In the following common emitter circuit,if be | vedclass

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(B) For the output loop of the common emitter transistor circuit,the Kirchhoff's voltage law equation is given by:$V_{CC} = I_C R_C + V_{CE}$Given $V_

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$0.04$

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(B) For the output loop of the common emitter transistor circuit,the Kirchhoff's voltage law equation is given by:$V_{CC} = I_C R_C + V_{CE}$Given $V_{CC} = 15\,V$,$V_{CE} = 7\,V$,and $R_C = 2\,k\Omega = 2000\,\Omega$.Substituting the values:$15 = I_C 	imes 2000 + 7$$15 - 7 = I_C 	imes 2000$$8 = I_C 	imes 2000$$I_C = \frac{8}{2000} = 4 	imes 10^{-3}\,A = 4\,mA$Now,using the relation between collector current $(I_C)$ and base current $(I_B)$:$\beta = \frac{I_C}{I_B}$$I_B = \frac{I_C}{\beta} = \frac{4\,mA}{100} = 0.04\,mA$

In the following common emitter circuit,if $\beta = 100$,$V_{CE} = 7\,V$,$V_{BE} = 0\,V$ (negligible),and $R_C = 2\,k\Omega$,then the value of $I_B$ is.......$mA$.

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