In the following common emitter circuit,if be | vedclass

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(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the output loop:$V_{CC} = I_C R_C + V_{CE}$Given $V_{CC} = 15 \, V$,$V_{CE} = 7 \, V$,and $R_C = 2 \,

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$0.04$

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(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the output loop:$V_{CC} = I_C R_C + V_{CE}$Given $V_{CC} = 15 \, V$,$V_{CE} = 7 \, V$,and $R_C = 2 \, k\Omega = 2000 \, \Omega$.$15 \, V = I_C (2000 \, \Omega) + 7 \, V$$I_C (2000 \, \Omega) = 15 \, V - 7 \, V = 8 \, V$$I_C = \frac{8 \, V}{2000 \, \Omega} = 0.004 \, A = 4 \, mA$Now,using the relationship 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 $I_B$ is.......$mA$.

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