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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` |
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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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