Question

Determine the current and quality factor at resonance for a series LCR circuit with $L=1.00mH,C=1.00nF,$ and $R=100\Omega$ connected to an ac source having peak voltage of $100V.$

Answer 1

Given:-

$L=1.00mH=1\times {10}^{-3}H,C=1.00\mu F=1\times {10}^{-9}F,R=100\Omega ,E_0=100V$

$I_0=\frac{E_0}{\sqrt{R^2+(\omega L-\frac{1}{\omega C}{)}^2}}=\frac{E_0}{Z}$ $\{atResonance\omega L=\frac{1}{\omega C},HenceZ=R\}$

$I_v=\frac{I_0}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{2}=\frac{1.44}{2}=0.707A$

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{1}{100}\sqrt{\frac{1.0\times {10}^{-3}}{1.0\times {10}^{-9}}}=\frac{1}{100}\times {10}^3=10$