Question

Find the current through the battery in each of the circuits shown in figure.

Answer 1

We know that when a diode is forward biassed, it has zero resistance ideally. So, it can be replaced by a short circuit. When a diode is reverse biassed, it has infinite resistance ideally. So, it can be replaced by an open circuit.

(a) In the given circuit diagram, both diodes are forward biassed. So, the resistance of both of them is zero. Thus, the diode resistance is zero.
$i=\frac{5}{{\displaystyle \frac{10\times 10}{10+10}}}=\frac{5}{5}=1\mathrm{A}$

(b) One diode is forward biassed and the other is reverse biassed. The reverse-biassed diode is replaced by an open circuit, so no current flows through this branch.
The current passes through the forward-biassed diode only.
$i=\frac{V}{R_{\mathrm{net}}}=\frac{5}{10}=0.5\mathrm{A}$