Question

A series resonant circuit contains $L=5/\pi mH,C=200/\pi ,R=100\Omega$. If a source of emf $e=200\sin 1000\pi t$ is applied with a rms current $1.41A$. Find the magnitude of voltage across the inductor.

Answer 1

$X_L=\omega L=1000\pi \times \frac{5}{\pi }\times {10}^{-3}=5$ $\Omega$

Equivalent Impedance $Z=\sqrt{(X_L-X_C{)}^2+R^2}$

$\Rightarrow X_C=\frac{1}{C\omega }=\frac{1}{200/\pi \times 1000\pi \times {10}^{-6}}=5$ $\omega$

$Z=\sqrt{(X_L-X_C{)}^2+R^2}$

$\Rightarrow R=1000$ $\Omega$

This is the condition of resonance

Voltage across inductor is $V_{Lmax}=I_{max}{X}_L=1.41\times \sqrt{2}\times 5=10$ $V$

$\therefore V_{rms}=I_{rms}\times 5=7.05$ $V$