A) \[\frac{5}{40}A\]
B) \[\frac{5}{50}A\]
C) \[\frac{5}{10}A\]
D) \[\frac{5}{20}A\]
Correct Answer: B
Solution :
The diode in lower branch is forward biased and diode in upper branch is reverse biased \[\therefore \] \[i=\frac{5}{20+30}=\frac{5}{50}A\]
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