Q. 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
A) 1.69 $mA$
B) 3.95 $mA$
C) 2.87
D) 7.02 $mA$
Explanation
(a) $i_{p} = k\left( V_{p} + \mu V_{g} \right)^{3/2}mA$
$\Rightarrow 4 = k(200 - 10 \times 4)^{3/2} = k \times (160)^{3/2}$ …(i)
and $i_{p} = k(160 - 10 \times 7)^{3/2} = k \times (90)^{3/2}$ …(ii)
From equation (i) and (ii), we get
$i_{p} = 4 \times \left( \frac{90}{160} \right)^{3/2} = 4 \times \left( \frac{3}{4} \right)^{3} = 1.69\ mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.