A triode whose mutual conductance is 2.5 text | vedclass

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

Question

(B) The amplification factor $\mu$ of a triode is given by the product of mutual conductance $(g_m)$ and anode resistance $(r_p)$: $\mu = g_m 	imes r

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The amplification factor $\mu$ of a triode is given by the product of mutual conductance $(g_m)$ and anode resistance $(r_p)$: $\mu = g_m 	imes r_p = 2.5 	imes 10^{-3} 	ext{ A/V} 	imes 20 	imes 10^3 	ext{ }\Omega = 50$. For a triode amplifier,the voltage gain $(A)$ is given by the formula: $A = \frac{\mu R_L}{r_p + R_L}$,where $R_L$ is the load resistance. Substituting the given values: $10 = \frac{50 	imes R_L}{20 + R_L}$. Multiplying both sides by $(20 + R_L)$: $10(20 + R_L) = 50 R_L$. $200 + 10 R_L = 50 R_L$. $200 = 40 R_L$. $R_L = \frac{200}{40} = 5 	ext{ k}\Omega$.

$A$ triode whose mutual conductance is $2.5 \text{ mA/V}$ and anode resistance is $20 \text{ k}\Omega$,is used as an amplifier whose amplification factor is $10$. The load resistance connected in the plate circuit will be.....$\text{k}\Omega$.

Similar Questions

In the mutual characteristic curve of a triode,which point represents the cut-off voltage?

The grid in a triode valve is used

Why is a vacuum tube called a valve in electronics?

In a triode,the cathode,grid,and plate are at potentials of $0 \, V$,$-2 \, V$,and $80 \, V$ respectively. Electrons are emitted from the cathode with an initial kinetic energy of $3 \, eV$. What is the energy of the electron when it reaches the plate in $eV$?

The transconductance of a triode amplifier is $2.5 \, \text{mS}$ (milli-mho) and its plate resistance is $20 \, \text{k}\Omega$. If the voltage amplification is $10$, find the load resistance in $\text{k}\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module