A 220-V, 50 Hz, ac generator is connected to an inductor and a `50 Omega` resistance in series. The current in the circuit is `1.0 A`. What is the PD across inductor?

Options

A. (A) `102.2 V`
B. (B) `186.4 V`
C. (C) `214 V`
D. (D) `170 V`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the Given Values
We are given:
- Voltage (V) = 220 V
- Frequency (f) = 50 Hz
- Resistance (R) = 50 Ω
- Current (I) = 1.0 A

### Step 2: Calculate the Impedance (Z)
The relationship between voltage (V), current (I), and impedance (Z) in an AC circuit is given by:
\[ I = \frac{V}{Z} \]
From this, we can express Z as:
\[ Z = \frac{V}{I} \]
Substituting the known values:
\[ Z = \frac{220 \, \text{V}}{1.0 \, \text{A}} = 220 \, \Omega \]

### Step 3: Relate Impedance to Resistance and Inductive Reactance
The impedance in a series circuit containing resistance (R) and inductive reactance (X_L) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
We can rearrange this to find \(X_L\):
\[ Z^2 = R^2 + X_L^2 \]
Substituting the values we have:
\[ 220^2 = 50^2 + X_L^2 \]
Calculating the squares:
\[ 48400 = 2500 + X_L^2 \]
Subtracting 2500 from both sides:
\[ X_L^2 = 48400 - 2500 = 45800 \]
Taking the square root:
\[ X_L = \sqrt{45800} \approx 214.24 \, \Omega \]

### Step 4: Calculate the Potential Difference Across the Inductor (V_L)
The potential difference across the inductor can be calculated using:
\[ V_L = I \times X_L \]
Substituting the known values:
\[ V_L = 1.0 \, \text{A} \times 214.24 \, \Omega = 214.24 \, \text{V} \]

### Step 5: Conclusion
The potential difference across the inductor is approximately:
\[ V_L \approx 214 \, \text{V} \]

### Final Answer
The potential difference across the inductor is **214 V**.

---