When `100 V` DC is applied across a solenoid, a current of `1.0 A` flows in it. When `100 V` AC is applied across the same coil. The current drops to `0.5A`. If the frequency of the ac source is `50 Hz`, the impedance and inductance of the solenoid are

Options

A. `200 Omega` and 0.55 H
B. `100 Omega` and 0.86 H
C. `200 Omega` and 1.0 H
D. 1100`Omega` and 0.93 H

Explanation

To solve the problem step by step, we will calculate the resistance, impedance, inductive reactance, and finally the inductance of the solenoid.

### Step 1: Calculate the Resistance (R) of the Solenoid
When a DC voltage of `100 V` is applied, the current flowing through the solenoid is `1.0 A`. We can calculate the resistance using Ohm's Law:

\[ R = \frac{V}{I} \]

Substituting the values:

\[ R = \frac{100 \, \text{V}}{1.0 \, \text{A}} = 100 \, \Omega \]

### Step 2: Calculate the Impedance (Z) of the Solenoid
When an AC voltage of `100 V` is applied, the current drops to `0.5 A`. The impedance can also be calculated using Ohm's Law:

\[ Z = \frac{V}{I} \]

Substituting the values:

\[ Z = \frac{100 \, \text{V}}{0.5 \, \text{A}} = 200 \, \Omega \]

### Step 3: Relate Impedance, Resistance, and Inductive Reactance
The relationship between impedance (Z), resistance (R), and inductive reactance (XL) is given by:

\[ Z^2 = R^2 + X_L^2 \]

Substituting the known values:

\[ 200^2 = 100^2 + X_L^2 \]

Calculating the squares:

\[ 40000 = 10000 + X_L^2 \]

### Step 4: Solve for Inductive Reactance (XL)
Rearranging the equation to find \(X_L\):

\[ X_L^2 = 40000 - 10000 = 30000 \]

Taking the square root:

\[ X_L = \sqrt{30000} \approx 173.2 \, \Omega \]

### Step 5: Calculate the Inductance (L) of the Solenoid
The inductive reactance is related to the inductance and frequency by the formula:

\[ X_L = 2 \pi f L \]

Rearranging to solve for L:

\[ L = \frac{X_L}{2 \pi f} \]

Substituting the values (where \(f = 50 \, \text{Hz}\)):

\[ L = \frac{173.2}{2 \pi \times 50} \]

Calculating:

\[ L = \frac{173.2}{314.16} \approx 0.551 \, \text{H} \]

### Final Results
- Impedance \(Z = 200 \, \Omega\)
- Inductance \(L \approx 0.551 \, \text{H}\)

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