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.
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
1 answer