Question
An incandescent lamp with a rating of 100Watts at 100Volts is connected in series with a capacitor across a 220V, 50Hz supply. Calculate
1) the capacitance of the capacitor
2) the phase angle between the voltage and current
1) the capacitance of the capacitor
2) the phase angle between the voltage and current
Answers
GPT-4o mini
To solve this problem, we will follow these steps:
1. Calculate the current through the circuit.
2. Determine the impedance of the series circuit.
3. Calculate the capacitance of the capacitor.
4. 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
1. **Capacitance of the Capacitor**: \(C \approx 26.5 \, \mu F\)
2. **Phase Angle between Voltage and Current**: \(\phi \approx 50.19^\circ\)
1. Calculate the current through the circuit.
2. Determine the impedance of the series circuit.
3. Calculate the capacitance of the capacitor.
4. 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
1. **Capacitance of the Capacitor**: \(C \approx 26.5 \, \mu F\)
2. **Phase Angle between Voltage and Current**: \(\phi \approx 50.19^\circ\)