The relation between $I_p$ and $V_p$ for a triode is $I_p = (0.125V_p - 7.5) \text{ mA}$. Keeping the grid potential constant at $1 \text{ V}$,the value of $r_p$ will be . . . . . . $\text{k}\Omega$.

$A$ triode has a mutual conductance of $2 \times 10^{-3} \text{ mho}$ and an amplification factor of $50$. The anode is connected through a resistance of $25 \times 10^3 \, \Omega$ to a $250 \, \text{V}$ supply. The voltage gain of this amplifier is

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

The unit of mutual conductance of a triode valve is

$14 \times 10^{15}$ electrons reach the anode per second. If the power consumed is $448 \text{ mW}$,then the plate (anode) voltage is .... $V$.

The current in a triode at anode potential $100 \, V$ and grid potential $-1.2 \, V$ is $7.5 \, mA$. If the grid potential is changed to $-2.2 \, V$, the current becomes $5.5 \, mA$. The value of transconductance $(g_m)$ will be .... $mili \, mho$.