When 100 V DC is applied across a solenoid, a steady current of 1 A flows in it. when 100V AC is applied across the same solenoid, the current drops to 0.5 A. If the frequency of the Ad source is `150// sqrt3pi Hz`, the impedance and inductance of the solenoid are

Options

A. `200Omega and 1//3H`
B. `100Omega and 1//16H`
C. `200Omega` and 1.0H
D. `1100Omega` and 3//117H

Answer

To solve the problem step by step, we will calculate the impedance and inductance of the solenoid based on the given information.

### Step 1: Calculate the Impedance (Z)
Given that when 100 V AC is applied, the current (I) is 0.5 A. The impedance (Z) can be calculated using Ohm's law for AC circuits:

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

Where:
- \( V = 100 \, V \) (AC voltage)
- \( I = 0.5 \, A \) (AC current)

Substituting the values:

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

### Step 2: Calculate the Resistance (R)
When 100 V DC is applied, a steady current of 1 A flows. The resistance (R) can be calculated using Ohm's law:

\[ R = \frac{V_{DC}}{I_{DC}} \]

Where:
- \( V_{DC} = 100 \, V \) (DC voltage)
- \( I_{DC} = 1 \, A \) (DC current)

Substituting the values:

\[ R = \frac{100 \, V}{1 \, A} = 100 \, \Omega \]

### Step 3: Calculate the Inductive Reactance (X_L)
Using the relationship between impedance, resistance, and inductive reactance, we can express the inductive reactance (X_L) as:

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

Rearranging gives:

\[ X_L = \sqrt{Z^2 - R^2} \]

Substituting the values we have:

\[ X_L = \sqrt{(200 \, \Omega)^2 - (100 \, \Omega)^2} \]

Calculating the squares:

\[ X_L = \sqrt{40000 - 10000} = \sqrt{30000} = 173.2 \, \Omega \]

### Step 4: Calculate the Angular Frequency (ω)
The angular frequency (ω) can be calculated using the formula:

\[ \omega = 2\pi f \]

Where the frequency \( f \) is given as:

\[ f = \frac{150}{\sqrt{3}\pi} \, Hz \]

Substituting this into the formula for ω:

\[ \omega = 2\pi \left(\frac{150}{\sqrt{3}\pi}\right) = \frac{300}{\sqrt{3}} \, rad/s \]

### Step 5: Calculate the Inductance (L)
The inductance (L) can be calculated using the formula:

\[ L = \frac{X_L}{\omega} \]

Substituting the values we have:

\[ L = \frac{173.2 \, \Omega}{\frac{300}{\sqrt{3}}} \]

Calculating the denominator:

\[ L = \frac{173.2 \, \Omega \cdot \sqrt{3}}{300} \]

Using \( \sqrt{3} \approx 1.732 \):

\[ L \approx \frac{173.2 \cdot 1.732}{300} \approx \frac{300.0}{300} \approx 1.0 \, H \]

### Final Results
- Impedance \( Z = 200 \, \Omega \)
- Inductance \( L \approx 1.0 \, H \)

### Summary
The impedance of the solenoid is \( 200 \, \Omega \) and the inductance is approximately \( 1.0 \, H \).