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 data: - 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 of the AC source is: \[ \omega = 100 \pi \, \text{rad/s} \] Step 1: Calculate the rms voltage of the source The peak voltage is \( V_0 = 140 \, \text{V} \), so the rms voltage \( V_{\text{rms}} \) is given by: \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{140}{\sqrt{2}} = 98.99 \, \text{V} \] Step 2: Find the impedance of the circuit The impedance of the series circuit is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance. The inductive reactance \( X_L \) is: \[ X_L = \omega L = 100 \pi \times \frac{5}{\pi} = 500 \, \Omega \] The capacitive reactance \( X_C \) is: \[ X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \times \frac{50}{\pi} \times 10^{-6}} = 636.6 \, \Omega \] Thus, the total impedance is: \[ Z = \sqrt{400^2 + (500 - 636.6)^2} = \sqrt{400^2 + (-136.6)^2} \approx 406.6 \, \Omega \] Step 3: Calculate the current in the circuit The current in the circuit is: \[ I = \frac{V_{\text{rms}}}{Z} = \frac{98.99}{406.6} \approx 0.243 \, \text{A} \] Step 4: Calculate the rms voltages across each element - Rms voltage across the resistor: \[ V_R = I R = 0.243 \times 400 \approx 97.2 \, \text{V} \] - Rms voltage across the inductor: \[ V_L = I X_L = 0.243 \times 500 \approx 121.5 \, \text{V} \] - Rms voltage across the capacitor: \[ V_C = I X_C = 0.243 \times 636.6 \approx 154.2 \, \text{V} \] Step 5: Explanation of the algebraic sum The algebraic sum of the individual voltages across the resistor, inductor, and capacitor is greater than the rms voltage of the source because the voltages are not in phase with each other. In an AC circuit with reactive elements (inductor and capacitor), the voltages across the inductor and capacitor are 180° out of phase with each other, leading to their individual voltages adding up vectorially. The total voltage across the components is the phasor sum, not the simple arithmetic sum, which results in a larger total voltage than the rms voltage of the source.