In a series LCR circuit, rms voltage across inductor and capacitor are found ot be 8 V and 5 V respectively. If applied voltage is 5 volts, then what is rms voltage across resistance in volts ?

Explanation

To solve the problem step by step, we will use the information provided about the voltages across the inductor (VL), capacitor (VC), and the applied voltage (V).

### Step-by-Step Solution:

1. **Identify Given Values:**
- RMS voltage across the inductor, \( V_L = 8 \, \text{V} \)
- RMS voltage across the capacitor, \( V_C = 5 \, \text{V} \)
- Applied voltage, \( V = 5 \, \text{V} \)

2. **Understand the Phase Relationships:**
- In a series LCR circuit, the voltage across the inductor leads the current by 90 degrees, while the voltage across the capacitor lags the current by 90 degrees. Therefore, the voltages across the inductor and capacitor can be treated as vectors in a phasor diagram.

3. **Calculate the Net Voltage Across the Inductor and Capacitor:**
- The effective voltage across the inductor and capacitor can be calculated as:
\[ V_{LC} = V_L - V_C \]
- Substituting the given values:
\[ V_{LC} = 8 \, \text{V} - 5 \, \text{V} = 3 \, \text{V} \]

4. **Apply the Pythagorean Theorem:**
- The relationship between the applied voltage, the voltage across the resistance (VR), and the net voltage across the inductor and capacitor can be expressed using the Pythagorean theorem:
\[ V^2 = V_R^2 + V_{LC}^2 \]
- Substituting the known values:
\[ 5^2 = V_R^2 + 3^2 \]
- This simplifies to:
\[ 25 = V_R^2 + 9 \]

5. **Solve for the Voltage Across the Resistance:**
- Rearranging the equation gives:
\[ V_R^2 = 25 - 9 = 16 \]
- Taking the square root:
\[ V_R = \sqrt{16} = 4 \, \text{V} \]

### Final Answer:
The RMS voltage across the resistance is \( V_R = 4 \, \text{V} \).

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