A 50 W, 100V lamp is to be connected to an ac mains of 200V, 50 Hz. What capacitor is essential to be put in series with the lamp?

Options

A. `(25)/(sqrt(2))(mu)F`
B. `(50)/( pi sqrt(2))(mu)F`
C. `(50)/(sqrt(2))(mu)F`
D. `(100)/( pi sqrt(3))(mu)F`

Explanation

To solve the problem of determining the necessary capacitor to be connected in series with a 50 W, 100 V lamp when connected to a 200 V AC mains supply, we can follow these steps:

### Step 1: Calculate the Resistance of the Lamp
The power (P) and voltage (V) ratings of the lamp are given. We can use the formula for power to find the resistance (R) of the lamp.

\[ P = \frac{V^2}{R} \implies R = \frac{V^2}{P} \]

Substituting the values:

\[ R = \frac{100^2}{50} = \frac{10000}{50} = 200 \, \Omega \]

### Step 2: Determine the Current through the Lamp
Next, we need to find the current (I) that flows through the lamp when it is operating at its rated voltage.

Using Ohm's Law:

\[ I = \frac{V}{R} \]

Substituting the values:

\[ I = \frac{100}{200} = 0.5 \, A \]

### Step 3: Set Up the AC Circuit
When the lamp is connected to the AC mains (200 V), we need to maintain the same current (0.5 A) through the lamp. The total voltage across the circuit will be the RMS voltage of the AC supply.

### Step 4: Calculate the Impedance (Z) of the Circuit
The current in an AC circuit is given by:

\[ I = \frac{V_{RMS}}{Z} \]

Setting the current equal to 0.5 A:

\[ 0.5 = \frac{200}{Z} \implies Z = \frac{200}{0.5} = 400 \, \Omega \]

### Step 5: Relate Impedance to Resistance and Reactance
In an RC circuit, the impedance (Z) is given by:

\[ Z = \sqrt{R^2 + X_C^2} \]

Substituting the known values:

\[ 400 = \sqrt{200^2 + X_C^2} \]

### Step 6: Solve for Reactance (X_C)
Squaring both sides:

\[ 400^2 = 200^2 + X_C^2 \]

Calculating:

\[ 160000 = 40000 + X_C^2 \implies X_C^2 = 160000 - 40000 = 120000 \]

Taking the square root:

\[ X_C = \sqrt{120000} = 200\sqrt{3} \, \Omega \]

### Step 7: Calculate the Capacitance (C)
The capacitive reactance (X_C) is related to capacitance (C) by the formula:

\[ X_C = \frac{1}{\omega C} \quad \text{where} \quad \omega = 2\pi f \]

Substituting for \( \omega \):

\[ X_C = \frac{1}{2\pi f C} \]

Rearranging gives:

\[ C = \frac{1}{2\pi f X_C} \]

Substituting \( f = 50 \, Hz \) and \( X_C = 200\sqrt{3} \):

\[ C = \frac{1}{2\pi \cdot 50 \cdot 200\sqrt{3}} = \frac{1}{10000\pi\sqrt{3}} \]

### Step 8: Convert to Microfarads
To convert capacitance to microfarads:

\[ C = \frac{1}{10000\pi\sqrt{3}} \times 10^6 \, \mu F = \frac{100000}{10000\pi\sqrt{3}} \, \mu F = \frac{10}{\pi\sqrt{3}} \, \mu F \]

### Final Answer
Thus, the required capacitance is:

\[ C \approx \frac{10}{\pi\sqrt{3}} \, \mu F \]