Q. A voltage of peak value 283 V and varying frequency is applied to a series L – C – R combination in which $R = 3\ \Omega$, $L = 25$ mH and $C = 400\mu F$. The frequency (in Hz)of the source at which maximum power is dissipated in the above, is
A) 51.5
B) 50.7
C) 51.1
D) 50.3
Explanation
(d) A series resonance circuit admits maximum current, as
$P = i^{2}R$
So, power dissipated is maximum at resonance.
So, frequency of the source at which maximum power is dissipated in the circuit is
$v = \frac{1}{2\pi\sqrt{LC}}$
$= \frac{1}{2 \times 3.14\sqrt{25 \times 10^{- 3} \times 400 \times 10^{- 6}}}$
$= \frac{1}{2 \times 3.14\sqrt{10^{- 5}}} = 50.3\ Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.