To solve this problem, we will follow these steps:
- Calculate the current through the circuit.
- Determine the impedance of the series circuit.
- Calculate the capacitance of the capacitor.
- Determine the phase angle between the voltage and current.
Step 1: Calculate the Current through the Circuit
The lamp is rated at \(100 , \text{W}\) at \(100 , \text{V}\). We first calculate the current through the lamp when it is connected to its rated voltage:
\[ P = VI \implies I = \frac{P}{V} = \frac{100 , \text{W}}{100 , \text{V}} = 1 , \text{A} \]
Step 2: Determine the Equivalent Circuit Voltage
Since the lamp is connected in series with the capacitor, the total voltage supply (\(V_s\)) is \(220 , \text{V}\).
Step 3: Calculate the Impedance of the Total Circuit
The load consists of a resistor (the lamp) and a capacitor, which means the total impedance (\(Z\)) can be found using:
\[ Z = \sqrt{R^2 + X_C^2} \]
Where:
- \(R\) is the resistance of the lamp.
- \(X_C\) is the capacitive reactance, given by \(X_C = \frac{1}{2\pi f C}\) where \(f\) is the frequency.
First, let's find the resistance \(R\) of the incandescent lamp:
Using the power formula: \[ R = \frac{V^2}{P} = \frac{100^2}{100} = 100 , \Omega \]
Now, we know: \[ I = 1 , A \quad \text{and} \quad V_s = 220 , V \]
The total series voltage across the lamp and capacitor is \(V_R + V_C = V_s\).
Step 4: Find the Voltage across the Resistor
Since we know the current (\(I\)): \[ V_R = IR = 1 , A \times 100 , \Omega = 100 , V \]
Step 5: Calculate the Voltage across the Capacitor
Now, we can find the voltage across the capacitor: \[ V_C = V_s - V_R = 220 , V - 100 , V = 120 , V \]
Step 6: Find the Capacitance
The voltage across the capacitor is \(V_C\) and the current through the circuit is \(I\). We use the relationship to find \(C\): \[ V_C = I X_C = I \frac{1}{2 \pi f C} \]
Rearranging gives: \[ C = \frac{I}{2 \pi f V_C} \]
Substituting values: \[ C = \frac{1 , A}{2 \pi (50 , \text{Hz}) (120 , V)} \] \[ C = \frac{1}{2 \pi \times 50 \times 120} \] \[ C = \frac{1}{37699.111843}\approx 2.65 \times 10^{-5} , F \quad \text{or} \quad 26.5 , \mu F \]
Step 7: Calculate the Phase Angle
The phase angle \(\phi\) in a series RL circuit is given by: \[ \tan(\phi) = \frac{X_C}{R} \]
Where \(X_C\) is the capacitive reactance: \[ X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (50) (2.65 \times 10^{-5})} \approx 120 , \Omega \]
Now, we can calculate the phase angle: \[ \tan(\phi) = \frac{X_C}{R} = \frac{120 , \Omega}{100 , \Omega} = 1.2 \]
Therefore, we find \(\phi\): \[ \phi = \tan^{-1}(1.2) \approx 50.19^\circ \]
Summary
- Capacitance of the Capacitor: \(C \approx 26.5 , \mu F\)
- Phase Angle between Voltage and Current: \(\phi \approx 50.19^\circ\)