A resistor an inductor and a capacitor are connected in series to an ac source An ac voltmeter measures the votage across them as `800 V, 30 V` and `90V` respectively The rms value of the supply voltage is .

Options

A. `100V`
B. `100sqrt2V`
C. `200V`
D. `200sqrt2V`

Explanation

To find the RMS value of the supply voltage in a series circuit with a resistor, inductor, and capacitor, we can use the following steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Voltage across the resistor, \( V_R = 80 \, \text{V} \)
- Voltage across the inductor, \( V_L = 30 \, \text{V} \)
- Voltage across the capacitor, \( V_C = 90 \, \text{V} \)

2. **Calculate the Voltage Difference Between Inductor and Capacitor**:
- Since the inductor and capacitor are out of phase, we need to find the net reactive voltage:
\[ V_{L - C} = V_L - V_C = 30 \, \text{V} - 90 \, \text{V} = -60 \, \text{V} \]
(This indicates that the capacitor's voltage is greater than the inductor's voltage.)

3. **Use the Formula for Total RMS Voltage**:
- The total RMS voltage across the series circuit can be calculated using the formula:
\[ V_{RMS} = \sqrt{V_R^2 + (V_L - V_C)^2} \]

4. **Substitute the Values**:
- Substitute the known values into the formula:
\[ V_{RMS} = \sqrt{(80 \, \text{V})^2 + (-60 \, \text{V})^2} \]

5. **Calculate the Squares**:
- Calculate \( V_R^2 \) and \( (V_L - V_C)^2 \):
\[ V_R^2 = 80^2 = 6400 \, \text{V}^2 \]
\[ (V_L - V_C)^2 = (-60)^2 = 3600 \, \text{V}^2 \]

6. **Add the Squares**:
- Now add the two results:
\[ V_{RMS}^2 = 6400 + 3600 = 10000 \, \text{V}^2 \]

7. **Take the Square Root**:
- Finally, take the square root to find \( V_{RMS} \):
\[ V_{RMS} = \sqrt{10000} = 100 \, \text{V} \]

### Final Answer:
The RMS value of the supply voltage is \( 100 \, \text{V} \).

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