Solution Electrical Numerical 21

Numerical 21: Find the frequency at which the following circuit resonant

Solution: $\displaystyle \mathrm{L\ = L_1 +L_2 =0.1+0.2\ =\ 0.3\, H}$ 

$\displaystyle \mathrm{\frac{1}{C}\ = \frac{1}{C_1} +\frac{1}{C_2} = \frac{1}{2}+\frac{1}{0.5} =\ 2.5\, H}$ 

C = 0.4μf = 0.4 x $\displaystyle \mathrm{10^{-6}}$ f

Resonating frequency,

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

$\displaystyle \mathrm{f=\frac{1}{2\pi \sqrt{0.3\times 0.4\times 10^{-6}}}}$ = 459.44 Hz