When 100 volt d.c. is applied across a solenoid, a current of 1.0 A flows in it. When 100 volt a.c. is applied across the same coil, the current drops to 0.5 A. If the frequency of a.c. source is 50 Hz the impedance and inductance of the solenoid is

Options

A. `100sqrt(3) Omega and 0.55 H`
B. `100 Omega and 0.86 H`
C. `200 Omega and 1.0 H`
D. `100 Omega and 0.93 H`

Answer

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

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

\[ V = I \times R \]

Given:
- \( V = 100 \, \text{V} \)
- \( I_1 = 1.0 \, \text{A} \)

Substituting the values into the equation:

\[ 100 = 1 \times R \]

Solving for R:

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

### Step 2: Calculate the Impedance of the Solenoid with AC
When the same solenoid is connected to a 100 V AC supply, the current drops to 0.5 A. We can calculate the impedance (Z) using the formula:

\[ Z = \frac{V}{I_2} \]

Given:
- \( V = 100 \, \text{V} \)
- \( I_2 = 0.5 \, \text{A} \)

Substituting the values:

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

### Step 3: Relate Impedance, Resistance, and Inductive Reactance
The total impedance (Z) in an AC circuit with inductance is given by:

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

Where \( X_L \) is the inductive reactance. We can rearrange this equation to find \( X_L \):

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

Substituting the known values:

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

Calculating:

\[ 40000 = 10000 + X_L^2 \]

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

Taking the square root:

\[ X_L = \sqrt{30000} = 100\sqrt{3} \, \Omega \]

### Step 4: Calculate the Inductance of the Solenoid
The inductive reactance \( X_L \) is related to the inductance (L) and the frequency (f) by the formula:

\[ X_L = \omega L \]

Where \( \omega = 2\pi f \). Given that \( f = 50 \, \text{Hz} \):

\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

Substituting \( X_L \) and \( \omega \):

\[ 100\sqrt{3} = (100\pi)L \]

Solving for L:

\[ L = \frac{100\sqrt{3}}{100\pi} = \frac{\sqrt{3}}{\pi} \approx 0.55 \, \text{H} \]

### Final Answer
The impedance of the solenoid is \( 200 \, \Omega \) and the inductance is approximately \( 0.55 \, \text{H} \).

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