Question

In a common emitter transistor amplifier β = 60, Ro = 5000 Ω and internal resistance of a transistor is 500 Ω. The voltage amplification of amplifier will be(a) 500(b) 460(c) 600(d) 560

Answer 1

question is ---> In a common emitter transistor amplifier β = 60, Ro = 5000 Ω and internal resistance of a transistor is 500 Ω. The voltage amplification of amplifier will be(a) 500(b) 460(c) 600(d) 560

solution : answer : option (c)

we know , voltage amplification of amplifier is given by, $A=\beta\frac{R_0}{R_i}$

here, $\beta=60$, $R_0=5000\Omega$ and $R_i=500\Omega$

so, A = 60 × 5000/500 = 600

hence, The voltage amplification of amplifier will be 600. option (c) is correct.

Answer 2

Answer:

C) 600

Explanation:

Amplifier of transistor = β = 60 (Given)

R0 =5000Ω (Given)

Internal resistance of a transistor = Ri=500Ω (Given)

The Potential energy of dipole in electric field is described by U = −PEcosθ; where q is the angle between electric field and dipole.

The voltage amplification of amplifier is given by-

A = β Ro/Ri

where, β = 60, R0 =5000Ω and Ri=500Ω

Thus, A = 60 × 5000/500

A = 600  

Hence, the voltage amplification of amplifier will be 600.