Question

When a $15V$ dc source was applied across a choke coild then a current of $5$ Amp flows in it. If the same coil is connected to a $15V$, $50$ rad/s ac source a current of $3Amp$ flows in the circuit and its resonance frequency if a $2500\mu f$ capacitor is connected in series with the coil.

Answer 1

For a coil, $Z=\sqrt{R^2+{\omega }^2{L}^2}$
$I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+{\omega }^2{L}^2}}$
for dc souce, $\omega =0$
$I=\frac{V}{R}\Rightarrow R=\frac{15}{5}=3\Omega ...\left(i\right)$
When ac is applied
$I=\frac{V}{Z}$
ie., $Z=\frac{15}{3.0}=5\Omega$
$\therefore R^2+X_L^2=25$
$X_L^2=25-9=16\Omega \Rightarrow X_L=4\Omega \Rightarrow L=\frac{4}{40}=0.08Henry$
Now when the capacitor is connected is series
$X_C=\frac{1}{\omega C}=\frac{1}{50\times 2500\times {10}^{-6}}=8\Omega$
$Z=Z=\sqrt{R^2+{(X_L-X_C)}^2}=\sqrt{3+{\left(48\right)}^2}=5\Omega$
$\therefore I=\frac{15}{5}=3A$
$\therefore P_{av}=v_{srm}{I}_{rms}\cos \varphi =I_{rms}^{2}R=3^2\times 3=27W$
Resonance frequency $f$
$f=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{\left(0.08\right)\times 2500\times {10}^{-6}}}=\frac{25\sqrt{2}}{\pi }=11.25Hz$