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In the circuit given below , when the input voltage of the base resistance is $10\;V,V_{BE}$ is zero. and $V_{CE}$ is also zero. Find the value of $I_b$

$\begin{array}{1 1} 2.5 \mu A \\ 25 \mu A \\ 250 \mu A \\ 0.25\;\mu A \end{array}$

given $V_i= 10\;V$
$V_{BE}=0$
$V_{CE}= 0$
$R_B=400 \;k \Omega= 400 \times 10^3\;\Omega$
$R_c=3 \;k \Omega= 3 \times 10^3 \Omega$
$V_i- V_{BE}= R_bI_b$
$10-0=(400 \times 10^3) I_b$
$I_b= \large\frac{10}{400 \times 10^3}$$=25 \times 10^{-6}A$
$\qquad= 25 \mu A$