Question

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

• A. 10 V
• B. 5 V
• C. 13 V
• D. 0 V

Answer 1

The correct option is C 13 V

$i_b=\frac{5-0.7}{8.6}=0.5mA\Rightarrow I_e=\beta I_b=100\times 0.5mA$
By using $V_{CE}=V_{CC}-I_e{R}_L=18-50\times {10}^{-3}\times 100=13V$.