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(C) Given: Input signal voltage change $\Delta V_{in} = 10 \ mV = 10 	imes 10^{-3} \ V$.Change in base current $\Delta I_{B} = 50 \ \mu A = 50 	imes

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$25$

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(C) Given: Input signal voltage change $\Delta V_{in} = 10 \ mV = 10 	imes 10^{-3} \ V$.Change in base current $\Delta I_{B} = 50 \ \mu A = 50 	imes 10^{-6} \ A$.Change in collector current $\Delta I_{C} = 5 \ mA = 5 	imes 10^{-3} \ A$.Load resistance $R_{L} = 5 \ k\Omega = 5 	imes 10^{3} \ \Omega$.The change in output voltage $\Delta V_{out}$ is given by the product of the change in collector current and the load resistance:$\Delta V_{out} = \Delta I_{C} 	imes R_{L}$Substituting the values:$\Delta V_{out} = (5 	imes 10^{-3} \ A) 	imes (5 	imes 10^{3} \ \Omega)$$\Delta V_{out} = 25 \ V$.Thus, the change in output voltage is $25 \ V$.

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