Q4 (a)

Series R-L-C: Determine R, L and C from Resonance Conditions

Problem. A 50 Ω resistor is connected in series with a coil having resistance $R$, inductance $L$, and a capacitor $C$, supplied from a 100 V variable-frequency supply. At a frequency of 200 Hz the maximum current of 0.7 A flows through the circuit, and the voltage across the capacitor is 200 V. Determine the values of R, L and C.

Fig. 1 — Series circuit: external 50 Ω, coil (R, L), capacitor C. Maximum current at f = 200 Hz indicates resonance.

Given: $R_1 = 50\ \Omega,\;\; V = 100\ \text{V},\;\; f_0 = 200\ \text{Hz (resonant)},\;\; I_{\max} = 0.7\ \text{A},\;\; V_C = 200\ \text{V}.$

"Maximum current" at f = 200 Hz ⇒ this is the resonant frequency $f_0$. At resonance $X_L = X_C$, so the impedance is purely resistive: $Z = R_1 + R$.

Step 1 — Total resistance and coil resistance R

At resonance, current is maximum and the impedance equals total resistance:

$$I_{\max} = \frac{V}{R_1 + R} \;\Rightarrow\; R_1 + R = \frac{V}{I_{\max}} = \frac{100}{0.7} = 142.857\ \Omega$$ $$R = 142.857 - 50 = 92.857\ \Omega$$

Coil resistance

$$\boxed{\,R \;\approx\; 92.86\ \Omega\,}$$

Step 2 — Capacitance C from voltage across C

Voltage across the capacitor:

$$V_C = I_{\max}\,X_C \;\Rightarrow\; X_C = \frac{V_C}{I_{\max}} = \frac{200}{0.7} = 285.71\ \Omega$$

With $\omega_0 = 2\pi f_0 = 2\pi(200) = 1256.64\ \text{rad/s}$:

$$C = \frac{1}{\omega_0\,X_C} = \frac{1}{1256.64 \times 285.71} = \frac{1}{359\,040}$$

Capacitance

$$\boxed{\,C \;\approx\; 2.785\times 10^{-6}\ \text{F} \;=\; 2.785\ \mu\text{F}\,}$$

Step 3 — Inductance L (using $X_L = X_C$ at resonance)

$$X_L = X_C = 285.71\ \Omega$$ $$L = \frac{X_L}{\omega_0} = \frac{285.71}{1256.64}$$

Inductance

$$\boxed{\,L \;\approx\; 0.2274\ \text{H} \;=\; 227.4\ \text{mH}\,}$$

Verification

Resonant frequency from computed L and C:

$$f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.2274 \times 2.785\times10^{-6}}} = \frac{1}{2\pi\sqrt{6.333\times10^{-7}}} \approx 200\ \text{Hz}\;\;\checkmark$$

Voltage magnification factor $= V_C / V = 200/100 = 2$. Compare with the quality factor:

$$Q = \frac{X_L}{R_1 + R} = \frac{285.71}{142.857} = 2 \;\;\checkmark$$

Final Answers

Quantity	Symbol	Value
Coil resistance	R	92.86 Ω
Coil inductance	L	0.2274 H (227.4 mH)
Capacitance	C	2.785 µF
Total circuit resistance	R₁ + R	142.86 Ω
Reactances at resonance	X_L = X_C	285.71 Ω
Quality factor	Q	2

Coil resistance

R ≈ 92.86 Ω

Inductance

L ≈ 227.4 mH

Capacitance

C ≈ 2.785 µF