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. `200 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, we need to find the impedance (Z) and the inductance (L) of the solenoid based on the given information. Let's break down the solution step by step.

### Step 1: Calculate the Resistance (R)
When a 100 V DC is applied across the solenoid, a current of 1.0 A flows through it. We can use Ohm's Law to calculate the resistance (R) of the solenoid.

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

### Step 2: Calculate the Impedance (Z)
When a 100 V AC is applied across the same solenoid, the current drops to 0.5 A. The impedance (Z) can be calculated using the formula:

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

### Step 3: Relate Impedance, Resistance, and Reactance
In an AC circuit with inductance, the impedance (Z) is related to the resistance (R) and the inductive reactance (X_L) by the formula:

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

Substituting the known values:

\[ 200 = \sqrt{100^2 + X_L^2} \]

### Step 4: Solve for Inductive Reactance (X_L)
Squaring both sides gives:

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

Calculating the squares:

\[ 40000 = 10000 + X_L^2 \]

Now, isolate \(X_L^2\):

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

Taking the square root:

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

### Step 5: Calculate the Angular Frequency (ω)
The angular frequency (ω) is given by the formula:

\[ \omega = 2\pi f \]

Given that the frequency (f) is 50 Hz:

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

### Step 6: Calculate Inductance (L)
The inductive reactance (X_L) is also related to the inductance (L) by the formula:

\[ X_L = \omega L \]

Substituting the values we have:

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

Dividing both sides by \(100\pi\):

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

Calculating the numerical value:

\[ L \approx \frac{1.732}{3.14} \approx 0.55 \, \text{H} \]

### Final Answers
- Impedance (Z) = 200 Ω
- Inductance (L) ≈ 0.55 H

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