Question

In the given circuit the peak voltage across C,L and R are 30V, 110V and 60V respectively. The rms value of the applied voltage is

• A. 100 V
• B. 200 V
• C. 70.7 V
• D. 141 V

Answer 1

The Correct Option is C

The RMS (Root Mean Square) value of a voltage is related to the peak value by the formula: \[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} \] For a series circuit, the applied voltage \(V_{\text{applied}}\) is the vector sum of the peak voltages across each component. The total peak voltage is calculated as: \[ V_{\text{total peak}} = \sqrt{V_C^2 + V_L^2 + V_R^2} \] Given that: - \(V_C = 30 \, \text{V}\), - \(V_L = 110 \, \text{V}\), - \(V_R = 60 \, \text{V}\), Substituting the values: \[ V_{\text{total peak}} = \sqrt{(30)^2 + (110)^2 + (60)^2} \] \[ V_{\text{total peak}} = \sqrt{900 + 12100 + 3600} = \sqrt{16600} \approx 128.8 \, \text{V} \] Now, we calculate the RMS value: \[ V_{\text{rms}} = \frac{128.8}{\sqrt{2}} \approx \frac{128.8}{1.414} \approx 70.7 \, \text{V} \] Thus, the RMS value of the applied voltage is approximately 70.7 V.

The correct option is (C): 70.7 V

Answer 2

The problem provides a series LCR circuit connected to an AC voltage source. The peak voltages across the capacitor (C), inductor (L), and resistor (R) are given as:

• Peak voltage across C, \( V_{C,peak} = 30 \) V
• Peak voltage across L, \( V_{L,peak} = 110 \) V
• Peak voltage across R, \( V_{R,peak} = 60 \) V

We need to find the RMS (root mean square) value of the applied voltage \( V_{rms} \).

Phasor Analysis

In a series LCR circuit, the applied voltage \( V \) is the phasor sum of the voltages across the individual components. The voltage across the resistor \( V_R \) is in phase with the current. The voltage across the inductor \( V_L \) leads the current by 90 degrees, and the voltage across the capacitor \( V_C \) lags the current by 90 degrees.

Therefore, \( V_L \) and \( V_C \) are 180 degrees out of phase. The peak value of the applied voltage \( V_{peak} \) can be calculated using the Pythagorean theorem on the phasor diagram: \[ V_{peak} = \sqrt{V_{R,peak}^2 + (V_{L,peak} - V_{C,peak})^2} \]

Calculation of Peak Applied Voltage

Substitute the given peak voltage values into the formula: \[ V_{peak} = \sqrt{(60)^2 + (110 - 30)^2} \] \[ V_{peak} = \sqrt{60^2 + (80)^2} \] \[ V_{peak} = \sqrt{3600 + 6400} \] \[ V_{peak} = \sqrt{10000} \] \[ V_{peak} = 100 \text{ V} \]

Calculation of RMS Applied Voltage

The RMS value of a sinusoidal voltage is related to its peak value by the formula: \[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \] Substitute the calculated peak voltage: \[ V_{rms} = \frac{100}{\sqrt{2}} \] \[ V_{rms} = \frac{100 \times \sqrt{2}}{2} \] \[ V_{rms} = 50\sqrt{2} \] Using \( \sqrt{2} \approx 1.414 \): \[ V_{rms} \approx 50 \times 1.414 \] \[ V_{rms} \approx 70.7 \text{ V} \]

The final answer is ${\text{70.7 V}}$ (Option C)