A coil having a resistance of `50.0 Omega` and an inductance of `0.5 H` is connected to an `AC` source of `110`, `v, 50Hz`.
The impedance of the coil is (appoximately)

Options

A. `165 Omega`
B. `55 Omega`
C. `330 Omega`
D. none of these

Answer

To find the impedance of the coil connected to an AC source, we can follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 50.0 Ω
- Inductance (L) = 0.5 H
- Frequency (f) = 50 Hz

### Step 2: Calculate the inductive reactance (X_L)
The inductive reactance (X_L) can be calculated using the formula:
\[ X_L = 2 \pi f L \]
Substituting the values:
\[ X_L = 2 \pi (50) (0.5) \]
\[ X_L = 2 \pi (25) \]
\[ X_L = 50 \pi \approx 157.08 \, \Omega \]

### Step 3: Calculate the impedance (Z)
The impedance (Z) of the coil in an AC circuit is given by the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the known values:
\[ Z = \sqrt{(50)^2 + (157.08)^2} \]
\[ Z = \sqrt{2500 + 24649.4464} \]
\[ Z = \sqrt{27149.4464} \]
\[ Z \approx 164.7 \, \Omega \]

### Final Answer
The impedance of the coil is approximately **164.7 Ω**.

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