For the transistor circuit shown below, if `beta=100`, voltage drop between emitter and base is `0.7V` then value of `V_(CE)` will be A. `10 V`B. `5 V`C. `13 V`D. `0 V`

For the transistor circuit shown below, if `beta=100`, voltage drop be

1.	For the transistor circuit shown below, if `beta=100`, voltage drop between emitter and base is `0.7V` then value of `V_(CE)` will be A. `10 V`B. `5 V`C. `13 V`D. `0 V`
Answer» Correct Answer - C `i_(b)=(5-0.7)/8.6=0.5 mA implies I_(c)=betaI_(b)=100xx0.5 mA` By using `V_(CE)=V_(C C)-I_(C)R_(L)=18-50xx10^(-3)xx100=13V`

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For the transistor circuit shown below, if `beta=100`, voltage drop between emitter and base is `0.7V` then value of `V_(CE)` will be A. `10 V`B. `5 V`C. `13 V`D. `0 V`

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