Electromagnetic Induction and Alternating Currents - Alternating Current (LCR Circuits, Resonance)

A series $LCR$ circuit contains a resistor of $R = 10 \, \Omega$, an inductor of $L = 2 \, H$, and a capacitor of $C = 18 \, \mu F$. Calculate the resonant frequency $f_r$ of the circuit.

Given: $L = 2 \, H$, $C = 18 \times 10^{-6} \, F$. Using the formula $f_r = \frac{1}{2\pi \sqrt{LC}}$: $f_r = \frac{1}{2 \times 3.14 \times \sqrt{2 \times 18 \times 10^{-6}}}$ $f_r = \frac{1}{6.28 \times \sqrt{36 \times 10^{-6}}}$ $f_r = \frac{1}{6.28 \times 6 \times 10^{-3}}$ $f_r = \frac{1000}{37.68} \approx 26.54 \, Hz$.

Resonant frequency is the frequency at which the inductive reactance equals capacitive reactance, causing the circuit to behave as purely resistive.

In an $LCR$ circuit, $R = 30 \, \Omega$, $X_L = 80 \, \Omega$, and $X_C = 40 \, \Omega$. Find the impedance $Z$ and the power factor.

$Z = \sqrt{R^2 + (X_L - X_C)^2}$ $Z = \sqrt{30^2 + (80 - 40)^2} = \sqrt{30^2 + 40^2}$ $Z = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega$ Power factor $\cos \phi = \frac{R}{Z} = \frac{30}{50} = 0.6$.

The impedance is the total resistance of the $LCR$ circuit. The power factor is the cosine of the phase angle, indicating how much of the supplied power is actually dissipated as heat.

An alternating voltage source $V = 200\sqrt{2} \sin(100t)$ is connected in series with a resistor $R = 40 \, \Omega$, an inductor $L = 0.5 \, H$, and a capacitor $C = 100 \, \mu F$. Calculate the peak current $I_0$ flowing through the circuit and the phase difference $\phi$ between the voltage and the current.

1. Identify given values: $V_0 = 200\sqrt{2} \, V$, $\omega = 100 \, rad/s$, $R = 40 \, \Omega$, $L = 0.5 \, H$, $C = 100 \times 10^{-6} \, F$.
2. Calculate Inductive Reactance: $X_L = \omega L = 100 \times 0.5 = 50 \, \Omega$
3. Calculate Capacitive Reactance: $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = \frac{1}{10^{-2}} = 100 \, \Omega$
4. Calculate Impedance ($Z$): $Z = \sqrt{R^2 + (X_L - X_C)^2}$ $Z = \sqrt{40^2 + (50 - 100)^2} = \sqrt{1600 + 2500} = \sqrt{4100} \approx 64.03 \, \Omega$
5. Calculate Peak Current ($I_0$): $I_0 = \frac{V_0}{Z} = \frac{200\sqrt{2}}{64.03} \approx \frac{282.84}{64.03} \approx 4.42 \, A$
6. Calculate Phase Difference ($\phi$): $\tan \phi = \frac{X_L - X_C}{R} = \frac{50 - 100}{40} = -\frac{50}{40} = -1.25$ $\phi = \tan^{-1}(-1.25) \approx -51.34^\circ$ (The negative sign indicates the current leads the voltage since $X_C > X_L$.)

First, we determine the individual reactances of the inductor and capacitor using the angular frequency provided in the voltage equation. Then, we find the total impedance by combining resistance and the net reactance. Finally, we use Ohm's law for AC to find peak current and use the tangent ratio for phase angle.

A series $LCR$ circuit is in resonance. The inductance $L = 10 \, mH$, the capacitance $C = 1 \, \mu F$, and the resistance $R = 2 \, \Omega$. Find the Quality Factor ($Q$-factor) of the circuit and calculate the bandwidth $\Delta \omega$.

1. Identify given values: $L = 10 \times 10^{-3} \, H$, $C = 1 \times 10^{-6} \, F$, $R = 2 \, \Omega$.
2. Calculate resonant angular frequency ($\omega_r$): $\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10 \times 10^{-3} \times 1 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-8}}} = 10^4 \, rad/s$
3. Calculate Q-factor: $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ $Q = \frac{1}{2}\sqrt{\frac{10^{-2}}{10^{-6}}} = \frac{1}{2}\sqrt{10^4} = \frac{100}{2} = 50$
4. Calculate Bandwidth ($\Delta \omega$): $Q = \frac{\omega_r}{\Delta \omega} \implies \Delta \omega = \frac{\omega_r}{Q}$ $\Delta \omega = \frac{10^4}{50} = 200 \, rad/s$

At resonance, the $Q$-factor represents the sharpness of the resonance peak. It is calculated using the circuit parameters $L, C,$ and $R$. The bandwidth is the range of frequencies over which the power is at least half of the maximum power, and it is inversely proportional to the $Q$-factor.