Amplification factor of a triode is 10. When the plate potential … - Vidyadip

MCQ

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 $4\,mA $ is observed. If plate potential is changed to $160 \,volt$ and grid potential is kept at $-7 \,volt$, then the plate current will be.....$mA$

✓
$1.69$
B
$3.95 $
C
$2.87$
D
$7.02$

✓

Answer

Correct option: A.

$1.69$

a
(a)${i_p} = k{({V_p} + \mu {V_g})^{3/2}}mA$
$==> 4 = k(200 -10 × 4)3/2 = k × (160)3/2 ….(i)$
and ${i_p} = k{(160 - 10 \times 7)^{3/2}} = k \times {(90)^{3/2}}$ ….$(ii)$
From equation $ (i) $ and $(ii) $ we get
${i_p} = 4 \times {\left( {\frac{{90}}{{160}}} \right)^{3/2}} = 4 \times {\left( {\frac{3}{4}} \right)^3} = 1.69\,mA$

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