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MCQ
BASIC ELECTRONICS
CONTROL SYSTEM ENGINEERING
DIGITAL ELECTRONICS
ELECTRICAL CIRCUIT ANALYSIS
ELECTRICAL ENERGY CONSERVATION AND AUDITING
ELECTRICAL ENGINEERING MATERIALS
ELECTRICAL MACHINES
ELECTRICAL WIRING
ELECTROMAGNETICS
MEASUREMENT & INSTRUMENTATION
MICROPROCESSOR AND MICROCONTROLLER INTERFACING
POWER ELECTRONICS & DRIVES
POWER SYSTEM
SIGNALS & SYSTEMS
UTILIZATION OF ELECTRICAL ENERGY
BASIC ELECTRONICS
Q1.
The change in collector emitter voltage from 6V to 9V causes increase in collector current from 6mA to 6.3mA. Determine the dynamic output resistance.
20kΩ
10kΩ
60kΩ
50kΩ
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Solution
Q2.
Negative feedback in an amplifier results in
more gain, more bandwidth
more gain, less bandwidth
less gain, more bandwidth
less gain, less bandwidth
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Solution
Q3.
A differential amplifier has an open circuit voltage gain of 100. The input signals are 3.35 V and 3.25 V. Determine the output voltage.
5 V
15 V
10 V
20 V
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Solution
Q4.
The forward resistance of the diode will be
small
high
zero
infinite
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Solution
Q5.
For the non-inverting amplifier shown in Figure, \(R_1 = 1\: k \Omega \) and \(R_f = 10\: k \Omega \). Find the output voltage for an input voltage of 1 V.
1 V
5 V
11 V
15
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Solution
Q6.
Introducing a resistor in the emitter of a common emitter amplifier stabilizes the dc operating point against variations in
Only temperature
Only the β of the transistor
Both temperature and β
None of the above
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Solution
Q7.
For a transistor the value of \(\beta \) is 50, then the value of current gain is
50/51
49/50
51/50
50/49
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Solution
Q8.
Current cannot flow to the ground through
A mechanical ground
An ac ground
A virtual ground
An a ordinary ground
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Solution
Q9.
An emitter follower has high input impedance because
large emitter resistance is used
large biasing resistance is used.
there is negative feedback in the base
the emitter-base junction is highly reverse
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Solution
Q10.
A dc power supply has no-load voltage of 30 V,and a full-load voltage of 25 V at full-load current of 1 A.Its output resistance & load regulation ,respectively are
5 Ω & 20 %
25 Ω & 20 %
5 Ω & 16.7 %
25 Ω & 16.7 %
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Solution
1
2
3
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