A choke coil is needed to oprated an arc lamp at 160V(rms)and 50Hz. The arc lamp has an effective resistance of `5 Omega` when running of 10A(rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on 160V(dc), what additional resistnce is requried? Compare the power losses in both the cases.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Voltage (V_rms) = 160 V
- Frequency (f) = 50 Hz
- Effective resistance of the arc lamp (R) = 5 Ω
- Current (I_rms) = 10 A

### Step 2: Calculate the voltage across the resistance (V_R)
Using Ohm's Law, the voltage across the resistance can be calculated as:
\[ V_R = I_{rms} \times R \]
Substituting the values:
\[ V_R = 10 \, \text{A} \times 5 \, \Omega = 50 \, \text{V} \]

### Step 3: Calculate the voltage across the choke coil (V_L)
Using the relationship for the total voltage in an AC circuit:
\[ V_{rms}^2 = V_R^2 + V_L^2 \]
We can rearrange this to find \( V_L \):
\[ V_L = \sqrt{V_{rms}^2 - V_R^2} \]
Substituting the values:
\[ V_L = \sqrt{(160 \, \text{V})^2 - (50 \, \text{V})^2} \]
\[ V_L = \sqrt{25600 - 2500} \]
\[ V_L = \sqrt{23100} \]
\[ V_L \approx 152 \, \text{V} \]

### Step 4: Calculate the inductive reactance (X_L)
Using the formula for inductive reactance:
\[ V_L = I_{rms} \times X_L \]
We can rearrange this to find \( X_L \):
\[ X_L = \frac{V_L}{I_{rms}} \]
Substituting the values:
\[ X_L = \frac{152 \, \text{V}}{10 \, \text{A}} = 15.2 \, \Omega \]

### Step 5: Calculate the inductance (L)
Using the formula for inductive reactance:
\[ X_L = \omega L \]
Where \( \omega = 2\pi f \). First, calculate \( \omega \):
\[ \omega = 2\pi \times 50 \, \text{Hz} \approx 314.16 \, \text{rad/s} \]
Now substituting to find \( L \):
\[ L = \frac{X_L}{\omega} = \frac{15.2 \, \Omega}{314.16 \, \text{rad/s}} \]
\[ L \approx 0.0484 \, \text{H} \]

### Step 6: Calculate the additional resistance required for DC operation
For DC operation, the total voltage across the circuit is:
\[ V = I \times (R + R_{additional}) \]
Where \( R_{additional} \) is the additional resistance needed. Rearranging gives:
\[ R_{additional} = \frac{V}{I} - R \]
Substituting the values:
\[ R_{additional} = \frac{160 \, \text{V}}{10 \, \text{A}} - 5 \, \Omega \]
\[ R_{additional} = 16 \, \Omega - 5 \, \Omega = 11 \, \Omega \]

### Step 7: Calculate power losses in both cases
**For AC:**
Power loss is given by:
\[ P_{AC} = I_{rms}^2 \times R \]
Substituting the values:
\[ P_{AC} = (10 \, \text{A})^2 \times 5 \, \Omega = 500 \, \text{W} \]

**For DC:**
Power loss is given by:
\[ P_{DC} = I^2 \times (R + R_{additional}) \]
Substituting the values:
\[ P_{DC} = (10 \, \text{A})^2 \times (5 \, \Omega + 11 \, \Omega) \]
\[ P_{DC} = 100 \times 16 = 1600 \, \text{W} \]

### Conclusion
- Inductance of the choke coil \( L \approx 0.0484 \, \text{H} \)
- Additional resistance required for DC operation \( R_{additional} = 11 \, \Omega \)
- Power loss in AC \( P_{AC} = 500 \, \text{W} \)
- Power loss in DC \( P_{DC} = 1600 \, \text{W} \)