A coil is series with a `20muF` capacitor across a `230V, 50Hz` supply. The current taken by the circuit is `8A` and the power consumed is `200W.` Calculate the inductance of the coil if the current in the circuit is
(a) leading
(b) lagging

Explanation

To solve the problem, we need to calculate the inductance of the coil in two scenarios: when the current is leading and when it is lagging. We will use the provided values and formulas step by step.

### Given Data:
- Capacitance, \( C = 20 \mu F = 20 \times 10^{-6} F \)
- Voltage, \( V_{RMS} = 230 V \)
- Frequency, \( f = 50 Hz \)
- Current, \( I_{RMS} = 8 A \)
- Power consumed, \( P = 200 W \)

### Step 1: Calculate the Power Factor
The power factor \( \cos \phi \) can be calculated using the formula:
\[ \cos \phi = \frac{P}{V_{RMS} \times I_{RMS}} \]
Substituting the values:
\[ \cos \phi = \frac{200}{230 \times 8} = \frac{200}{1840} \approx 0.1087 \]

### Step 2: Calculate the Phase Angle \( \phi \)
To find the phase angle \( \phi \):
\[ \phi = \cos^{-1}(0.1087) \approx 83.3^\circ \]

### Step 3: Calculate the Resistance \( R \)
Using the formula for resistance:
\[ R = \frac{P}{I_{RMS}^2} \]
Substituting the values:
\[ R = \frac{200}{8^2} = \frac{200}{64} = 3.125 \, \Omega \]

### Step 4: Calculate the Capacitive Reactance \( X_C \)
The capacitive reactance \( X_C \) can be calculated using:
\[ X_C = \frac{1}{2 \pi f C} \]
Substituting the values:
\[ X_C = \frac{1}{2 \pi \times 50 \times 20 \times 10^{-6}} \approx 159.15 \, \Omega \]

### Case (a): Current Leading Voltage
#### Step 5: Calculate Inductive Reactance \( X_L \)
For the case where the current leads the voltage:
\[ \tan \phi = \frac{X_C - X_L}{R} \]
Rearranging gives:
\[ X_L = X_C - R \tan \phi \]
Calculating \( \tan \phi \):
\[ \tan(83.3^\circ) \approx 9.5 \]
Now substituting:
\[ X_L = 159.15 - 3.125 \times 9.5 \approx 159.15 - 29.69 \approx 129.46 \, \Omega \]

#### Step 6: Calculate Inductance \( L \)
Using the relation:
\[ X_L = 2 \pi f L \implies L = \frac{X_L}{2 \pi f} \]
Substituting the values:
\[ L = \frac{129.46}{2 \pi \times 50} \approx \frac{129.46}{314.16} \approx 0.412 \, H \]

### Case (b): Current Lagging Voltage
#### Step 7: Calculate Inductive Reactance \( X_L \)
For the case where the current lags the voltage:
\[ \tan \phi = \frac{X_L - X_C}{R} \]
Rearranging gives:
\[ X_L = X_C + R \tan \phi \]
Substituting:
\[ X_L = 159.15 + 3.125 \times 9.5 \approx 159.15 + 29.69 \approx 188.84 \, \Omega \]

#### Step 8: Calculate Inductance \( L \)
Using the relation:
\[ L = \frac{X_L}{2 \pi f} \]
Substituting the values:
\[ L = \frac{188.84}{2 \pi \times 50} \approx \frac{188.84}{314.16} \approx 0.601 \, H \]

### Final Results:
- (a) Inductance when current is leading: \( L \approx 0.412 \, H \)
- (b) Inductance when current is lagging: \( L \approx 0.601 \, H \)