Alternating Current - LCR Series Circuit

In a series LCR circuit, $R = 20 \Omega$, $L = 1.5 \text{ H}$, and $C = 35 \mu\text{F}$ are connected to a $220 \text{ V}$, $50 \text{ Hz}$ AC source. Calculate (i) the reactance of the circuit and (ii) the impedance.

Given: $f = 50 \text{ Hz}$, $L = 1.5 \text{ H}$, $C = 35 \times 10^{-6} \text{ F}$, $R = 20 \Omega$.

1. Inductive Reactance: $X_L = 2\pi f L = 2 \times 3.14 \times 50 \times 1.5 = 471 \Omega$.
2. Capacitive Reactance: $X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times 3.14 \times 50 \times 35 \times 10^{-6}} \approx 90.95 \Omega$.
3. Net Reactance: $X = X_L - X_C = 471 - 90.95 = 380.05 \Omega$.
4. Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + (380.05)^2} \approx 380.57 \Omega$.

We first calculate the individual reactances using the frequency provided. Since $X_L > X_C$, the circuit is predominantly inductive. The impedance is then found using the Pythagorean relationship between resistance and net reactance.

Calculate the resonant frequency and the Quality factor ($Q$) of a series LCR circuit with $L = 2.0 \text{ H}$, $C = 32 \mu\text{F}$, and $R = 10 \Omega$.

Given: $L = 2.0 \text{ H}$, $C = 32 \times 10^{-6} \text{ F}$, $R = 10 \Omega$.

1. Resonant angular frequency: $\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2 \times 32 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125 \text{ rad/s}$.
2. Quality factor: $Q = \frac{\omega_r L}{R} = \frac{125 \times 2}{10} = \frac{250}{10} = 25$.

The resonant frequency is the frequency at which $X_L = X_C$, making the circuit purely resistive. The $Q$ factor is a dimensionless quantity that characterizes the circuit's bandwidth and damping.

A series LCR circuit has $R = 40 \Omega$, $X_L = 100 \Omega$, and $X_C = 70 \Omega$ connected to an AC source of $V = 250 \sin(100\pi t)$. Determine the phase angle $\phi$ between the voltage and the current, and calculate the power factor of the circuit.

1. Calculate the phase angle $\phi$ using the formula: $\tan \phi = \frac{X_L - X_C}{R}$ $\tan \phi = \frac{100 - 70}{40} = \frac{30}{40} = 0.75$ $\phi = \tan^{-1}(0.75) \approx 36.87^\circ$

2. The power factor is given by $\cos \phi$: $\text{Power Factor} = \cos \phi = \frac{R}{Z}$ First, find Impedance $Z$: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $Z = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \Omega$

3. Calculate Power Factor: $\cos \phi = \frac{40}{50} = 0.8$ Since $X_L > X_C$, the current lags the voltage.

The phase angle represents the time lag or lead between current and voltage. Since the inductive reactance is greater than the capacitive reactance, the circuit is inductive, meaning the current lags the voltage by $36.87^\circ$. The power factor of 0.8 indicates that 80% of the apparent power is being converted into real power.

A series LCR circuit is connected to an AC source of $V = 200\text{ V}$, $50\text{ Hz}$. The circuit contains a resistor $R = 50 \Omega$, an inductor $L = 0.4\text{ H}$, and a capacitor $C = 100 \mu\text{F}$. Determine the RMS current ($I_{rms}$) flowing through the circuit and the power factor.

$f = 50\text{ Hz}, \omega = 2\pi f = 100\pi \approx 314.16\text{ rad/s}$

$X_L = \omega L = 314.16 \times 0.4 = 125.66 \Omega$

$X_C = \frac{1}{\omega C} = \frac{1}{314.16 \times 100 \times 10^{-6}} \approx 31.83 \Omega$

$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + (125.66 - 31.83)^2}$

$Z = \sqrt{2500 + (93.83)^2} = \sqrt{2500 + 8804.07} = \sqrt{11304.07} \approx 106.32 \Omega$

$I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{106.32} \approx 1.88\text{ A}$

$\text{Power Factor } (\cos \phi) = \frac{R}{Z} = \frac{50}{106.32} \approx 0.47$

First, calculate the inductive and capacitive reactances using the source frequency. Then, find the total impedance ($Z$) of the series combination. The RMS current is found using Ohm's law for AC ($I = V/Z$). The power factor is the ratio of resistance to impedance.