1.

When the plate voltage of a triode is 150 V, its cut-ff voltage is `-5V`. On increasing the plate voltate to 200V, the cut-off voltage can beA. `-4.5`VB. `-5.0`VC. 2.3 VD. `-6.66 V`

Answer» Correct Answer - D
`V_(p_(1))=150V,V_(P_(2)) = 200V, V_(g_(1)) = -5V, V_(g_(2))=?`
`mu`= amplification factor of a triode valve, it is a characteristic of valve remains constant.
So long as `l_(p)` is constant.
`thereforemu=-V_(P_(1))/(V_(q_(1)))=-V_(p_(2))/V_(g_(2))`
`(150)/(-5) = (200)/(V_(g_(2)))`
`V_(g_(2)) = -20/3 = -6.66V`


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