The relation between dynamic plate resistance $(r_p)$ of a vacuum diode and plate current $(I_p)$ in the space charge limited region is:

The value of the amplification factor from the following graph is:

The slope of the anode characteristic and the mutual characteristic of a triode are $0.02 \text{ mA/V}$ and $1 \text{ mA/V}$ respectively. The amplification factor of the valve is:

The plate current $i_p$ in a triode valve is given by $i_p = K(V_p + \mu V_g)^{3/2}$,where $i_p$ is in $mA$ and $V_p$ and $V_g$ are in $V$. If $r_p = 10^4 \, \Omega$ and $g_m = 5 \times 10^{-3} \, \text{mho}$,then for $i_p = 8 \, mA$ and $V_p = 300 \, V$,what are the values of $K$ and the grid cut-off voltage?

If the amplification factor of a triode $(\mu)$ is $22$ and its plate resistance is $6600 \, \Omega$,then the mutual conductance of this valve in mho is:

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 ..... $mA$.