MCQ
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......$V$
- A$10$
- B$5$
- ✓$13$
- D$0$
✓
Answer
Correct option: C.
$13$
c
(c)${i_b} = \frac{{5 - 0.7}}{{8.6}} = 0.5\,mA$ ==> ${I_c} = \beta \,{I_b} = 100 \times 0.5\,mA$
By using ${V_{CE}} = {V_{CC}} - {I_c}{R_L} = 18 - 50 \times {10^{ - 3}} \times 100 = 13\,V$
(c)${i_b} = \frac{{5 - 0.7}}{{8.6}} = 0.5\,mA$ ==> ${I_c} = \beta \,{I_b} = 100 \times 0.5\,mA$
By using ${V_{CE}} = {V_{CC}} - {I_c}{R_L} = 18 - 50 \times {10^{ - 3}} \times 100 = 13\,V$
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