Physics - EMI/AC Question with Solution | TestHub

A series resonance circuit admits maximum current, as

$P=i_{\text{rms}}^{2}z\cos 0^o$(Z = Impedance)

So, power dissipated is maximum at resonance.

So, frequency of the source at which maximum power is dissipated in the circuit is

$f=\frac{1}{2\pi \sqrt{LC}}$

$=\frac{1}{2\times 3.14\sqrt{25\times {10}^{-3}\times 400\times {10}^{-6}}}$

$=\frac{1}{2\times 3.14\sqrt{{10}^{-5}}}=50.3 \mathrm{H}\mathrm{z}$