An inductor 200mH, a capacitor $100\mu\text{F}$ and a resistor $10\Omega$ are connected in series to an a.c. source of 100V, having variable frequency.
- At what frequency of the applied voltage will the power factor of the circuit be 1?
- What will be the current amplitude at this frequency?
- Calculate the Q-factor of the circuit.
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- Since the power factor $\varphi=1$ it means the phase difference between voltage and the current is zero.
$\text{wL}=\frac{1}{\omega\text{C}}$
$\omega^2=\frac{1}{\text{LC}}\Rightarrow\text{v}=\frac{1}{2\pi\sqrt{\text{LC}}}=35.58\text{Hz}$
- Current Amplitude, $\text{I}=\frac{\text{V}_{\text{eff}}}{\text{Z}}=\frac{\text{V}_{\text{eff}}}{\text{R}}=\frac{100}{10}=10\text{A}$ $(\therefore\text{Z = R})$
- Quality factor, $\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{}\text{C}}=4.47$
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