1. **Problem Statement:** A coil with resistance $R=10\ \Omega$ and inductance $L=0.1\ \text{H}$ is connected in parallel with a capacitor of capacitance $C=50\ \mu\text{F}=50\times10^{-6}\ \text{F}$ across a variable frequency source. We need to find: - The frequency $f$ at which the resultant current is in phase with the supply voltage. - The Q-factor at this frequency. - For supply voltage $V=100$, calculate: (a) Current in each branch, (b) Resultant current, (c) Power factor of the whole circuit. 2. **Key Formulas and Concepts:** - Inductive reactance: $X_L = 2\pi f L$ - Capacitive reactance: $X_C = \frac{1}{2\pi f C}$ - At resonance (current in phase with voltage), the inductive and capacitive reactances are equal: $$X_L = X_C$$ - Q-factor for coil: $$Q = \frac{X_L}{R}$$ - Current in resistor-inductor branch: $$I_L = \frac{V}{\sqrt{R^2 + X_L^2}}$$ - Current in capacitor branch: $$I_C = \frac{V}{X_C}$$ - Resultant current in parallel branches: $$I = \sqrt{I_L^2 + I_C^2}$$ - Power factor: $$\text{pf} = \frac{P}{VI} = \frac{I_R}{I}$$ where $I_R = \frac{V}{R}$ is the current through the resistor (in phase with voltage). 3. **Calculate Resonant Frequency $f$:** At resonance, $$X_L = X_C \Rightarrow 2\pi f L = \frac{1}{2\pi f C}$$ Multiply both sides by $2\pi f$: $$ (2\pi f)^2 L = \frac{1}{C} $$ Solve for $f$: $$ f = \frac{1}{2\pi \sqrt{LC}} $$ Substitute values: $$ f = \frac{1}{2\pi \sqrt{0.1 \times 50 \times 10^{-6}}} = \frac{1}{2\pi \sqrt{5 \times 10^{-6}}} $$ Calculate inside the root: $$ \sqrt{5 \times 10^{-6}} = \sqrt{5} \times 10^{-3} \approx 2.236 \times 10^{-3} $$ So, $$ f = \frac{1}{2\pi \times 2.236 \times 10^{-3}} = \frac{1}{0.01404} \approx 71.2\ \text{Hz} $$ 4. **Calculate Q-factor:** At resonance, $$ X_L = 2\pi f L = 2\pi \times 71.2 \times 0.1 = 44.7\ \Omega $$ Then, $$ Q = \frac{X_L}{R} = \frac{44.7}{10} = 4.47 $$ 5. **Calculate Currents at Resonance:** - Current in coil branch (resistor + inductor): $$ I_L = \frac{V}{\sqrt{R^2 + X_L^2}} = \frac{100}{\sqrt{10^2 + 44.7^2}} = \frac{100}{\sqrt{100 + 1998}} = \frac{100}{\sqrt{2098}} $$ $$ \sqrt{2098} \approx 45.8 $$ So, $$ I_L = \frac{100}{45.8} \approx 2.18\ \text{A} $$ - Current in capacitor branch: $$ I_C = \frac{V}{X_C} = \frac{100}{44.7} = 2.24\ \text{A} $$ 6. **Resultant Current:** Since the coil branch current lags voltage and capacitor branch current leads voltage by equal amounts at resonance, the reactive components cancel, and the resultant current is the sum of the resistor current and capacitor current in phase. The resistor current (in phase) is: $$ I_R = \frac{V}{R} = \frac{100}{10} = 10\ \text{A} $$ The inductive and capacitive reactive currents cancel, so the resultant current magnitude is: $$ I = I_R = 10\ \text{A} $$ 7. **Power Factor:** Power factor is the ratio of real power current to total current: $$ \text{pf} = \frac{I_R}{I} = \frac{10}{10} = 1 $$ **Final answers:** - Resonant frequency: $f \approx 71.2\ \text{Hz}$ - Q-factor: $4.47$ - (a) Currents: $I_L \approx 2.18\ \text{A}$, $I_C \approx 2.24\ \text{A}$ - (b) Resultant current: $10\ \text{A}$ - (c) Power factor: $1$