Q. Two resistances when connected in parallel across a cell of negligible internal resistance consume 4 times the power they would consume when connected in series. If one resistance is 5$\ \Omega$ , the other is
A) 1$\ \Omega$
B) 2.5$\ \Omega$
C) 5$\ \Omega$
D) 10$\ \Omega$
Explanation
(c) $\frac{V^{2}}{R} + \frac{V^{5}}{5} = 4 \times \left( \ \frac{V^{2}}{R + 5} \right)\ \ or\ \frac{R + 5}{5R} = \frac{4}{R + 5}$
On solving, we get $R = 5\Omega.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.