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 $



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