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)
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} )}^{3/2}}=1.69mA $
BETA
