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 find the applied voltage in a series circuit with resistance \( R \) and inductance \( L \), where the potential drop across the resistance is \( V_R = 200V \) and across the inductance is \( V_L = 150V \), we can follow these steps:

### Step-by-step Solution:

1. **Understand the Circuit Configuration**:
The circuit consists of a resistor \( R \) and an inductor \( L \) connected in series with an alternating voltage source. In a series circuit, the same current flows through both components.

2. **Identify Voltage Drops**:
The voltage drop across the resistor \( R \) is given as \( V_R = 200V \) and the voltage drop across the inductor \( L \) is \( V_L = 150V \).

3. **Use Phasor Representation**:
In AC circuits, the voltage across the resistor \( V_R \) and the voltage across the inductor \( V_L \) can be represented as phasors. The voltage across the resistor is in phase with the current, while the voltage across the inductor lags the current by \( 90^\circ \).

4. **Apply the Pythagorean Theorem**:
Since the voltage drops are perpendicular to each other (due to the phase difference), the total applied voltage \( V \) can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_R^2 + V_L^2} \]

5. **Substitute the Values**:
Plugging in the values:
\[ V = \sqrt{(200)^2 + (150)^2} \]
\[ V = \sqrt{40000 + 22500} \]
\[ V = \sqrt{62500} \]

6. **Calculate the Result**:
Now, calculating the square root:
\[ V = 250V \]

7. **Conclusion**:
The applied voltage across the RL circuit is \( 250V \).

### Final Answer:
The applied voltage is \( 250V \).

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