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In the circuit shown in figure, when the input voltage of thebase resistanceis $10 \mathrm{~V}, \mathrm{~V}_{\mathrm{BE}}$ is zero and $\mathrm{V}_{\mathrm{CE}}$ is also zero. Find the values of $\mathrm{I}_{\mathrm{B}}, \mathrm{I}_{\mathrm{c}}$ and $\beta$.
Solution:
1437 Upvotes
Verified Answer
As given that,
voltage across $R_B=10 \mathrm{~V}$
Resistance $R_B=400 \mathrm{k} \Omega$
$$
\begin{aligned}
&\mathrm{V}_{\mathrm{BE}}=0, \mathrm{~V}_{\mathrm{CE}}=0 \\
&\mathrm{R}_{\mathrm{C}}=3 \mathrm{k} \Omega
\end{aligned}
$$
$$
\begin{gathered}
\mathrm{I}_{\mathrm{B}}=\frac{\text { voltage across } \mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}} \\
=\frac{10}{400 \times 10^3}=25 \times 10^{-6} \mathrm{~A}=25 \mu \mathrm{A} \\
\mathrm{I}_{\mathrm{C}}=\frac{\left(\text { voltage acrossR }_{\mathrm{C}}\right)}{\mathrm{R}_{\mathrm{C}}}=\frac{10}{3 \times 10^3}
\end{gathered}
$$
$$
=3.33 \times 10^{-3} \mathrm{~A}=3.33 \mathrm{~mA}
$$
Current gain $\beta=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=\frac{3.33 \times 10^{-3}}{25 \times 10^{-6}}$
$$
=1.33 \times 10^2=133
$$
voltage across $R_B=10 \mathrm{~V}$
Resistance $R_B=400 \mathrm{k} \Omega$
$$
\begin{aligned}
&\mathrm{V}_{\mathrm{BE}}=0, \mathrm{~V}_{\mathrm{CE}}=0 \\
&\mathrm{R}_{\mathrm{C}}=3 \mathrm{k} \Omega
\end{aligned}
$$
$$
\begin{gathered}
\mathrm{I}_{\mathrm{B}}=\frac{\text { voltage across } \mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}} \\
=\frac{10}{400 \times 10^3}=25 \times 10^{-6} \mathrm{~A}=25 \mu \mathrm{A} \\
\mathrm{I}_{\mathrm{C}}=\frac{\left(\text { voltage acrossR }_{\mathrm{C}}\right)}{\mathrm{R}_{\mathrm{C}}}=\frac{10}{3 \times 10^3}
\end{gathered}
$$
$$
=3.33 \times 10^{-3} \mathrm{~A}=3.33 \mathrm{~mA}
$$
Current gain $\beta=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=\frac{3.33 \times 10^{-3}}{25 \times 10^{-6}}$
$$
=1.33 \times 10^2=133
$$
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