First slide

Diodes

Question

The current (in A) in the network is

Moderate

A

0.25
B

0.225
C

0.35
D

0.45

Solution

Both diodes are in reverse biasing, so can be treated as infinite resistance or broken wires
Now all resistors are in series
$\begin{array}{l} I = \frac{V}{R_{e f f}} = \frac{9}{(5 + 10 + 5 + 20)} = 0.225 A \end{array}$

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