Calculate the amplification factor of a triode valve that has plate resistance of 2 kΩ and transconductance of 2 millimho.

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Solution

Given:
Plate resistance, $r_{p} = 2 kΩ$

Transconductance of plate, $g_{m} = 2 mili mho = 2 \times 10^{- 3} mho$

Therefore, the amplification factor is given as:
$μ = r_{P} \times g_{m} μ = 2 \times 10^{3} \times 2 \times 10^{- 3} = 4$
That is, amplification factor is 4.

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Calculate the amplification factor of a triode valve that has plate resistance of 2 kΩ and transconductance of 2 millimho.

Given: Plate resistance, rp = 2 kΩ Transconductance of plate, gm=2 mili mho=2×10-3 mho Therefore, the amplification factor is given as: μ=rP×gmμ=2×103×2×10-3=4 That is, amplification factor is 4.

Given:
Plate resistance, $r_{p} = 2 kΩ$

Transconductance of plate, $g_{m} = 2 mili mho = 2 \times 10^{- 3} mho$

Therefore, the amplification factor is given as:
$μ = r_{P} \times g_{m} μ = 2 \times 10^{3} \times 2 \times 10^{- 3} = 4$
That is, amplification factor is 4.