For the transistor circuit shown in figure, if β =100, voltage drop across emitter and base is 0.7 V, then the value of VCE will be

Question

For the transistor circuit shown in figure, if β =100, voltage drop across emitter and base is 0.7 V, then the value of VCE will be (A) Zero (B) 5 V (C) 10 V (D) 13 V

Tardigrade Admin · Accepted Answer

We have, ib =(5-0.7/8.6)=0.5 mA ∴ ic =βi b=100 × 0.5 mA =50 mA By using VC E =VC C-iC RL =18-50 × 10-3 × 100 =18-5=13 V

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

Zero

5 V

10 V

13 V

We have, $i_{b} = \frac{5 - 0.7}{8.6} = 0.5 m A$
$∴ i_{c} = β_{ib} = 100 \times 0.5 m A$
$= 50 m A$
By using
$V_{CE} = V_{CC} - i_{C} R_{L}$
$= 18 - 50 \times 1 0^{- 3} \times 100$
$= 18 - 5 = 13 V$

Q. For the transistor circuit shown in figure, if β=100, voltage drop across emitter and base is 0.7V, then the value of VCE​ will be

Zero

5 V

10 V

13 V

We have, ib​=8.65−0.7​=0.5mA ∴ic​=βib​=100×0.5mA =50mA By using VCE​=VCC​−iC​RL​ =18−50×10−3×100 =18−5=13V

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

We have, $i_{b} = \frac{5 - 0.7}{8.6} = 0.5 m A$
$∴ i_{c} = β_{ib} = 100 \times 0.5 m A$
$= 50 m A$
By using
$V_{CE} = V_{CC} - i_{C} R_{L}$
$= 18 - 50 \times 1 0^{- 3} \times 100$
$= 18 - 5 = 13 V$