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:

ip=k(Vp+μVg)3/2mA

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

nd ip=k(16010×7)3/2=k×(90)3/2 -.(ii)

From equation (i) and (ii) we get ip=4×(90160)3/2=4×(34)3=1.69mA



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