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 capacitor needed to be connected in series with a 50 W, 100 V lamp when connected to an AC mains of 200 V, 50 Hz, we can follow these steps:

### Step 1: Calculate the Resistance of the Lamp
Using the formula for power:
\[ P = \frac{V^2}{R} \]
We can rearrange it to find resistance \( R \):
\[ R = \frac{V^2}{P} \]
Substituting the values:
\[ R = \frac{100^2}{50} = \frac{10000}{50} = 200 \, \Omega \]

### Step 2: Calculate the Current through the Lamp
Using Ohm's law:
\[ I = \frac{V}{R} \]
Substituting the values:
\[ I = \frac{100}{200} = 0.5 \, A \]

### Step 3: Analyze the AC Circuit with the Capacitor
In the AC circuit, the total voltage \( V_{rms} \) is given as 200 V. The current through the circuit must remain the same as the current through the lamp, which is 0.5 A.

### Step 4: Calculate the Impedance of the Circuit
Using the formula for current in an R-C circuit:
\[ I = \frac{V_{rms}}{Z} \]
Where \( Z \) is the impedance given by:
\[ Z = \sqrt{R^2 + X_C^2} \]
Substituting the known values:
\[ 0.5 = \frac{200}{Z} \]
Thus,
\[ Z = \frac{200}{0.5} = 400 \, \Omega \]

### Step 5: Set Up the Equation for Impedance
Now we have:
\[ Z^2 = R^2 + X_C^2 \]
Substituting the values:
\[ 400^2 = 200^2 + X_C^2 \]
Calculating:
\[ 160000 = 40000 + X_C^2 \]
\[ X_C^2 = 160000 - 40000 = 120000 \]
\[ X_C = \sqrt{120000} = 100\sqrt{12} \, \Omega \]

### Step 6: Relate Capacitive Reactance to Capacitance
The capacitive reactance \( X_C \) is given by:
\[ X_C = \frac{1}{\omega C} \]
Where \( \omega = 2\pi f \) and \( f = 50 \, Hz \):
\[ \omega = 2\pi \times 50 = 100\pi \, rad/s \]
Thus,
\[ 100\sqrt{12} = \frac{1}{100\pi C} \]
Rearranging gives:
\[ C = \frac{1}{100\pi \times 100\sqrt{12}} = \frac{1}{10000\pi\sqrt{12}} \]

### Step 7: Calculate the Value of Capacitance
Simplifying:
\[ C = \frac{1}{10000\pi\sqrt{12}} = \frac{1}{10000\pi \times 3.464} \approx \frac{1}{34640} \approx 2.89 \times 10^{-5} \, F \]
Converting to microfarads:
\[ C \approx 28.9 \, \mu F \]

### Final Answer
The required capacitance to be put in series with the lamp is approximately:
\[ C \approx \frac{50}{\pi\sqrt{3}} \, \mu F \]