A) 0mA
B) 5mA
C) 10mA
D) 15mA
Correct Answer: B
Solution :
[b] \[1K\Omega \] Parallel to diode hence 5v across \[1K\Omega \]remaining 5V drop across 5000. \[V=IR\] \[5=1\times 500\] \[I={{10}^{-2}}A\] Apply ohm law across \[1K\Omega \] \[5=1\times 500\] \[I=5mA\]You need to login to perform this action.
You will be redirected in
3 sec