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In a diode vaccum tube, the plate curren...

In a diode vaccum tube, the plate currentis 5 mA, when the plate voltage is 160 V, a grid is introduced between the plate and cathode and a voltage of -2V is applied to it. The plate current will becomes
(if `g_(m) = 5 xx 10^(-3)Omega^(-1)`

A

20 mA

B

10 mA

C

4 mA

D

7.5 mA

Text Solution

Verified by Experts

The correct Answer is:
C

`V_(g_(1))=0`
`l_(p)` = plate current =5mA
`V_(p)` = plate voltage =150 mV
`V_(g)` = grid voltage=`-2V`
When `V_(g)` = 0, it is a diode
When `V_(g)=-2`, It is a triode
`therefore g_(m) = ((Deltal_(p))/(DeltaV_(g)))_(V_(p)="constant")`
`5 xx 10^(-4) = (l_(2)-l_(1))/(V_(2)-V_(1))(l_(2)-5 xx 10^(-3))/(-2-0)`
`-10 xx 10^(-4) + 5 xx 10^(-3) = l_(2)`
`-1 xx 10^(-3) + 5 xx 10^(-3) = l_(2)`
`l_(2)=4mA`
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