In the following common emitter circuit,if be | vedclass

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(B) For the output loop of the common emitter transistor circuit,applying Kirchhoff's Voltage Law $(KVL)$:$V_{CC} = I_C R_C + V_{CE}$Given $V_{CC} = 1

Vedclass · Accepted Answer

$0.04$

Vedclass · Answer

(B) For the output loop of the common emitter transistor circuit,applying Kirchhoff's Voltage Law $(KVL)$:$V_{CC} = I_C R_C + V_{CE}$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$$I_C 	imes 2000 = 15 - 7 = 8 \, V$$I_C = \frac{8}{2000} = 4 	imes 10^{-3} \, A = 4 \, mA$Now,using the current gain relation $\beta = \frac{I_C}{I_B}$:$I_B = \frac{I_C}{\beta} = \frac{4 \, mA}{100} = 0.04 \, mA$.

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 find $I_B$ in $mA$.

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