An alternating voltage is connected in series with a resistance `R` and inductance `L` if the potential drop across the resistance is `200V` and across the inductance is `150V`, then the applied voltage is

Options

A. 350 V
B. 250 V
C. 500 V
D. 300 V

Explanation

To solve the problem, we need to find the applied voltage in a series LR circuit where the potential drops across the resistance (R) and the inductance (L) are given. Here are the steps to find the solution:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Potential drop across the resistance, \( V_R = 200 \, V \)
- Potential drop across the inductance, \( V_L = 150 \, V \)

2. **Understand the Relationship**:
- In a series LR circuit, the total applied voltage \( V \) is given by the formula:
\[ V = \sqrt{V_R^2 + V_L^2} \]
- This formula arises from the fact that the voltage across the resistance and the voltage across the inductance are perpendicular to each other in the phasor representation.

3. **Substitute the Values into the Formula**:
- Substitute \( V_R \) and \( V_L \) into the formula:
\[ V = \sqrt{(200)^2 + (150)^2} \]

4. **Calculate the Squares**:
- Calculate \( (200)^2 \):
\[ (200)^2 = 40000 \]
- Calculate \( (150)^2 \):
\[ (150)^2 = 22500 \]

5. **Add the Squares**:
- Now add the two results:
\[ 40000 + 22500 = 62500 \]

6. **Take the Square Root**:
- Finally, take the square root of the sum:
\[ V = \sqrt{62500} = 250 \, V \]

7. **Final Answer**:
- The applied voltage is \( V = 250 \, V \).