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  • The value of current in a triode valve is given by <img alt="\mathrm{I_p=0.004\left(V_p+10 V_{\mathrm{g}}\right)^{3 / 2} \mathrm{~mA}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5C

The value of current in a triode valve is given by \mathrm{I_p=0.004\left(V_p+10 V_{\mathrm{g}}\right)^{3 / 2} \mathrm{~mA}} . When plate potential and grid potential are \mathrm{120\ V} and \mathrm{-2\ V} respectively, then the value of mutual conductance will be

Option: 1

\mathrm{600\ mho}


Option: 2

\mathrm{60\ mho}


Option: 3

\mathrm{6\ mho}


Option: 4

\mathrm{6 \times 10^{-4} mho}


Answers (1)

best_answer

\mathrm{I_p=0.004\left(V_p+10 V_{\mathrm{g}}\right)^{3 / 2} \times 10^{-3} \ldots(\mathrm{i})}
Differentiating it w.r.t. \mathrm{V_{g}}, keeping \mathrm{V_{p}} constant, we have
\mathrm{\frac{\Delta I_p}{\Delta V_{\mathrm{g}}}=0.004 \times \frac{3}{2}\left(V_p+10 V_{\mathrm{g}}\right)^{1 / 2} \times 10 \times 10^{-3}}
\mathrm{\begin{aligned} & \text { or } g_m=0.004 \times \frac{3}{2}[120+10 \times(-2)]^{1 / 2} \times 10^{-2} \\ & =6 \times 10^{-4} \mathrm{mho} \end{aligned}}

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himanshu.meshram

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