The grid voltage of any triode valve is changed from -1 volt to -3 volt and the mutual conductance is 3 $\times$ 10 $^{- 4}$ mho. The change in plate circuit current will be

0.8 mA

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0.6 mA

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0.4 mA

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1 mA

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Solution

The correct option is B 0.6 mA
$g_{m} = \frac{Δ I_{p}}{Δ V_{g}}$
or $Δ I_{p} = g_{m} \times Δ V_{g} = 3 \times 10^{- 4} \times [- 3 - (- 1)]$
$= - 0.6 \times 10^{- 3} A$ = shortage of $0.6 \times 10^{- 3} A$

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The grid voltage of any triode valve is changed from -1 volt to -3 volt and the mutual conductance is 3× 10−4 mho. The change in plate circuit current will be

The correct option is B 0.6 mAgm=ΔIpΔVgor ΔIp=gm×ΔVg=3×10−4×[−3−(−1)]=−0.6×10−3A = shortage of 0.6×10−3A

The grid voltage of any triode valve is changed from -1 volt to -3 volt and the mutual conductance is 3 $\times$ 10 $^{- 4}$ mho. The change in plate circuit current will be

The correct option is B 0.6 mA
$g_{m} = \frac{Δ I_{p}}{Δ V_{g}}$
or $Δ I_{p} = g_{m} \times Δ V_{g} = 3 \times 10^{- 4} \times [- 3 - (- 1)]$
$= - 0.6 \times 10^{- 3} A$ = shortage of $0.6 \times 10^{- 3} A$