When the load resistance of an amplifier is modified from 4 k to 8 k , its gain factor increases from 10 to 12 . Determine the plate resistance.

Correct answer: D. 2 k The voltage gain of an amplifier is given by A = (μ R_L)/(R_p + R_L) where μ is the amplification factor, R_p is the plate resistance, and R_L is the load resistance. For the first case: 10 = (μ × 4)/(R_p + 4) For the second case: 12 = (μ × 8)/(R_p + 8) Dividing the first equation by the second equation: (10)/(12) = (4(R_p + 8))/(8(R_p + 4)) (5)/(6) = (R_p + 8)/(2(R_p + 4)) 10(R_p + 4) = 6(R_p + 8) 10R_p + 40 = 6R_p + 48 4R_p = 8 R_p = 2 k

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Current through Gases

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Current through Gases

When the load resistance of an amplifier is modified from 4 k to 8...

When the load resistance of an amplifier is modified from

4 \text{ k}\Omega

8 \text{ k}\Omega

, its gain factor increases from

10

12

. Determine the plate resistance.

Options

A $6 \text{ k}\Omega$
B $4 \text{ k}\Omega$
C $8 \text{ k}\Omega$
D $2 \text{ k}\Omega$

Correct answer

D. $2 \text{ k}\Omega$

Step-by-step solution

The voltage gain of an amplifier is given by

A = \dfrac{\mu R_L}{R_p + R_L}

where

\mu

is the amplification factor,

R_p

is the plate resistance, and

R_L

is the load resistance.

For the first case:

10 = \dfrac{\mu \times 4}{R_p + 4}

For the second case:

12 = \dfrac{\mu \times 8}{R_p + 8}

Dividing the first equation by the second equation:

\dfrac{10}{12} = \dfrac{4(R_p + 8)}{8(R_p + 4)}

\dfrac{5}{6} = \dfrac{R_p + 8}{2(R_p + 4)}

10(R_p + 4) = 6(R_p + 8)

10R_p + 40 = 6R_p + 48

4R_p = 8

R_p = 2 \text{ k}\Omega

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