Question

Current in the circuit will be

• A. $\frac{5}{40}A$
• B. $\frac{5}{50}A$
• C. $\frac{5}{10}A$
• D. $\frac{5}{20}A$

Answer 1

Answer: (B)

The diode in lower branch is forward and diode in upper branch is reversed biased,
so $i=\frac{5}{20+30}=\frac{5}{50}A$