Question

Find the current through the battery in each of the circuits shown in figure (45-E5).

Answer 1

Both diodes are forward biased. Thus the net diode resistance is 0.

$i=\frac{5}{\frac{10\times 10}{10+10}}=\frac{5}{5}=1A$

One diode is forward biased and other is reverse biased.

Current passes through the forward biased diode only.

$i=\frac{V}{R_{net}}=\frac{5}{10+0}=0.5A$