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In a series LCR circuit the rms voltage ...

In a series LCR circuit the rms voltage across the inductance , capacitance and resistance are respectively 4 V, 8 V and 5 V . The RMS voltage of the AC source in the circuit is

A

17 V

B

13 V

C

5 V

D

6.4 V

Text Solution

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The correct Answer is:
To find the RMS voltage of the AC source in a series LCR circuit where the RMS voltages across the inductance (VL), capacitance (VC), and resistance (VR) are given as 4 V, 8 V, and 5 V respectively, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Given Values:** - Voltage across the resistance (VR) = 5 V - Voltage across the capacitance (VC) = 8 V - Voltage across the inductance (VL) = 4 V 2. **Understand the Phasor Relationships:** - In a series LCR circuit, the current (I) is the same through all components. - The voltage across the resistor (VR) is in phase with the current. - The voltage across the inductor (VL) leads the current by 90 degrees. - The voltage across the capacitor (VC) lags behind the current by 90 degrees. 3. **Set Up the Equation for the RMS Voltage of the AC Source (V):** - The total voltage (V) can be calculated using the formula: \[ V = \sqrt{VR^2 + (VC - VL)^2} \] - This formula accounts for the phase differences between the voltages. 4. **Substitute the Given Values:** - Substitute VR = 5 V, VC = 8 V, and VL = 4 V into the equation: \[ V = \sqrt{5^2 + (8 - 4)^2} \] 5. **Calculate the Values:** - Calculate \( VR^2 \): \[ VR^2 = 5^2 = 25 \] - Calculate \( (VC - VL)^2 \): \[ (VC - VL) = (8 - 4) = 4 \quad \Rightarrow \quad (VC - VL)^2 = 4^2 = 16 \] 6. **Combine the Results:** - Now substitute back into the equation: \[ V = \sqrt{25 + 16} = \sqrt{41} \] 7. **Calculate the Final Result:** - Find the square root: \[ V \approx 6.4 \, \text{V} \] ### Conclusion: The RMS voltage of the AC source in the circuit is approximately **6.4 V**.
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