The plate current in a triode can be written as i_p = k\!(V_g +...

The plate current in a triode is given by:

$i_p = k \left(V_g + \dfrac{V_p}{\mu}\right)^{3/2}$

The mutual conductance $g_m$ is defined as the rate of change of plate current with respect to grid voltage at a constant plate voltage:

$g_m = \dfrac{\partial i_p}{\partial V_g}$

Differentiating the given expression with respect to $V_g$:

$g_m = \dfrac{3}{2} k \left(V_g + \dfrac{V_p}{\mu}\right)^{1/2}$

From the equation for plate current, we can write:

\left(V_g + \dfrac{V_p}{\mu}\right)^{3/2} = \dfrac{i_p}{k}$<br><br>\left(V_g + \dfrac{V_p}{\mu}\right)^{1/2} = \left(\dfrac{i_p}{k}\right)^{1/3}$

Substituting this back into the expression for mutual conductance:

$g_m = \dfrac{3}{2} k \left(\dfrac{i_p}{k}\right)^{1/3}$

$g_m = \dfrac{3}{2} k^{2/3} i_p^{1/3}$

Thus, $g_m \propto i_p^{1/3}$

The mutual conductance is proportional to the $1/3$ power of the plate current.