The current I in the circuit shown below is : (All diodes are ideal and identical) [figure]
- 13 A
- 152 A
- 53 A
- 59 A
Answer
(2) 152 A
Solution
Current leaves the + terminal, rises in the left wire and enters every branch from the left.
∵ The diodes in the 4 Ω and 2 Ω branches point left to right ⇒ forward biased (ideal diode = zero resistance).
∵ The diodes in the 3 Ω and 5 Ω branches point right to left ⇒ reverse biased (no current).
⇒ 4 Ω and 2 Ω are in parallel: Req = 4 24 + 2 = 43 Ω
∴ I = 104/3 = 152 A
Answer: (2)