14 times 10^{15} electrons reach the anode pe | vedclass

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

Question

(B) The current $i$ is given by the rate of flow of charge: $i = n 	imes e$,where $n = 14 	imes 10^{15} 	ext{ electrons/s}$ and $e = 1.6 	imes 10^

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) The current $i$ is given by the rate of flow of charge: $i = n 	imes e$,where $n = 14 	imes 10^{15} 	ext{ electrons/s}$ and $e = 1.6 	imes 10^{-19} 	ext{ C}$.$i = (14 	imes 10^{15}) 	imes (1.6 	imes 10^{-19}) = 22.4 	imes 10^{-4} 	ext{ A} = 2.24 	imes 10^{-3} 	ext{ A}$.The power consumed is $P = 448 	ext{ mW} = 448 	imes 10^{-3} 	ext{ W}$.Using the formula $P = V 	imes i$,we find the voltage $V = \frac{P}{i}$.$V = \frac{448 	imes 10^{-3}}{2.24 	imes 10^{-3}} = \frac{448}{2.24} = 200 	ext{ V}$.

$14 \times 10^{15}$ electrons reach the anode per second. If the power consumed is $448 \text{ mW}$,then the plate (anode) voltage is .... $V$.

Similar Questions

The value of plate current $I_P$ in the given circuit diagram will be in $mA$.

In a triode amplifier,the value of maximum gain is equal to

The amplification factor of a triode is $50$. If the grid potential is decreased by $0.20\, V$,what increase in plate potential will keep the plate current unchanged?

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?

If ${R_p} = 7 \text{ k}\Omega$ and ${g_m} = 2.5 \text{ millimho}$, then on increasing the plate voltage by $50 \text{ V}$, how much must the grid voltage be changed so that the plate current remains the same? (in $V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module