Q3 (a) — [8 marks]

Parallel R-L-C Resonance — Ideal & Practical (Theory) + Series Resonance (Numerical)

Problem. Define resonance in parallel R-L-C circuit with the help of a phasor diagram. A 220 V, 100 Hz AC source supplies a series circuit with a capacitor and a coil. If the coil has 50 mΩ resistance and 5 mH inductance, find the value of capacitor to create resonance. Also calculate:

Voltage across R, L and C
Quality factor

Part A · Resonance in a Parallel R-L-C Circuit

Fig. 1 — Ideal parallel R-L-C circuit. The same voltage V appears across each branch.

Each branch sees the same supply voltage $V$. The branch currents are:

$$I_R = \frac{V}{R}\;(\text{in phase with }V),\quad I_L = \frac{V}{\omega L}\;(\text{lags }V \text{ by } 90^\circ),\quad I_C = V\omega C\;(\text{leads }V \text{ by } 90^\circ)$$

The total line current is

$$\bar{I} = \bar{I}_R + \bar{I}_L + \bar{I}_C = \frac{V}{R} + j\!\left(V\omega C - \frac{V}{\omega L}\right)$$

Resonance (parallel R-L-C): The condition at which the inductive susceptance equals the capacitive susceptance, so the reactive components of $I_L$ and $I_C$ cancel. The line current $I$ is then in phase with $V$ (unity p.f.) and is minimum.

$$I_C = I_L \;\Rightarrow\; V\omega_0 C = \frac{V}{\omega_0 L} \;\Rightarrow\; \boxed{\;\omega_0 = \dfrac{1}{\sqrt{LC}}\;}$$

Fig. 2 — At resonance $|I_L|=|I_C|$ but oppositely directed, so they cancel. Net current $I = I_R$, minimum and in phase with V.

Properties at parallel resonance

Net susceptance is zero ⇒ admittance is minimum, equal to $1/R$.
Impedance is maximum (≈ R for ideal case): often called a rejector circuit.
Line current $I = V/R$ is minimum.
Power factor is unity, current in phase with voltage.
Branch currents $I_L$ and $I_C$ can be much larger than the line current $I$ (current magnification by $Q$).

Part A · 2 — Practical Parallel Resonant Circuit (Tank Circuit)

In a real-world parallel resonant circuit, the inductor is not pure — its winding has a small resistance $R$ in series with $L$. The capacitor is taken as essentially loss-free. This R–L branch is placed in parallel with C and is the practical parallel (or "tank") circuit.

Fig. 3 — Practical parallel resonant ("tank") circuit: the inductor's winding resistance R is in series with L; this branch is in parallel with the capacitor C.

Derivation of the resonant frequency $f_r$

Admittance of the coil branch:

$$Y_L = \frac{1}{R + j\omega L} = \frac{R - j\omega L}{R^2 + \omega^2 L^2}$$

Admittance of the capacitor branch:

$$Y_C = j\omega C$$

Total admittance of the parallel combination:

$$Y = Y_L + Y_C = \underbrace{\frac{R}{R^2 + \omega^2 L^2}}_{\text{conductance } G} + j\!\left[\omega C - \frac{\omega L}{R^2 + \omega^2 L^2}\right]$$

Resonance condition: The circuit is at resonance when the supply current is in phase with the supply voltage, i.e. when the imaginary part of $Y$ (the net susceptance) is zero.

Setting the imaginary part to zero:

$$\omega_r C = \frac{\omega_r L}{R^2 + \omega_r^2 L^2}$$ $$R^2 + \omega_r^2 L^2 = \frac{L}{C}$$ $$\omega_r^2 L^2 = \frac{L}{C} - R^2$$ $$\omega_r^2 = \frac{1}{LC} - \frac{R^2}{L^2}$$

Resonant frequency — practical parallel circuit

$$\boxed{\;\omega_r = \sqrt{\dfrac{1}{LC} - \dfrac{R^2}{L^2}}\;}$$ $$\boxed{\;f_r = \dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{L^2}}\;}$$

This can also be written as $\;f_r = \dfrac{1}{2\pi\sqrt{LC}}\sqrt{1 - \dfrac{R^2 C}{L}}\;$ — the ideal resonant frequency multiplied by the "practical" correction factor.

Dynamic resistance (impedance at resonance)

At $\omega = \omega_r$ the susceptance vanishes, so the admittance is purely real:

$$Y(\omega_r) = \frac{R}{R^2 + \omega_r^2 L^2} = \frac{R}{L/C} = \frac{RC}{L}$$ $$\boxed{\;Z_r \;=\; \dfrac{1}{Y(\omega_r)} \;=\; \dfrac{L}{RC}\;}$$

$Z_r$ is called the dynamic resistance. It is large but finite, unlike the ideal case (∞). The line current at resonance is the minimum:

$$I_{\min} = \frac{V}{Z_r} = \frac{V R C}{L}$$

How does the practical circuit differ from the ideal?

