The amplification factor of a triode is 20. I | vedclass

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(B) The amplification factor $\mu$ of a triode is defined as the ratio of the change in plate voltage to the change in grid voltage for a constant pla

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$4$

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(B) The amplification factor $\mu$ of a triode is defined as the ratio of the change in plate voltage to the change in grid voltage for a constant plate current: $\mu = -\frac{\Delta V_p}{\Delta V_G}$.Given: $\mu = 20$ and $\Delta V_G = -0.2\, V$ (since the potential is reduced).To keep the plate current constant,we rearrange the formula: $\Delta V_p = -\mu 	imes \Delta V_G$.Substituting the values: $\Delta V_p = -20 	imes (-0.2\, V) = 4\, V$.Therefore,the plate voltage must be increased by $4\, V$.

The amplification factor of a triode is $20$. If the grid potential is reduced by $0.2\, V$,then to keep the plate current constant,its plate voltage must be increased by how many volts?

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