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:

In the grid circuit of a triode,a signal $E = 2\sqrt{2} \cos(\omega t)$ is applied. If the amplification factor $\mu = 14$ and the plate resistance $r_p = 10\,k\Omega$,then the root mean square $(RMS)$ current flowing through the load resistance $R_L = 12\,k\Omega$ will be.......$mA$.

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 $V$)

The amplification factor of a triode is $20$. If the grid potential is reduced by $0.2\, V$,then to keep the plate current constant,its plate voltage must be increased by how many volts?

The plate resistance of a triode is $2.5 \times 10^{4} \ \Omega$ and mutual conductance is $2 \times 10^{-3} \ \text{mho}$. What will be the value of the amplification factor?

The introduction of a grid in a triode valve affects plate current by