain–source current is controlled mainly by VGS.
2Compared with a bipolar junction transistor (BJT), a major advantage of a JFET is its:
A.Much lower input impedance
B.Very high input impedance
C.Need for large continuous base current
D.Inability to operate as a voltage amplifier
Explanation: The JFET gate is reverse-biased (essentially drawing negligible DC gate current), so input impedance is very high compared with a BJT’s base input.
3The three terminals of an n-channel JFET are:
A.Base, collector and emitter
B.Anode, cathode and gate only as in an SCR
C.Gate, drain and source
D.Plate, grid and cathode only as in a vacuum triode
Explanation: An n-channel JFET has gate (control), drain and source terminals. Base/collector/emitter are BJT terminals.
4Pinch-off voltage VP of a JFET is the gate-to-source voltage at which, for a given drain–source voltage in the saturation description,
A.The gate is forward-biased into heavy conduction
B.Drain current becomes infinite
C.The device turns into a pure short circuit between drain and source
D.The channel is effectively pinched and ID becomes nearly independent of further VDS increase (saturation behaviour)
Explanation: At pinch-off, the depletion regions close the channel enough that ID saturates (approximately constant with VDS in the active/saturation region description used in textbooks).
5The drain current of an n-channel JFET in the saturation (pinch-off) region is often modelled as ID = IDSS(1 − VGS/VP)². If IDSS = 8 mA and VP = −4 V, the drain current at VGS = −2 V is approximately:
A.2 mA
B.4 mA
C.6 mA
D.8 mA
Explanation: ID = 8(1 − (−2)/(−4))² mA = 8(1 − 0.5)² = 8(0.5)² = 8(0.25) = 2 mA.
6In an n-channel enhancement-mode MOSFET, a conducting inversion channel forms when:
A.VGS is always zero
B.VGS exceeds the positive threshold voltage VT
C.The gate is reverse-biased like a JFET p–n junction gate
D.Drain and source are shorted externally
Explanation: Enhancement n-MOSFETs are off at VGS = 0. Applying VGS > VT (positive for n-channel) creates an n-type inversion layer connecting source and drain.
7The parameter gm (transconductance) of a FET is defined as:
A.gm = ∂VGS/∂ID only
B.gm = VDS/ID
C.gm = ∂ID/∂VGS (at constant VDS)
D.gm = ID × VGS
Explanation: Transconductance quantifies how much drain current changes for a change in gate–source voltage at fixed drain–source voltage: gm = ∂ID/∂VGS|VDS constant.
8Which statement correctly compares JFET and MOSFET gate structures?
A.Both always use a forward-biased p–n gate junction
B.A JFET gate is oxide-insulated while a MOSFET gate is a p–n junction
C.Neither device has a gate terminal
D.A MOSFET gate is insulated by an oxide layer; a JFET gate is a reverse-biased p–n junction
Explanation: JFET control is by a reverse-biased gate–channel junction. MOSFET control is through an insulated gate (metal–oxide–semiconductor stack), yielding extremely high DC input resistance.
9The main purpose of DC biasing a BJT amplifier is to:
A.Establish a stable quiescent operating point (Q-point) in the active region
B.Remove the need for any power supply
C.Force the transistor permanently into cutoff only
D.Convert the BJT into a pure AC generator with no DC
Explanation: Biasing sets collector current and VCE so the transistor sits at a designed Q-point, allowing linear amplification of AC signals without clipping.
10In the common relation IC = βIB for an NPN BJT in the active region, if β = 100 and IB = 20 μA, IC is:
A.0.2 mA
B.2 mA
C.20 mA
D.200 μA
Explanation: IC = βIB = 100 × 20 μA = 2000 μA = 2 mA.

About the Karnataka II PUC Electronics (KSEAB) Practice Questions

Verified exam format metadata for Karnataka School Examination and Assessment Board (KSEAB) II PUC Class 12 Electronics (Code 40) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.