A bulb of resistance `10Omega` , connected to an inductor L, is in series with an ac source marked 100V, 50 Hz. If the phase angle between the voltage and current is `pi//4` radian, calculate the value of L.

Explanation

To find the value of the inductor \( L \) in the given circuit, we will follow these steps:

### Step 1: Identify the given values
- Resistance \( R = 10 \, \Omega \)
- Voltage \( V = 100 \, V \)
- Frequency \( f = 50 \, Hz \)
- Phase angle \( \phi = \frac{\pi}{4} \, \text{radians} \)

### Step 2: Use the relationship between phase angle, resistance, and inductive reactance
The phase angle \( \phi \) in an R-L circuit is given by:
\[ \tan(\phi) = \frac{X_L}{R} \]
where \( X_L \) is the inductive reactance.

### Step 3: Calculate \( \tan(\phi) \)
Given \( \phi = \frac{\pi}{4} \):
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]

### Step 4: Set up the equation
From the relationship:
\[ 1 = \frac{X_L}{R} \]
Substituting \( R = 10 \, \Omega \):
\[ 1 = \frac{X_L}{10} \]
This implies:
\[ X_L = 10 \, \Omega \]

### Step 5: Relate inductive reactance to inductance
The inductive reactance \( X_L \) is given by:
\[ X_L = \omega L \]
where \( \omega = 2 \pi f \).

### Step 6: Calculate \( \omega \)
Substituting \( f = 50 \, Hz \):
\[ \omega = 2 \pi \times 50 = 100 \pi \, \text{rad/s} \]

### Step 7: Substitute \( X_L \) into the equation
Now we have:
\[ 10 = (100 \pi) L \]

### Step 8: Solve for \( L \)
Rearranging gives:
\[ L = \frac{10}{100 \pi} = \frac{1}{10 \pi} \, H \]

### Step 9: Calculate the numerical value of \( L \)
Using \( \pi \approx 3.14 \):
\[ L \approx \frac{1}{10 \times 3.14} \approx \frac{1}{31.4} \approx 0.0318 \, H \]

### Final Answer
Thus, the value of the inductor \( L \) is approximately:
\[ L \approx 0.0318 \, H \quad \text{or} \quad 0.032 \, H \, (\text{to 3 significant figures}) \]

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