Question

In the given circuit, ideal diodes are used. Power delivered by battery is (in watt)

• A. 24 W
• B. 36 W
• C. 12 W
• D. $\frac{15}{9}\text{W}$

Answer 1

The correct option is B 36 W

$D_1$ is reverse biased, so there is no current through it. $D_2$ is forward biased, and the resistance seen by the battery is $6+3=9\Omega$.
Threfore the current through the battery is $i=\frac{18}{6+3}=\frac{18}{9}=2\text{A}$
$P=Vi=18\times 2=36\text{W}$