A) \[\frac{1}{15}A\]
B) \[\frac{1}{7}A\]
C) \[\frac{1}{25}A\]
D) \[\frac{1}{180}A\]
Correct Answer: B
Solution :
The junction diode is forward biased. Therefore, the effective resistance \[=25+10=35\,\Omega \] \[\therefore \] Current in diode \[I=\frac{5V}{35\,\Omega }=\frac{1}{7}A\]You need to login to perform this action.
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