A triode possesses a mutual conductance of 2.0 millimho and a plate resistance of 20 k . To amplify a signal by a factor of 30 , what load resistance must be incorporated into the circuit?

Correct answer: B. 60 k Given g_m = 2.0 millimho = 2 × 10^ -3 mho and r_p = 20 k = 20 × 10^3 . The amplification factor μ is given by: μ = r_p × g_m μ = (20 × 10^3) × (2 × 10^ -3 ) = 40 The voltage gain A_v of the amplifier is given by: A_v = (μ R_L)/(r_p + R_L) Substituting the given values: 30 = (40 R_L)/(20 + R_L) 600 + 30 R_L = 40 R_L 10 R_L = 600 R_L = 60 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

A triode possesses a mutual conductance of 2.0 millimho and a plate...

A triode possesses a mutual conductance of

2.0 \text{ millimho}

and a plate resistance of

20 \text{ k}\Omega

. To amplify a signal by a factor of

30

, what load resistance must be incorporated into the circuit?

Options

A $30 \text{ k}\Omega$
B $60 \text{ k}\Omega$
C $45 \text{ k}\Omega$
D $15 \text{ k}\Omega$

Correct answer

B. $60 \text{ k}\Omega$

Step-by-step solution

Given

g_m = 2.0 \text{ millimho} = 2 \times 10^{-3} \text{ mho}

and

r_p = 20 \text{ k}\Omega = 20 \times 10^3 \Omega

.

The amplification factor

\mu

is given by:

\mu = r_p \times g_m

\mu = (20 \times 10^3) \times (2 \times 10^{-3}) = 40

The voltage gain

A_v

of the amplifier is given by:

A_v = \dfrac{\mu R_L}{r_p + R_L}

Substituting the given values:

30 = \dfrac{40 R_L}{20 + R_L}

600 + 30 R_L = 40 R_L

10 R_L = 600

R_L = 60 \text{ k}\Omega

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