The amplification factor of a triode valve is | vedclass

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(C) The amplification factor $\mu$ of a triode valve is defined as the ratio of the change in plate voltage to the change in grid voltage for a consta

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$4.5$

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(C) The amplification factor $\mu$ of a triode valve is defined as the ratio of the change in plate voltage to the change in grid voltage for a constant plate current.Mathematically,$\mu = \left| \frac{\Delta V_p}{\Delta V_g} ight|$.Given: $\mu = 15$ and $\Delta V_g = 0.3 \,V$.To keep the plate current constant,the change in plate voltage $\Delta V_p$ is given by:$\Delta V_p = \mu 	imes \Delta V_g$$\Delta V_p = 15 	imes 0.3 = 4.5 \,V$.Therefore,the change in plate voltage is $4.5 \,V$.

The amplification factor of a triode valve is $15$. If the grid voltage is changed by $0.3 \,V$,what is the change in plate voltage required to keep the plate current constant (in $V$)?

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