Q. A combination of two resistance of 2 W and 2/3 W connected in parallel is joined across a battery of emf of 3 V and of negligible internal resistance. The energy given out per sec will be
A) $\frac{1}{2} \times 3 \times$3 J
B) $\frac{1}{2} \times \frac{1}{3} \times 3 \times 3$ J
C) $2 \times 3\ $J
D) $3 \times 3 \times 2$ J
Explanation
(d) Total power spend across two resistors connected
in parallel to battery $= \frac{V^{2}}{R_{1}} + \frac{V^{2}}{R_{2}}$
$= \frac{3 \times 3}{2} + \frac{3 \times 3}{2/3} = \frac{36}{2} = 18$
$\ = 3 \times 3 \times 2\ $J
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.