Question

The current i in the network is:

• A. 0.2 A
• B. 0.6 A
• C. 0.3 A
• D. 0 A

Answer 1

Answer: (B)

0.6 A

Answer 2

Assuming ideal diodes with zero forward voltage drop, we set the negative terminal of the battery to 0V and the positive terminal to 9V. The current $i$ flows through a 5$\Omega$ resistor. By applying Kirchhoff's current law and diode forward bias conditions, we find the potential at node C to be 6V. Therefore, the current $i = (9V - 6V) / 5\Omega = 0.6A$.