Question

A sinusoidal voltage with a frequency of $50Hz$ is applied to a series $LCR$ circuit with a resistance of $5\Omega$, inductance of $20mH$ 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, $f=50Hz$, resistance $R=5\Omega$, inductance of indutor, $L=20mH$ and capacitance, $C=500\mu F$
In an AC series $LCR$ 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, $\omega =2\pi f=2\pi \times 50=100\pi )$
$Z\approx \sqrt{25}=5\Omega$