Question

A resistor of 400 \( \Omega \), an inductor of \( \frac{5}{\pi} \) H, and a capacitor of \( \frac{50}{\pi} \) µF are joined in series across an AC source \( v = 140 \sin (100 \pi t) \) V. Find the rms voltages across these three circuit elements. The algebraic sum of these voltages is more than the rms voltage of source. Explain.

Hint:

In a series RLC circuit, the total voltage is not simply the sum of the voltages across each element because the voltages are out of phase with each other. The voltages across the inductor and capacitor tend to cancel each other out, while the voltage across the resistor is in phase with the current.

Answer 1

Given: - Resistor: \( R = 400 \, \Omega \) - Inductor: \( L = \frac{5}{\pi} \, \text{H} \) - Capacitor: \( C = \frac{50}{\pi} \, \mu\text{F} \) - Source Voltage: \( v = 140 \sin (100 \pi t) \, \text{V} \) The angular frequency is: \[ \omega = 100 \pi \, \text{rad/s} \] 1. RMS Source Voltage: The peak voltage is \( V_0 = 140 \, \text{V} \). The RMS voltage is: \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{140}{\sqrt{2}} \approx 98.99 \, \text{V} \] 2. Circuit Impedance: The impedance \( Z \) is calculated using \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Inductive Reactance: \[ X_L = \omega L = 100 \pi \times \frac{5}{\pi} = 500 \, \Omega \] Capacitive Reactance: \[ X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times \frac{50}{\pi} \times 10^{-6}} \approx 636.6 \, \Omega \] Total Impedance: \[ Z = \sqrt{400^2 + (500 - 636.6)^2} = \sqrt{400^2 + (-136.6)^2} \approx 406.6 \, \Omega \] 3. Circuit Current: The RMS current is: \[ I = \frac{V_{\text{rms}}}{Z} = \frac{98.99}{406.6} \approx 0.243 \, \text{A} \] 4. RMS Voltages Across Components: - Resistor: \( V_R = I R = 0.243 \times 400 \approx 97.2 \, \text{V} \) - Inductor: \( V_L = I X_L = 0.243 \times 500 \approx 121.5 \, \text{V} \) - Capacitor: \( V_C = I X_C = 0.243 \times 636.6 \approx 154.2 \, \text{V} \) 5. Explanation of Voltage Sum: The algebraic sum of voltages across the resistor, inductor, and capacitor exceeds the source's RMS voltage because these voltages are out of phase. In AC circuits with inductors and capacitors, the voltages across them are 180° out of phase, resulting in a phasor sum. This vector addition means the total voltage across components can be greater than the source voltage.