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 ..... $mA$.

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:

$A$ change of $0.8 \,mA$ in the anode current of a triode occurs when the anode potential is changed by $10 \,V$. If $\mu = 8$ for the triode,then what change in the grid voltage would be required to produce a change of $4 \,mA$ in the anode current?

The plate current in a triode is given by ${I_p} = 0.004 ({V_p} + 10{V_g})^{3/2} \text{ mA}$,where ${I_p}, {V_p}$ and ${V_g}$ are the values of plate current,plate voltage,and grid voltage,respectively. What are the triode parameters $\mu$,${r_p}$ and ${g_m}$ for the operating point at ${V_p} = 120 \text{ V}$ and ${V_g} = -2 \text{ V}$?

The current conduction in a discharge tube is due to