A) 1.71 A
B) 2.00 A
C) 2.31 A
D) 1.33A
Correct Answer: B
Solution :
We consider the diodes whether it is forward biased or reversed biased. Now, for a closed circuit (loop), KVL can be applied. In the given circuit, diode\[{{D}_{1}}\]is reverse biased while\[{{D}_{2}}\]is forward biased, so the circuit can be redrawn as (\[\because \]for ideal, diodes, reverse biased means open circuited and forward biased means short circuited) Apply KVL to get current flowing through the circuit \[-12+4i-2i=0\] \[\Rightarrow \] \[i=\frac{12}{6}=2A\]You need to login to perform this action.
You will be redirected in
3 sec