The value of plate current $I_P$ in the given circuit diagram will be in $mA$.

In the mutual characteristic curve of a triode,which point represents the cut-off voltage?

$A$ 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 $i_p$ 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?

The linear portions of the characteristic curves of a triode valve give the following readings. The plate resistance is ....... $\Omega$.

$V_g$ (volt)	$0$	$-2$	$-4$	$-6$
$I_p$ (mA) for $V_p = 150$ $V$	$15$	$12.5$	$10$	$7.5$
$I_p$ (mA) for $V_p = 120$ $V$	$10$	$7.5$	$5$	$2.5$

The point representing the cut-off grid voltage on the mutual characteristic of a triode is: