Semiconducting Devices Question 65
Question: In the grid circuit of a triode a signal $ E=2\sqrt{2}\cos \omega t $ is applied. If m = 14 and rp =10 kW then root mean square current flowing through $ R _{L}=12k\Omega $ will be
Options:
A) 1.27 mA
B) 10 mA
C) 1.5 mA
D) 12.4 mA
Show Answer
Answer:
Correct Answer: A
Solution:
$ A=\frac{\mu R _{L}}{r _{p}+R _{L}}=\frac{14\times 12}{10+12}=\frac{84}{11} $ .
Peak value of output signal $ V _{0}=\frac{84}{11}\times 2\sqrt{2}V $
Therefore $ V _{rms}=\frac{V _{0}}{\sqrt{2}}=\frac{84\times 2}{11}V $
Therefore r.m.s.
value of current through the load $ =\frac{84\times 2}{11\times 12\times 10^{3}}A=1.27mA $
