capacitors

By shyamlal • 05:50

Capacitance Calculation for a Lamp-Capacitor Series Circuit

Problem Statement

A circuit consists of a \(110\text{ V}\), \(40\text{ W}\) incandescent lamp connected in series with a capacitor across a \(230\text{ V}\), \(50\text{ Hz}\) AC supply. Determine the required capacitance of the capacitor so that the lamp operates at its rated voltage and power.

Given Data

  • Rated lamp voltage (\(V_R\)) = \(110\text{ V}\) (non-inductive / purely resistive load)
  • Rated lamp power (\(P\)) = \(40\text{ W}\)
  • Total AC supply voltage (\(V\)) = \(230\text{ V}\)
  • Supply frequency (\(f\)) = \(50\text{ Hz}\)

Circuit and Phasor Representation

Series R-C Connection Lamp (110V, 40W) C = ? 230V, 50Hz Voltage Phasor Triangle VR = 110 V VC = 201.99 V V = 230 V

Step-by-Step Solution

Step 1: Calculate Rated Lamp Current (\(I\))

For an incandescent lamp (purely resistive), the rated current is:

$$I = \frac{P}{V_R} = \frac{40}{110} = \frac{4}{11} \approx \mathbf{0.3636\text{ A}}$$

Step 2: Determine Voltage Across the Capacitor (\(V_C\))

In a series R-C circuit, the total supply voltage \(V\) is the vector sum of \(V_R\) and \(V_C\):

$$V = \sqrt{V_R^2 + V_C^2} \implies V^2 = V_R^2 + V_C^2$$

Solving for \(V_C\):

$$V_C = \sqrt{V^2 - V_R^2} = \sqrt{(230)^2 - (110)^2}$$

$$V_C = \sqrt{52900 - 12100} = \sqrt{40800} \approx \mathbf{201.99\text{ V}}$$

Step 3: Calculate Capacitive Reactance (\(X_C\))

Since the circuit is connected in series, the same current \(I\) flows through both components:

$$X_C = \frac{V_C}{I} = \frac{201.99}{\frac{4}{11}} = \frac{201.99 \times 11}{4} \approx \mathbf{555.47\ \Omega}$$

Step 4: Calculate the Capacitance (\(C\))

The capacitive reactance formula is:

$$X_C = \frac{1}{2\pi f C} \implies C = \frac{1}{2\pi f X_C}$$

Substituting \(f = 50\text{ Hz}\) and \(X_C = 555.47\ \Omega\):

$$C = \frac{1}{2 \times \pi \times 50 \times 555.47} = \frac{1}{174505.5} \approx \mathbf{5.73 \times 10^{-6}\text{ F}}$$

$$C \approx \mathbf{5.73\ \mu F}$$

Alternative Verification (Impedance Method)

  • Lamp resistance: \(R = \frac{V_R^2}{P} = \frac{(110)^2}{40} = 302.5\ \Omega\)
  • Total circuit impedance: \(Z = \frac{V}{I} = \frac{230}{4/11} = 632.5\ \Omega\)
  • Reactance: \(X_C = \sqrt{Z^2 - R^2} = \sqrt{(632.5)^2 - (302.5)^2} = \sqrt{400056.25 - 91506.25} = \sqrt{308550} \approx 555.47\ \Omega\)
  • Capacitance: \(C = \frac{1}{2\pi \times 50 \times 555.47} \approx \mathbf{5.73\ \mu F}\)
Final Result:

The capacitance of the capacitor required is \(5.73\ \mu\text{F}\).