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 . . . . . . $\text{k}\Omega$.

For a given triode,$\mu = 20$. The load resistance is $1.5$ times the anode resistance. The voltage gain will be:

Before the saturation state of a diode,at plate voltages of $400 \, V$ and $200 \, V$,the currents are $i_1$ and $i_2$ respectively. The ratio $i_1/i_2$ will be:

In a triode,the cathode,grid,and plate are at potentials of $0 \, V$,$-2 \, V$,and $80 \, V$ respectively. Electrons are emitted from the cathode with an initial kinetic energy of $3 \, eV$. What is the energy of the electron when it reaches the plate in $eV$?

$A$ 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?

In an amplifier,the load resistance ${R_L}$ is equal to the plate resistance ${r_p}$. The voltage amplification is equal to: