The given circuit has two ideal diodes connected as shown in the figure. The current flowing through the resistance R1 will be [figure]
- 2.5 A
- 10.0 A
- 1.43 A
- 3.13 A
Answer
(1) 2.5 A
Solution
The upper (long-plate) terminal of the cell is positive, so the top wire is at the higher potential.
D1 has its p-side at the lower potential reverse biased no current in R2.
D2 is forward biased (ideal, zero resistance) R1 and R3 are in series.
I = 10R1 + R3 = 102 + 2 = 2.5 A
Answer: (1)