The amplification factor of a triode is 20. I | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The relationship between amplification factor $\mu$, plate resistance $r_p$, and mutual conductance $g_m$ is given by the formula: $\mu = r_p 	im

Vedclass · Accepted Answer

$2 	imes 10^{-3} \, 	ext{mho}$

Vedclass · Answer

(D) The relationship between amplification factor $\mu$, plate resistance $r_p$, and mutual conductance $g_m$ is given by the formula: $\mu = r_p 	imes g_m$.Given: $\mu = 20$ and $r_p = 10 \, k\Omega = 10 	imes 10^3 \, \Omega = 10^4 \, \Omega$.Rearranging the formula to solve for $g_m$: $g_m = \frac{\mu}{r_p}$.Substituting the values: $g_m = \frac{20}{10 	imes 10^3} = \frac{20}{10000} = 2 	imes 10^{-3} \, 	ext{mho}$ (or Siemens).Therefore, the correct option is $D$.

The amplification factor of a triode is $20$. Its plate resistance is $10 \, k\Omega$. The mutual conductance is:

Similar Questions

The correct relation for a triode is

How many electrodes are in the tetrode?

Which of the following are triode constants?

The current in a triode at anode potential $100 \, V$ and grid potential $-1.2 \, V$ is $7.5 \, mA$. If the grid potential is changed to $-2.2 \, V$, the current becomes $5.5 \, mA$. The value of transconductance $(g_m)$ will be .... $mili \, mho$.

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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module