A circuit containing an inductance and a resistance connected in series, has an AC source of 200V, 50Hz connected across it. An AC current of 10A rms flows through th ecircuit and the power loss is measured to be 1W.

Options

A. The inductance of the circuit is `(sqrt3)/(10pi)H`
B. The frequency of the AC when the phase difference between the current and emf becomes `pi/4`, with the above components is `(50)/(sqrt3)Hz`
C. The frequency of the AC when the phase difference between the current and emf becomes `pi//3`, with the above components is `(25)/(sqrt3)Hz`
D. The frequency of the AC when the phase difference between the current and emf becomes `pi//4`, with the above components is `(25)/(sqrt3)Hz`.

Answer

To solve the problem step by step, we will analyze the given circuit containing an inductance (L) and a resistance (R) connected in series with an AC source. We will use the provided information to calculate the required parameters.

### Step 1: Identify Given Values
- AC Voltage (V_rms) = 200 V
- Frequency (f) = 50 Hz
- AC Current (I_rms) = 10 A
- Power Loss (P) = 1 W

### Step 2: Calculate the Power Factor
The power factor (cos φ) can be calculated using the formula for power in an AC circuit:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos φ \]
Rearranging gives:
\[ \cos φ = \frac{P}{V_{rms} \cdot I_{rms}} \]
Substituting the known values:
\[ \cos φ = \frac{1}{200 \cdot 10} = \frac{1}{2000} \]
Thus,
\[ \cos φ = 0.0005 \]

### Step 3: Calculate the Phase Angle (φ)
Using the cosine value, we can find the phase angle:
\[ φ = \cos^{-1}(0.0005) \]
This angle is very close to 90 degrees (or π/2 radians), indicating that the circuit is predominantly inductive.

### Step 4: Calculate the Impedance (Z)
Using Ohm's law for AC circuits:
\[ Z = \frac{V_{rms}}{I_{rms}} \]
Substituting the known values:
\[ Z = \frac{200}{10} = 20 \, \Omega \]

### Step 5: Relate Impedance to Resistance and Reactance
In a series R-L circuit, the impedance is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Where \( X_L \) is the inductive reactance. Rearranging gives:
\[ R^2 + X_L^2 = Z^2 \]
Substituting the value of Z:
\[ R^2 + X_L^2 = 20^2 = 400 \]

### Step 6: Calculate the Resistance (R)
We know that the power loss (P) in a resistor is given by:
\[ P = I_{rms}^2 \cdot R \]
Rearranging gives:
\[ R = \frac{P}{I_{rms}^2} = \frac{1}{10^2} = \frac{1}{100} = 0.01 \, \Omega \]

### Step 7: Calculate the Inductive Reactance (X_L)
Substituting R back into the impedance equation:
\[ 0.01^2 + X_L^2 = 400 \]
\[ 0.0001 + X_L^2 = 400 \]
\[ X_L^2 = 400 - 0.0001 = 399.9999 \]
Taking the square root gives:
\[ X_L \approx 20 \, \Omega \]

### Step 8: Calculate the Inductance (L)
Inductive reactance is given by:
\[ X_L = 2 \pi f L \]
Rearranging gives:
\[ L = \frac{X_L}{2 \pi f} \]
Substituting the values:
\[ L = \frac{20}{2 \pi \cdot 50} \]
Calculating gives:
\[ L \approx \frac{20}{314.16} \approx 0.0637 \, H \]

### Summary of Results
- Resistance (R) = 0.01 Ω
- Inductive Reactance (X_L) = 20 Ω
- Inductance (L) ≈ 0.0637 H