Property	Ideal (R = 0)	Practical (R ≠ 0)
Resonant frequency	$\omega_0 = \dfrac{1}{\sqrt{LC}}$	$\omega_r = \sqrt{\dfrac{1}{LC} - \dfrac{R^2}{L^2}} < \omega_0$
Impedance at resonance	$Z \to \infty$	$Z_r = \dfrac{L}{RC}$ (finite — "dynamic resistance")
Line current at resonance	$I = 0$	$I_{\min} = \dfrac{V R C}{L}$ (small but non-zero)
Power factor at resonance	Unity (cos φ = 1)	Unity (cos φ = 1) — same condition, by definition
Quality factor	Q → ∞ (no losses)	$Q = \dfrac{\omega_r L}{R} = \dfrac{1}{R}\sqrt{\dfrac{L}{C} - R^2}$ (finite)
Bandwidth	Zero (infinitely sharp peak)	BW $= \dfrac{R}{L}$ (finite, non-zero)
Selectivity	Perfect — passes only $f_0$	Limited by R; broader peak as R increases
Energy loss	None — sustained oscillation	Resistor dissipates $I^2 R$ each cycle

Key insight: The presence of coil resistance does two important things:

Shifts the resonant frequency down (by the $R^2/L^2$ term), and
Keeps the line current finite, giving the circuit a measurable bandwidth and a finite Q-factor.

If $\dfrac{R^2 C}{L} \geq 1$, the inner expression becomes zero or negative — no resonance is possible. Practical circuits are designed so that $R \ll \sqrt{L/C}$, keeping $f_r$ very close to the ideal $f_0$.

Part B · Numerical — Series R-L-C Resonance

Given: $V = 220\ \text{V},\ f = 100\ \text{Hz},\ R = 50\ \text{m}\Omega = 0.05\ \Omega,\ L = 5\ \text{mH} = 5\times10^{-3}\ \text{H}.$

The numerical describes a series connection of coil and capacitor (a series R-L-C). The resonance condition for a series circuit is also $\omega_0 = 1/\sqrt{LC}$.

Step 1 — Capacitance for resonance

At resonance $X_L = X_C$:

$$\omega_0 L = \frac{1}{\omega_0 C} \;\Rightarrow\; C = \frac{1}{\omega_0^{2}\,L}$$

With $\omega_0 = 2\pi f = 2\pi(100) = 628.32\ \text{rad/s}$:

$$C = \frac{1}{(628.32)^2 \times 5\times10^{-3}} = \frac{1}{394\,784 \times 5\times10^{-3}} = \frac{1}{1973.92}$$

Resonating capacitance

$$\boxed{\,C \;\approx\; 5.066\times 10^{-4}\ \text{F} \;=\; 506.6\ \mu\text{F}\,}$$

Step 2 — Reactances and resonant current

$$X_L = \omega_0 L = 628.32 \times 0.005 = 3.1416\ \Omega$$ $$X_C = \frac{1}{\omega_0 C} = X_L = 3.1416\ \Omega \;(\text{by resonance})$$

Net impedance at resonance is purely resistive: $Z = R = 0.05\ \Omega$.

$$I_0 = \frac{V}{R} = \frac{220}{0.05} = 4400\ \text{A}$$

Note (sanity check): The 4.4 kA current is enormous because $R$ is only 50 mΩ — a textbook idealization to show the dramatic voltage magnification at resonance, not a practical circuit.

Step 3 (i) — Voltages across R, L and C

$$V_R = I_0\,R = 4400 \times 0.05 = 220\ \text{V}$$ $$V_L = I_0\,X_L = 4400 \times 3.1416 = 13\,823\ \text{V} \;\approx\; 13.82\ \text{kV}$$ $$V_C = I_0\,X_C = 4400 \times 3.1416 = 13\,823\ \text{V} \;\approx\; 13.82\ \text{kV}$$

$V_L$ and $V_C$ are equal in magnitude but $180°$ out of phase, so they cancel — only $V_R = V$ appears across the series combination, as expected at resonance.

Fig. 3 — Series resonance: $V_L$ and $V_C$ are equal & opposite, leaving only $V_R = V$. Note $V_L$ and $V_C$ here are ~63× the supply voltage — voltage magnification.

Step 3 (ii) — Quality factor

$$Q = \frac{\omega_0 L}{R} = \frac{X_L}{R} = \frac{3.1416}{0.05}$$

Quality factor

$$\boxed{\,Q \;\approx\; 62.83\,}$$

Cross-check using $Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$:

$$Q = \frac{1}{0.05}\sqrt{\frac{5\times 10^{-3}}{506.6\times 10^{-6}}} = 20 \times \sqrt{9.869} = 20 \times 3.1416 = 62.83\ \checkmark$$

Consistency with voltage magnification: $V_L / V = 13823 / 220 \approx 62.83 = Q$ ✓

Final Answers

Quantity	Value
Capacitance for resonance, $C$	506.6 µF
Resonant (line) current, $I_0$	4400 A
Voltage across $R$	220 V
Voltage across $L$	≈ 13.82 kV
Voltage across $C$	≈ 13.82 kV
Quality factor, $Q$	62.83

Capacitor

C ≈ 506.6 µF

V across R

220 V

V across L = V across C

≈ 13.82 kV

Quality factor

Q ≈ 62.83