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(D) Given:Collector supply voltage $V_{CC} = 8\, V$Load resistance $R_L = 800\,\Omega$Voltage drop across load $V_{out} = 0.8\, V$Current amplificatio

Vedclass · Accepted Answer

$V_{CE} = 7.2\, V; I_B = 40\,\mu A$

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(D) Given:Collector supply voltage $V_{CC} = 8\, V$Load resistance $R_L = 800\,\Omega$Voltage drop across load $V_{out} = 0.8\, V$Current amplification factor $\alpha = 25/26$$1$. Collector-emitter voltage $V_{CE}$:$V_{CE} = V_{CC} - V_{out} = 8\, V - 0.8\, V = 7.2\, V$$2$. Collector current $I_C$:$I_C = \frac{V_{out}}{R_L} = \frac{0.8\, V}{800\,\Omega} = 10^{-3}\, A = 1\, mA$$3$. Base current $I_B$:We know that $\beta = \frac{\alpha}{1 - \alpha} = \frac{25/26}{1 - 25/26} = \frac{25/26}{1/26} = 25$Since $I_C = \beta I_B$,we have:$I_B = \frac{I_C}{\beta} = \frac{1\, mA}{25} = 0.04\, mA = 40\,\mu A$Thus,$V_{CE} = 7.2\, V$ and $I_B = 40\,\mu A$.

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