A coil draws a current of `1.0` ampere and a power of `100` watt from an `A.C.` source of `110` volts and `(5sqrt22)/pi` hertz. Find the inductance and resistance of the coil.

Explanation

To solve the problem, we need to find the inductance (L) and resistance (R) of the coil given the current (I), power (P), and voltage (V). Here are the steps to find the solution:

### Step 1: Calculate the Resistance (R)
Using the formula for power in an AC circuit:
\[ P = V \cdot I \cdot \cos(\phi) \]
Where \( \phi \) is the phase angle. However, we can also express the power in terms of resistance and current:
\[ P = I^2 R \]
Given:
- \( P = 100 \) watts
- \( I = 1.0 \) ampere

Rearranging the formula to solve for \( R \):
\[ R = \frac{P}{I^2} = \frac{100}{(1.0)^2} = 100 \, \Omega \]

### Step 2: Calculate the Impedance (Z)
The impedance (Z) can be calculated using the formula:
\[ Z = \frac{V}{I} \]
Given:
- \( V = 110 \) volts
- \( I = 1.0 \) ampere

Calculating \( Z \):
\[ Z = \frac{110}{1.0} = 110 \, \Omega \]

### Step 3: Relate Impedance, Resistance, and Reactance
The relationship between impedance (Z), resistance (R), and inductive reactance (X_L) is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the known values:
\[ 110 = \sqrt{(100)^2 + X_L^2} \]
Squaring both sides:
\[ 110^2 = 10000 + X_L^2 \]
\[ 12100 = 10000 + X_L^2 \]
\[ X_L^2 = 12100 - 10000 = 2100 \]
\[ X_L = \sqrt{2100} = 10\sqrt{21} \, \Omega \]

### Step 4: Calculate Inductive Reactance (X_L)
The inductive reactance (X_L) is given by:
\[ X_L = \omega L \]
Where \( \omega = 2\pi f \) and \( f \) is the frequency. Given:
- \( f = \frac{5\sqrt{22}}{\pi} \)

Calculating \( \omega \):
\[ \omega = 2\pi \left(\frac{5\sqrt{22}}{\pi}\right) = 10\sqrt{22} \]

### Step 5: Solve for Inductance (L)
Now, substituting \( X_L \) into the equation:
\[ X_L = \omega L \]
\[ 10\sqrt{21} = 10\sqrt{22} \cdot L \]
Dividing both sides by \( 10\sqrt{22} \):
\[ L = \frac{\sqrt{21}}{\sqrt{22}} = \sqrt{\frac{21}{22}} \, \text{H} \]

### Final Answers:
- Resistance \( R = 100 \, \Omega \)
- Inductance \( L = \sqrt{\frac{21}{22}} \, \text{H} \)

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