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JEE PHYSICS MCQ: SEMI CONDUCTOR DEVICES

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Application +4 | -1
On applying a potential of −1  volt- 1\;volt at the grid of a triode, the following relation between plate voltage Vp(volt){V_p}\left( {volt} \right) and plate current Ip{I_p} (in mAmA) is found Ip=0.125Vp−7.5{I_p} = 0.125{V_p} - 7.5. If on applying −3  volt- 3\;volt potential at grid and 300  V300\;V potential at plate, the plate current is found to be 5mA,5mA, then amplification factor of the triode is
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Correct option: A
100
Option A (100) is correct because the amplification factor is calculated as the ratio of change in plate voltage to change in grid voltage while keeping the plate current constant.

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