A 20 watt electric lamp can be operated at a 50 V DC supply . Calculate the value of capacitance of the capacitor required to run the given lamp at 220 V , 50 Hz AC supply .

Answer

To find the value of capacitance required to run a 20-watt electric lamp at 220 V, 50 Hz AC supply, we can follow these steps:

### Step 1: Calculate the resistance of the lamp
The power (P) of the lamp is given as 20 watts, and it operates at a voltage (V) of 50 V DC. We can use the formula for power:

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

Rearranging the formula to find resistance (R):

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

Substituting the values:

\[ R = \frac{50^2}{20} = \frac{2500}{20} = 125 \, \Omega \]

### Step 2: Determine the required voltage drop across the lamp
To ensure the lamp operates efficiently, we need a voltage drop of 50 V across it when connected to the AC supply.

### Step 3: Calculate the total impedance of the circuit
The total impedance (Z) in the circuit, which includes the resistance (R) and capacitive reactance (Xc), is given by:

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

### Step 4: Calculate the RMS current in the circuit
The RMS voltage of the AC supply is 220 V. The RMS current (I_rms) can be calculated using:

\[ I_{rms} = \frac{V_{rms}}{Z} \]

### Step 5: Relate the voltage drop across the lamp to the current
The voltage drop across the lamp (V_rms) is given by:

\[ V_{rms} = I_{rms} \times R \]

Since we want V_rms to be 50 V, we can set up the equation:

\[ 50 = I_{rms} \times 125 \]

### Step 6: Substitute for I_rms
From the previous step, we can express I_rms in terms of Z:

\[ I_{rms} = \frac{220}{\sqrt{125^2 + X_c^2}} \]

Substituting this into the equation for V_rms:

\[ 50 = \frac{220}{\sqrt{125^2 + X_c^2}} \times 125 \]

### Step 7: Solve for Xc
Rearranging gives:

\[ \sqrt{125^2 + X_c^2} = \frac{220 \times 125}{50} \]

Calculating the right side:

\[ \sqrt{125^2 + X_c^2} = 550 \]

Squaring both sides:

\[ 125^2 + X_c^2 = 550^2 \]

Calculating \(550^2\):

\[ 125^2 + X_c^2 = 302500 \]

Now substituting \(125^2 = 15625\):

\[ 15625 + X_c^2 = 302500 \]

So,

\[ X_c^2 = 302500 - 15625 = 286875 \]

Taking the square root:

\[ X_c = \sqrt{286875} \approx 535.6 \, \Omega \]

### Step 8: Calculate the capacitance
The capacitive reactance (Xc) is related to capacitance (C) by the formula:

\[ X_c = \frac{1}{\omega C} \]

Where \(\omega = 2\pi f\) and \(f = 50 \, Hz\):

\[ X_c = \frac{1}{2\pi \times 50 \times C} \]

Rearranging to find C:

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

Substituting the values:

\[ C = \frac{1}{2\pi \times 50 \times 535.6} \]

Calculating this gives:

\[ C \approx 6 \times 10^{-6} \, F \]

### Final Answer
The value of capacitance required to run the lamp at 220 V, 50 Hz AC supply is approximately:

\[ C \approx 6 \, \mu F \]