⇦ | ![]() | ⇨ |
A common emitter amplifier is designed with n-p-n transistor (α=0.99). The input impedence is 1kΩ and load is 10kΩ. The voltage gain will be
Options
(a) 9900
(b) 99
(c) 9.9
(d) 990
Correct Answer:
990
Explanation:
Voltage gain = β × Resistance gain
β = α/(1-α) = 0.99 / (1-0.99) = 99
Resistance gain = (10 × 10³) = 10
Voltage gain = 99 × 10 = 990
Related Questions:
- In an ac circuit an alternating voltage e = 200 √2 sin 100 t volts is connected
- An electron in potentiometer experiences a force 2.4×10⁻¹⁹N. The length of potentiometer
- A particle executing SHM with amplitude of 0.1 m. At a certain instant,
- Huygen’s principle of secondary wavelets may be used to
- The maximum and minimum intensities of two sources is 4:1. The ratio of amplitude is
Topics: Electronic Devices
(124)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply