A change of 0.04 V takes place between the b | vedclass

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(B) Given: $\Delta I_b = 20 \ \mu A = 20 	imes 10^{-6} \ A$,$\Delta I_c = 2 \ mA = 2 	imes 10^{-3} \ A$,and $\Delta V_{BE} = 0.04 \ V$.The input res

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$2 \ k\Omega \ \& \ 100$

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(B) Given: $\Delta I_b = 20 \ \mu A = 20 	imes 10^{-6} \ A$,$\Delta I_c = 2 \ mA = 2 	imes 10^{-3} \ A$,and $\Delta V_{BE} = 0.04 \ V$.The input resistance $R_{	ext{input}}$ is defined as the ratio of the change in base-emitter voltage to the change in base current:$R_{	ext{input}} = \frac{\Delta V_{BE}}{\Delta I_b} = \frac{0.04}{20 	imes 10^{-6}} = \frac{0.04 	imes 10^6}{20} = 2 	imes 10^3 \ \Omega = 2 \ k\Omega$.The $AC$ current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current:$\beta = \frac{\Delta I_c}{\Delta I_b} = \frac{2 	imes 10^{-3}}{20 	imes 10^{-6}} = \frac{2000}{20} = 100$.Thus,the input resistance is $2 \ k\Omega$ and the $AC$ current gain is $100$.

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