Question

A triode has mutual conductance of 2.0 millimho and plate resistance of 20 kΩ. It is desired to amplify a signal by a factor of 30. What load resistance should be added in the circuit?

Answer 1

The value of load resistance is  RL = 60 KΩ

Explanation:

Given data:

gm = 2 millimho = 2 × 10^-3 mho

rP  = 20 kΩ

A = 30

Here μ = rp x gm

Where rp is the plate resistance and gm is the trans resistance.

20 x 10^3 x 2  10^3 = 40

We know:

A =  μ / 1+rp / RL

30 = 40 / 1 + 20000 / RL

=> 1 + 20000 / RL  = 4/3

=> 20000 / RL  = 1/3

RL = 60 KΩ

Thus the value of load resistance is  RL = 60 KΩ

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What is the resistance in a circuit that has a voltage of 60 v and a current of 2 a?

a. 60  b. 10  c. 120  d. 30

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Answer 2

In the circuit, the load resistance to be added is 60 kΩ.

Explanation:

Resistance of Plate, rp = 20 kΩ

Amplification factor, µ = 30

Mutual conductance of triode, gm = 2.0 milli mho

$g m=2 \times 10^{-3} \mathrm{mho}$

Load Resistance = RL = ?

It is known that $A=r p R L+\mu I$,

Here, the voltage amplification factor is denoted by A, which is $\mu=g m \times r P$

$A=g m 1 \times r p+r p R L$

When the values are substituted, the values given become:  

$30=20 \times 103 \times 2 \times 10^{-31}+20000 R L$

$\Rightarrow 1+20000 R L$

$\Rightarrow R L=60000 \Omega=60 \mathrm{k} \Omega$