If ${R_p} = 7 \text{ k}\Omega$ and ${g_m} = 2.5 \text{ millimho}$, then on increasing the plate voltage by $50 \text{ V}$, how much must the grid voltage be changed so that the plate current remains the same? (in $V$)

Why is a vacuum tube called a valve in electronics?

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 amplification factor of a triode is $10$. When the plate potential is $200 \, V$ and the grid potential is $-4 \, V$,a plate current of $4 \, mA$ is observed. If the plate potential is changed to $160 \, V$ and the grid potential is kept at $-7 \, V$,then the plate current will be.....$mA$.

The slope of the plate characteristic of a vacuum diode is $2 \times 10^{-2} \, mA/V$. The plate resistance of the diode will be:

$A$ serious drawback of semiconductor devices is: