An electric lamp which runs at `100 V DC` and consumes `10 A` current is connected to `AC` mains at `150 V, 50 Hz` cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect the resistance of choke.

Explanation

To solve the problem step-by-step, we will follow the given information and apply the relevant formulas.

### Step 1: Understand the given values
- The electric lamp operates at a voltage of \( V_R = 100 \, V \) (DC).
- The current consumed by the lamp is \( I = 10 \, A \).
- The AC mains voltage is \( V = 150 \, V \).
- The frequency of the AC supply is \( f = 50 \, Hz \).

### Step 2: Calculate the voltage drop across the choke
We can use the formula for the voltage across the choke \( V_L \):
\[ V_L = \sqrt{V^2 - V_R^2} \]
Substituting the known values:
\[ V_L = \sqrt{150^2 - 100^2} \]
Calculating the squares:
\[ 150^2 = 22500 \quad \text{and} \quad 100^2 = 10000 \]
Now, substituting these values into the equation:
\[ V_L = \sqrt{22500 - 10000} = \sqrt{12500} \]
Calculating the square root:
\[ V_L \approx 111.8 \, V \]

### Step 3: Calculate the inductance \( L \)
We know that the voltage across the choke can also be expressed as:
\[ V_L = I \cdot X_L \]
Where \( X_L = \omega L \) and \( \omega = 2 \pi f \).

First, calculate \( \omega \):
\[ \omega = 2 \pi f = 2 \cdot 3.14 \cdot 50 \approx 314 \, rad/s \]

Now, substituting into the equation for \( V_L \):
\[ V_L = I \cdot \omega L \]
Rearranging to find \( L \):
\[ L = \frac{V_L}{I \cdot \omega} \]
Substituting the known values:
\[ L = \frac{111.8}{10 \cdot 314} \]
Calculating the denominator:
\[ 10 \cdot 314 = 3140 \]
Now substituting back:
\[ L = \frac{111.8}{3140} \approx 0.0357 \, H \quad \text{or} \quad 0.036 \, H \]

### Final Answers
- The inductance \( L \approx 0.036 \, H \)
- The voltage drop across the choke \( V_L \approx 111.8 \, V \)