Electronic Devices Question 264

Question: Amplification factor of a triode is 10. When the plate potential is 200 volt and grid potential is - 4 volt, then the plate current of 4mA is observed. If plate potential is changed to 160 volt and grid potential is kept at - 7 volt, then the plate current will be

Options:

A) 1.69 mA

B) 3.95 mA

C) 2.87

D) 7.02 mA

Show Answer

Answer:

Correct Answer: A

Solution:

$ i _{p}=k{{(V _{p}+\mu V _{g})}^{3/2}}mA $

Therefore 4 = k(200 - 10 x 4)3/2 = k x (160)3/2 -.(i) a

nd $ i _{p}=k{{(160-10\times 7)}^{3/2}}=k\times {{(90)}^{3/2}} $ -.(ii)

From equation (i) and (ii) we get $ i _{p}=4\times {{( \frac{90}{160} )}^{3/2}}=4\times {{( \frac{3}{4} )}^{3}}=1.69mA $



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