Q140 / 155 · WBJEE 2016 · Physics
In case of a bipolar transistor $\beta=45$. The potential drop across the collector resistance $\begin{array}{llllll}\text { of } 1 \mathrm{k} \Omega & \text { is } 5 \mathrm{V} . & \text { The base } & \text { current } & \text { is }\end{array}$ approximately
- A$222 \mu \mathrm{A}$
- B$55 \mu \mathrm{A}$
- Correct answerC$111 \mu \mathrm{A}$
- D$45 \mu \mathrm{A}$
Solution
Current gain $\beta=45$
We know that, $\quad \beta=\frac{\Delta I_{c}}{\Delta I_{b}}$
$\therefore$
$$
\frac{\Delta I_{c}}{\Delta I_{b}}=45
$$
Potential drop across $1 \mathrm{k} \Omega$ or $1 \times 10^{3} \Omega$
Collector resistor is $5 \mathrm{V}$.
$$
\begin{aligned}
V &=I_{C} R \\
I_{C} &=\frac{V}{R}=\frac{5}{1 \times 10^{3}} \\
&=5 \times 10^{-3} \mathrm{A}
\end{aligned}
$$
From Eqs. (i) and (ii).
$$
\begin{aligned}
\Delta I_{B} &-\frac{\Delta I_{C}}{45} \\
&=\frac{5 \times 10^{-3}}{45}=\frac{1}{9} \times 10^{-3} \\
&=0.111 \times 10^{-3} \mathrm{A} \\
&=111 \times 10^{-6} \mathrm{A}=111 \mu \mathrm{A}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.