In the following common emitter circuit,if be | vedclass

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(B) From the collector-emitter loop,applying Kirchhoff's Voltage Law $(KVL)$:$V_{CC} = I_C R_C + V_{CE}$Given $V_{CC} = 15 	ext{ V}$,$V_{CE} = 7 	ex

Vedclass · Accepted Answer

$0.04$

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(B) From the collector-emitter loop,applying Kirchhoff's Voltage Law $(KVL)$:$V_{CC} = I_C R_C + V_{CE}$Given $V_{CC} = 15 	ext{ V}$,$V_{CE} = 7 	ext{ V}$,and $R_C = 2 	ext{ k}\Omega = 2000 \, \Omega$.$15 = I_C 	imes 2000 + 7$$8 = I_C 	imes 2000$$I_C = \frac{8}{2000} 	ext{ A} = 4 	imes 10^{-3} 	ext{ A} = 4 	ext{ mA}$.Now,using the current gain relation $\beta = \frac{I_C}{I_B}$:$I_B = \frac{I_C}{\beta} = \frac{4 	ext{ mA}}{100} = 0.04 	ext{ mA}$.

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

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