An `AC` voltage is applied acrss a series combination of `L` and `R`. If the voltage drop across the resistor and inductor are `20 V` and `15 V` respectiely, then applied peak voltage is

Options

A. `25V`
B. `35V`
C. `25sqrt2V`
D. `5sqrt7V`

Explanation

To find the applied peak voltage across a series combination of an inductor (L) and a resistor (R) when given the voltage drops across each component, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Voltage drop across the resistor, \( V_R = 20 \, V \)
- Voltage drop across the inductor, \( V_L = 15 \, V \)

2. **Calculate the total voltage (RMS voltage) using the Pythagorean theorem:**
- In a series circuit, the total voltage (RMS voltage) can be calculated as:
\[ V_{rms} = \sqrt{V_R^2 + V_L^2} \]
- Substitute the known values:
\[ V_{rms} = \sqrt{(20)^2 + (15)^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \, V \]

3. **Convert RMS voltage to peak voltage:**
- The relationship between RMS voltage and peak voltage is given by:
\[ V_{peak} = V_{rms} \times \sqrt{2} \]
- Substitute the calculated RMS voltage:
\[ V_{peak} = 25 \times \sqrt{2} \, V \]

4. **Final answer:**
- Therefore, the applied peak voltage is:
\[ V_{peak} = 25\sqrt{2} \, V \]

### Summary:
The applied peak voltage across the series combination of the inductor and resistor is \( 25\sqrt{2} \, V \).