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

The plate current $i_p$ in a triode valve is given by $i_p = K(V_p + \mu V_g)^{3/2}$,where $i_p$ is in $mA$ and $V_p$ and $V_g$ are in $V$. If $r_p = 10^4 \, \Omega$ and $g_m = 5 \times 10^{-3} \, \text{mho}$,then for $i_p = 8 \, mA$ and $V_p = 300 \, V$,what are the values of $K$ and the grid cut-off voltage?

$A$ 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

In the space charge limited region,the plate current in a diode is $10\, mA$ for a plate voltage of $150\, V$. If the plate voltage is increased to $600\, V$,then the plate current will be.......$mA$.

In a diode,when there is saturation current,the plate resistance $({r_p})$ is

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: