Physics - Semiconductors Question with Solution | TestHub

Current gain $\beta =45$ We know that, $\beta =\frac{\Delta I_c}{\Delta I_b}$ $\therefore$ $$\frac{\Delta I_c}{\Delta I_b}=45$$ Potential drop across $1\mathrm{k}\Omega$ or $1\times 10^3\Omega$ Collector resistor is $5\mathrm{V}$. $$\begin{array}{rl}V& =I_{C}R\\ I_C& =\frac{V}{R}=\frac{5}{1\times 10^3}\\ & =5\times 10^{-3}\mathrm{A}\end{array}$$ From Eqs. (i) and (ii). $$\begin{array}{rl}\Delta I_B& -\frac{\Delta I_C}{45}\\ & =\frac{5\times 10^{-3}}{45}=\frac{1}{9}\times 10^{-3}\\ & =0.111\times 10^{-3}\mathrm{A}\\ & =111\times 10^{-6}\mathrm{A}=111\mu \mathrm{A}\end{array}$$