The grid voltage of any triode valve is chang | vedclass

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Question

(B) The mutual conductance $(g_m)$ of a triode valve is defined as the ratio of the change in plate current $(\Delta i_p)$ to the change in grid volta

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(B) The mutual conductance $(g_m)$ of a triode valve is defined as the ratio of the change in plate current $(\Delta i_p)$ to the change in grid voltage $(\Delta v_g)$ at a constant plate voltage.Formula: $g_m = \frac{\Delta i_p}{\Delta v_g}$Given:$g_m = 3 	imes 10^{-4} \, 	ext{mho}$$\Delta v_g = |(-3) - (-1)| = |-2| = 2 \, V$Substituting the values:$3 	imes 10^{-4} = \frac{\Delta i_p}{2}$$\Delta i_p = 3 	imes 10^{-4} 	imes 2 = 6 	imes 10^{-4} \, A$Converting to milliamperes $(mA)$:$\Delta i_p = 6 	imes 10^{-4} 	imes 10^3 \, mA = 0.6 \, mA$Therefore, the change in plate circuit current is $0.6 \, mA$.

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

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