Question

Using a phasor diagram, derive an expression for the impedance of a series LCR circuit connected to an ac source. What is meant by resonance in this circuit? Obtain expression for the resonant frequency of this circuit.

Hint:

Draw V_R along the current, V_L up and V_C down. Add by Pythagoras for Z. Resonance is when X_L = X_C.

Answer 1

Step 1: Draw the phasor diagram on axes:
Take the current phasor along the $+x$ axis. All three elements carry this same current, so it is the natural reference.
The voltage across $R$ lies along $+x$, with length $IR$.
The voltage across $L$ points along $+y$, with length $I\omega L$. Voltage across an inductor leads the current by a quarter cycle.
The voltage across $C$ points along $-y$, with length $I/\omega C$. Voltage across a capacitor lags the current by a quarter cycle.

Step 2: Add the phasors by components:
Horizontal part of the total voltage: $V_x = IR$.
Vertical part: $V_y = I\omega L - \dfrac{I}{\omega C} = I (X_L - X_C)$.
The source voltage is the arrow from the origin to the point $(V_x, V_y)$. Its length is:
\[ V = \sqrt{V_x^2 + V_y^2} = I \sqrt{R^2 + (X_L - X_C)^2} \]

Step 3: Read off the impedance:
Impedance is the ratio of voltage to current in the same units (peak or rms).
\[ Z = \frac{V}{I} = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]
The angle of the resultant above the current axis satisfies $\tan\phi = V_y / V_x = (X_L - X_C)/R$.

Step 4: What resonance means:
Look at $Z$ as the frequency changes. The reactance term $(X_L - X_C)^2$ is never negative, so $Z$ is smallest when this term is zero.
That happens when $\omega L = \dfrac{1}{\omega C}$. The two reactive voltages are equal and opposite, the resultant points along the current, and $\phi = 0$.
At that frequency the circuit behaves as a plain resistor, $Z = R$, and the current peaks.

Step 5: Resonant frequency and a quick check:
From $\omega_0^2 LC = 1$, we get $\omega_0 = \dfrac{1}{\sqrt{LC}}$ and $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$.
Check with numbers: $L = 1\ \text{mH}$ and $C = 10\ \mu\text{F}$ give $\omega_0 = 1 / \sqrt{10^{-3} \times 10^{-5}} = 10^{4}\ \text{rad/s}$. Then $f_0 = 10^4 / 2\pi \approx 1.6\ \text{kHz}$.
At this $\omega_0$: $X_L = 10\ \Omega$ and $X_C = 1/(10^4 \times 10^{-5}) = 10\ \Omega$. They are equal, as required.

Final Answer:
The impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$. Resonance occurs when $X_L = X_C$, and $f_0 = \dfrac{1}{2\pi\sqrt{LC}}$. \[ \boxed{f_0 = \frac{1}{2\pi\sqrt{LC}}} \]