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

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 plate resistance of two triode valves is $2 \, k\Omega$ and $4 \, k\Omega$. The amplification factor of each valve is $40$. The ratio of voltage amplification,when used with a $4 \, k\Omega$ load resistance,will be:

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 ..... $mA$.

$A$ triode has a plate resistance of $10 \, k\Omega$ and an amplification factor of $24$. If the input signal voltage is $0.4 \, V$ $(r.m.s.)$ and the load resistance is $10 \, k\Omega$,then the output voltage $(r.m.s.)$ is ...... $V$.

In a diode valve,the cathode temperature must be ($\phi$ = work function):