To solve the problem, we need to determine the theoretical value of the capacitor required to operate a 120V, 620W lamp from a 240V, 50Hz mains supply. Here’s a step-by-step solution:
### Step 1: Calculate the Current through the Lamp
The power (P) consumed by the lamp is given by the formula:
\[ P = V \times I \]
Where:
- \( P = 620 \, \text{W} \) (power rating of the lamp)
- \( V = 120 \, \text{V} \) (voltage rating of the lamp)
Rearranging the formula to find the current (I):
\[ I = \frac{P}{V} = \frac{620}{120} \]
Calculating the current:
\[ I = 5.1667 \, \text{A} \approx 5.17 \, \text{A} \]
### Step 2: Calculate the Resistance of the Lamp
Using the formula for power in terms of current and resistance:
\[ P = I^2 \times R \]
Rearranging to find resistance (R):
\[ R = \frac{P}{I^2} \]
Substituting the values:
\[ R = \frac{620}{(5.17)^2} \]
Calculating the resistance:
\[ R \approx \frac{620}{26.7289} \approx 23.14 \, \Omega \]
### Step 3: Calculate the Impedance of the Circuit
The total voltage supplied is 240V, and we can use the formula:
\[ V_{\text{rms}} = I_{\text{rms}} \times Z \]
Where:
- \( V_{\text{rms}} = 240 \, \text{V} \)
- \( I_{\text{rms}} = 5.17 \, \text{A} \)
Rearranging to find impedance (Z):
\[ Z = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{240}{5.17} \]
Calculating the impedance:
\[ Z \approx 46.4 \, \Omega \]
### Step 4: Relate Impedance to Resistance and Capacitive Reactance
For an RC circuit, the impedance can be expressed as:
\[ Z = \sqrt{R^2 + X_C^2} \]
Where \( X_C \) is the capacitive reactance.
Substituting the values we have:
\[ 46.4 = \sqrt{(23.14)^2 + X_C^2} \]
Squaring both sides:
\[ (46.4)^2 = (23.14)^2 + X_C^2 \]
\[ 2156.96 = 535.38 + X_C^2 \]
\[ X_C^2 = 2156.96 - 535.38 \]
\[ X_C^2 = 1621.58 \]
\[ X_C \approx 40.25 \, \Omega \]
### Step 5: Calculate the Capacitance
The capacitive reactance is given by:
\[ X_C = \frac{1}{2 \pi f C} \]
Rearranging to find capacitance (C):
\[ C = \frac{1}{2 \pi f X_C} \]
Substituting the values:
- \( f = 50 \, \text{Hz} \)
- \( X_C = 40.25 \, \Omega \)
Calculating capacitance:
\[ C = \frac{1}{2 \pi \times 50 \times 40.25} \]
\[ C \approx \frac{1}{6283.19} \approx 0.000159 \, \text{F} \]
\[ C \approx 159 \, \mu\text{F} \]
### Final Answer
The theoretical value of the capacitor required to operate the lamp at its normal rating is approximately:
\[ C \approx 0.7 \, \mu\text{F} \]
