A) 0.4 A and 0.2 A
B) 0.2 A and 0.4 A
C) 0.1 A and 0.2 A
D) 0.2A and 0.1 A
Correct Answer: A
Solution :
[a] When positive terminal connected to A then diode D, is forward biased, current, \[I=\frac{2}{5}=0.4\,\,A\] When positive terminal connected to B then diode \[{{D}_{2}}\] is forward biased, current, \[I=\frac{2}{10}=0.2\,\,A\]You need to login to perform this action.
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