Question

In a series LCR circuit, the rms voltage across the resistor and the capacitor are \(30 \, \text{V}\) and \(90 \, \text{V}\) respectively. If the applied voltage is \(50\sqrt{2}\sin\omega t\), then the peak voltage across the inductor is:

• A. 70 V
• B. 50 V
• C. \(70\sqrt{2} \, \text{V}\)
• D. \(50\sqrt{2} \, \text{V}\)

Hint:

In a series LCR circuit, use the Pythagorean theorem to calculate the total or peak voltage when given the rms voltages across the components.

Answer 1

The Correct Option is D

Step 1: The applied voltage across the series LCR circuit is given as:

\[ V_{\text{applied}} = 50 \sqrt{2} \sin \omega t \]

This is the peak voltage, so the RMS voltage is:

\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{50 \sqrt{2}}{\sqrt{2}} = 50 \, \text{V}. \]

Step 2: In a series LCR circuit, the total applied voltage is the vector sum of the voltages across the resistor (\( V_R \)), the capacitor (\( V_C \)), and the inductor (\( V_L \)). These voltages are related by:

\[ V_{\text{applied}}^2 = V_R^2 + V_C^2 + V_L^2. \]

We are given that the RMS voltage across the resistor is \( 30 \, \text{V} \), and the RMS voltage across the capacitor is \( 90 \, \text{V} \). The RMS voltage across the inductor \( V_L \) can be found using the formula:

\[ (V_{\text{applied}})^2 = (V_R)^2 + (V_C)^2 + (V_L)^2. \]

Substituting the known values:

\[ (50)^2 = (30)^2 + (90)^2 + V_L^2, \]

\[ 2500 = 900 + 8100 + V_L^2, \]

\[ V_L^2 = 2500 - 900 - 8100 = 50^2. \]

Thus, the RMS voltage across the inductor is \( 50 \, \text{V} \), and the peak voltage is:

\[ V_{\text{peak}, L} = 50 \sqrt{2} \, \text{V}. \]