Question

For the transistor circuit shown in figure, if $\beta =100,$ voltage drop across emitter and base is 0.7 V, then the value of $V_{CE}$ will be :

• A. Zero
• B. 5 V
• C. 10 V
• D. 13 V

Answer 1

The correct option is D 13 V

We have, $i_b=\frac{5-0.7}{8.6}=0.5mA$
$\therefore i_c={\beta }_{ib}=100\times 0.5mA$
$=50mA$
By using
$V_{CE}=V_{CC}-i_C{R}_L$
$=18-50\times {10}^{-3}\times 100$
$=18-5=13V$