The plate resistance of two triode valves is | vedclass

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

Question

(B) The voltage amplification $A_V$ for a triode valve is given by the formula: $A_V = \frac{\mu R_L}{r_p + R_L}$,where $\mu$ is the amplification fac

Vedclass · Accepted Answer

$4/3$

Vedclass · Answer

(B) The voltage amplification $A_V$ for a triode valve is given by the formula: $A_V = \frac{\mu R_L}{r_p + R_L}$,where $\mu$ is the amplification factor,$r_p$ is the plate resistance,and $R_L$ is the load resistance.Given $\mu = 40$ and $R_L = 4 \, k\Omega$.For the first valve,$r_{p1} = 2 \, k\Omega$,so $A_1 = \frac{40 	imes 4}{2 + 4} = \frac{160}{6} = \frac{80}{3}$.For the second valve,$r_{p2} = 4 \, k\Omega$,so $A_2 = \frac{40 	imes 4}{4 + 4} = \frac{160}{8} = 20$.The ratio of voltage amplification is $\frac{A_1}{A_2} = \frac{80/3}{20} = \frac{80}{3 	imes 20} = \frac{4}{3}$.Wait,re-evaluating the ratio: $\frac{A_1}{A_2} = \frac{r_{p2} + R_L}{r_{p1} + R_L} = \frac{4 + 4}{2 + 4} = \frac{8}{6} = \frac{4}{3}$.

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:

Similar Questions

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?

In a triode amplifier,$\mu = 25$,$r_p = 40 \text{ k}\Omega$,and load resistance $R_L = 10 \text{ k}\Omega$. If the input signal voltage is $0.5 \text{ V}$,then the output signal voltage will be.....$\text{V}$.

In a diode valve,the cathode temperature must be ($\phi$ = work function):

In a triode valve,which of the following statements is correct?

$A$ triode has a plate resistance of $10 \, k\Omega$ and an amplification factor of $24$. If the input signal voltage is $0.4 \, V$ $(r.m.s.)$ and the load resistance is $10 \, k\Omega$,then the output voltage $(r.m.s.)$ is ...... $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module