Question

A sinusoidal voltage with a frequency of $50\, Hz$ is applied to a series $L C R$ circuit with a resistance of $5\, \Omega$, inductance of $20\, mH$ and a capacitance of $500\, \mu F$. The magnitude of impedance of the circuit is closed to

• A. $19.2 \Omega$
• B. $14.4 \Omega$
• C. $9.6 \Omega$
• D. $5 \Omega$

Answer 1

Answer: (D)

Given, frequency of sinusoidal voltage, $f=50\, Hz$, resistance $R=5\, \Omega$, inductance of indutor, $L=20\, mH$ and capacitance, $C=500\, \mu F$
In an AC series $L C R$ circuit, the circuit impedance is given as
$Z=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}$
Hence, putting the given values, we get
$Z=\sqrt{5^{2}+\left(100 \pi \times 20 \times 10^{-3}-\frac{1}{100 \pi \times 500 \times 10^{-6}}\right)^{2}}$
$(\because$ angular frequency, $\omega=2 \pi f=2 \pi \times 50=100 \pi)$
$Z \approx \sqrt{25}=5 \Omega$