Question

The gain factor of an amplifier in increased from 10 to 12 as the load resistance is changed from 4 kΩ to 8 kΩ. Calculate (a) the amplification factor and (b) the plate resistance.

Answer 1

The resistance of the plate is rp = 2 KΩ

Explanation:

Voltage gain = μ / 1 + rp / RL ---- (1)

Voltage amplification factor A  = 10

RL = 4KΩ

Now put the values:

10 =  μ / 1 + rp / 4 x 10^3

10 =  μ x 4 x 10^3 / 4 x 10^3 +rp

4 x 10^4 + 10 rp = 4000 μ

Now increased gain A = 12

Substituting this value in equation (1) we get;

12 = μ / 1 + rp / RL

12 = μ / 1 + rp / 8 x 10^3

12 = μ  x 8000 / 8000 + rp

96000 + 12 rp = 8000  μ

On solving the equations we get:

μ = 15

rp = 2000 Ω = 2 KΩ

Thus the resistance of the plate is rp = 2 KΩ

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Answer 2

Explanation:

It is known that:

Gain of voltage =  $r p R L+\mu 1$                   … (1)

When voltage amplification factor, A = 10,

$R_{L}=4 \mathrm{k} \Omega$

$10=\mu 1+103 \times r p 4$

$\Rightarrow 10=\mu \times 103 \times 4 \times 1034+r p$

$\Rightarrow 104 \times 4+10 r P=4000 \mu$             … (2)

Now, the gain increase, A = 12

When this value is substituted in equation (1), we obtain:

$12=r P R L 12+\mu 1=\mu 1+103 \times r p 8$

$\Rightarrow 12=80008000 \times \mu+r P$

$\Rightarrow 12 r P+96000=8000 \mu$            …(3)

When the equations (2) and (3) are solved, we can obtain:

$\mu=15 \mathrm{rP}=2000 \Omega=2 \mathrm{k} \Omega$