Question
An incandescent lamp with a rating of 100Watts at 100Volts is connected in series with a capacitor across a 220 Volts, 50Hz. Calculate
a)The capacitance of the capacitor
b) the phase angle between the voltage and the current
a)The capacitance of the capacitor
b) the phase angle between the voltage and the current
Answers
GPT-4o
To solve this problem, we need to consider the voltage and current characteristics in an AC circuit containing a capacitor and a resistor (the incandescent lamp) in series.
Let's breakdown the steps:
### Given:
- Incandescent lamp rating (Power, P) = 100 W
- Voltage across lamp when at rated power (V) = 100 V
- Supply voltage (V_s) = 220 V
- Frequency (f) = 50 Hz
### Understand the Problem:
1. **Calculate the resistance (R) of the lamp:**
The lamp is rated for 100W at 100V. Using the power formula:
\(P = \frac{V^2}{R}\)
Rearrange to solve for \(R\):
\(R = \frac{V^2}{P} = \frac{100^2}{100} = 100 \ \Omega\)
2. **Impedance in series circuit:**
The impedance (Z) in the series circuit with resistor (R) and capacitor (X_C) should be such that the overall voltage drop is 220V.
### Approach:
**a) Calculate the Capacitance (C):**
- The overall impedance \(Z\) of the series R-C circuit is given by:
\[
Z = \sqrt{R^2 + X_C^2}
\]
- Since we know \(V_s = 220 \text{V}\) and \(V_R = 100 \text{V}\), the current (I) through the circuit is:
\[
I = \frac{V_R}{R} = \frac{100}{100} = 1 \text{A}
\]
- The voltage across the capacitor \(V_C\) is given by:
\[
V_s^2 = V_R^2 + V_C^2
\]
Solving for \(V_C\):
\[
220^2 = 100^2 + V_C^2
\]
\[
48400 = 10000 + V_C^2
\]
\[
V_C^2 = 38400
\]
\[
V_C = \sqrt{38400} \approx 196 \text{V}
\]
- The reactance of the capacitor \(X_C\) is:
\[
X_C = \frac{V_C}{I} = \frac{196}{1} = 196 \ \Omega
\]
- Using the capacitive reactance formula:
\[
X_C = \frac{1}{2 \pi f C}
\]
Rearrange to solve for \(C\):
\[
C = \frac{1}{2 \pi f X_C} = \frac{1}{2 \pi \cdot 50 \cdot 196}
\]
\[
C \approx \frac{1}{61544} \approx 1.624 \times 10^{-5} \text{F}
\]
\[
C \approx 16.24 \ \mu\text{F}
\]
**b) Calculate the Phase Angle (\(\phi\)) between the Voltage and the Current:**
- In an R-C circuit, the phase angle \(\phi\) is given by:
\[
\tan(\phi) = \frac{X_C}{R}
\]
\[
\phi = \tan^{-1}\left(\frac{196}{100}\right)
\]
\[
\phi = \tan^{-1}(1.96) \approx 63.74^\circ
\]
### Results:
a) The capacitance of the capacitor is approximately \(16.24 \ \mu\text{F}\).
b) The phase angle between the voltage and the current is approximately \(63.74^\circ\).
These calculations assume ideal components and ignore potential real-world imperfections like the internal resistance of the capacitor or inductive effects in the circuit.
Let's breakdown the steps:
### Given:
- Incandescent lamp rating (Power, P) = 100 W
- Voltage across lamp when at rated power (V) = 100 V
- Supply voltage (V_s) = 220 V
- Frequency (f) = 50 Hz
### Understand the Problem:
1. **Calculate the resistance (R) of the lamp:**
The lamp is rated for 100W at 100V. Using the power formula:
\(P = \frac{V^2}{R}\)
Rearrange to solve for \(R\):
\(R = \frac{V^2}{P} = \frac{100^2}{100} = 100 \ \Omega\)
2. **Impedance in series circuit:**
The impedance (Z) in the series circuit with resistor (R) and capacitor (X_C) should be such that the overall voltage drop is 220V.
### Approach:
**a) Calculate the Capacitance (C):**
- The overall impedance \(Z\) of the series R-C circuit is given by:
\[
Z = \sqrt{R^2 + X_C^2}
\]
- Since we know \(V_s = 220 \text{V}\) and \(V_R = 100 \text{V}\), the current (I) through the circuit is:
\[
I = \frac{V_R}{R} = \frac{100}{100} = 1 \text{A}
\]
- The voltage across the capacitor \(V_C\) is given by:
\[
V_s^2 = V_R^2 + V_C^2
\]
Solving for \(V_C\):
\[
220^2 = 100^2 + V_C^2
\]
\[
48400 = 10000 + V_C^2
\]
\[
V_C^2 = 38400
\]
\[
V_C = \sqrt{38400} \approx 196 \text{V}
\]
- The reactance of the capacitor \(X_C\) is:
\[
X_C = \frac{V_C}{I} = \frac{196}{1} = 196 \ \Omega
\]
- Using the capacitive reactance formula:
\[
X_C = \frac{1}{2 \pi f C}
\]
Rearrange to solve for \(C\):
\[
C = \frac{1}{2 \pi f X_C} = \frac{1}{2 \pi \cdot 50 \cdot 196}
\]
\[
C \approx \frac{1}{61544} \approx 1.624 \times 10^{-5} \text{F}
\]
\[
C \approx 16.24 \ \mu\text{F}
\]
**b) Calculate the Phase Angle (\(\phi\)) between the Voltage and the Current:**
- In an R-C circuit, the phase angle \(\phi\) is given by:
\[
\tan(\phi) = \frac{X_C}{R}
\]
\[
\phi = \tan^{-1}\left(\frac{196}{100}\right)
\]
\[
\phi = \tan^{-1}(1.96) \approx 63.74^\circ
\]
### Results:
a) The capacitance of the capacitor is approximately \(16.24 \ \mu\text{F}\).
b) The phase angle between the voltage and the current is approximately \(63.74^\circ\).
These calculations assume ideal components and ignore potential real-world imperfections like the internal resistance of the capacitor or inductive effects in the circuit.