Q.Figure 7.17 shows a series LCR circuit connected to a variable frequency $230\ \text{V}$ source. $L = 5.0\ \text{H}$, $C = 80\ \mu\text{F}$, $R = 40\ \Omega$.

Concept: Resonance in AC Circuits — at resonance, the inductive reactance equals the capacitive reactance ($X_L = X_C$), making the impedance purely resistive and minimum.

Reasoning:

1. Resonant frequency:

   $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{0.02} = 50\ \text{rad/s}$

   Frequency $f_0 = \frac{\omega_0}{2\pi} = \frac{50}{2\pi} \approx 7.96\ \text{Hz}$.

2. Impedance and current at resonance:

   $Z = R = 40\ \Omega$ (since $X_L = X_C$).

   Peak voltage $V_0 = 230\sqrt{2}\ \text{V}$, so $I_0 = \frac{V_0}{R} = \frac{230\sqrt{2}}{40} \approx 8.13\ \text{A}$.

   RMS current $I_{\text{rms}} = \frac{230}{40} = 5.75\ \text{A}$.

3. RMS potential drops:

   $V_R = I_{\text{rms}} R = 5.75 \times 40 = 230\ \text{V}$.

   $X_L = \omega_0 L = 50 \times 5.0 = 250\ \Omega$, so $V_L = I_{\text{rms}} X_L = 5.75 \times 250 = 1437.5\ \text{V}$. …

(a) Resonant frequency

Resonance occurs when $X_L = X_C$, i.e. $\omega_0 = \dfrac{1}{\sqrt{LC}}$:

(b) Impedance and current amplitude

At resonance $X_L = X_C$, so

The current amplitude uses the peak source voltage $V_0 = \sqrt{2}\times230 = 325.3\ \text{V}$:

(c) RMS voltage drops

The rms current is $I_{\mathrm{rms}} = \dfrac{V_{\mathrm{rms}}}{Z} = \dfrac{230}{40} = 5.75\ \text{A}$, and at resonance $X_L = X_C = \omega_0 L = 50\times5.0 = 250\ \Omega$.

$V_R = I_{\mathrm{rms}}R = 5.75\times40 = 230\ \text{V},$ …

Method: Solving a Series LCR Resonance Problem End-to-End

This is the standard three-stage method for any series LCR circuit problem that asks for the resonant frequency, the current/impedance at resonance, and the individual element voltage drops.

Steps

Step 1: Find the resonant frequency from $L$ and $C$ alone.

Resonance occurs when the two reactances cancel, $X_L = X_C$, which gives

Notice this depends only on $L$ and $C$ — never on $R$ or on the source voltage.

Step 2: Use the fact that impedance is purely resistive at resonance.

Because $X_L = X_C$ at $\omega_0$, they cancel in $Z = \sqrt{R^2 + (X_L - X_C)^2}$, leaving $Z = R$. This lets you get the peak or rms current directly from Ohm's law for AC:

Always convert between peak and rms with a factor of $\sqrt{2}$ ($V_0 = \sqrt{2}\,V_{\text{rms}}$), and be careful which one the question is actually asking for.

Step 3: Find each element's own reactance at resonance, then its voltage drop. …