First slide

Diodes

Question

The current i in the network is:

Easy

A

0.6 A
B

0 A
C

0.2 A
D

0.3 A

Solution

Both diodes are in reverse bias conditions. Hence diodes do not allow current to pass.
$\begin{array}{l} i = \frac{v}{R_{e f f}} = \frac{9}{5 + 10 + 5 + 10} \\ \Rightarrow i = \frac{9}{30} = 0.3 A \end{array}$

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