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

The relation between dynamic plate resistance $(r_p)$ of a vacuum diode and plate current $(I_p)$ in the space charge limited region is:

For a triode, ${r_p} = 10 \text{ k}\Omega$ and ${g_m} = 3 \text{ m}\mho$. If the load resistance is double the plate resistance, then the value of the voltage gain will be:

In a triode,${g_m} = 2 \times 10^{-3} \, \Omega^{-1}$,$\mu = 42$,and the resistance load $R_L = 50 \, \text{k}\Omega$. The voltage amplification obtained from this triode will be:

The variation of anode current $(I_p)$ in a triode corresponding to a change in grid potential $(V_g)$ at three different values of the plate potential $(V_p)$ is shown in the diagram. The mutual conductance $(g_m)$ of the triode is.........$ m \text{ mho}$.

The plate voltage of a triode is increased from $200 \, V$ to $225 \, V$. To maintain the plate current,a change in grid voltage from $5 \, V$ to $5.75 \, V$ is required. The amplification factor is: