In a CE transistor,when the base current incr | vedclass

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(D) The input resistance $R_{	ext{in}}$ of a transistor is defined as the ratio of the change in base-emitter voltage $(\Delta V_{BE})$ to the change

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$20000$

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(D) The input resistance $R_{	ext{in}}$ of a transistor is defined as the ratio of the change in base-emitter voltage $(\Delta V_{BE})$ to the change in base current $(\Delta I_b)$.Given:$\Delta I_b = 60 \mu A = 60 	imes 10^{-6} \ A$$\Delta V_{BE} = 1.2 \ V$Using the formula:$R_{	ext{in}} = \frac{\Delta V_{BE}}{\Delta I_b}$$R_{	ext{in}} = \frac{1.2}{60 	imes 10^{-6}}$$R_{	ext{in}} = \frac{1.2 	imes 10^6}{60}$$R_{	ext{in}} = \frac{1200000}{60} = 20000 \ \Omega$Therefore,the input resistance is $20000 \ \Omega$.

In a $CE$ transistor,when the base current increases by $60 \mu A$,the change in base-emitter voltage is $1.2 \ V$. Then the input resistance is: (in $Omega$)

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