Voltage and Phase Angle

Sample Answer

Experiment 5: R-C Series Circuit – Voltage and Phase Angle

Aim

The aim of this experiment is to investigate the voltages and phase angle of a series R-C circuit. The purpose is to calculate and measure the component voltages (across the resistor and capacitor), supply voltage, and the overall impedance of the circuit. The supply phase angle is also calculated and compared with experimental measurements to confirm theoretical formulae.

Objectives

1. To calculate and measure the voltages across the resistor and the capacitor.

2. To measure the total supply voltage and compare it with theoretical calculations.

3. To verify the impedance formula for a series R-C circuit.

4. To calculate and measure the supply phase angle between current and voltage.

5. To confirm the vector (phasor) relationship between voltages and phase angle in an R-C series circuit.

Theory

An R-C circuit consists of a resistor (R) and a capacitor (C) connected in series with an alternating current (AC) supply. When AC flows through the circuit:

• The resistor voltage (VR) is in phase with the current.

• The capacitor voltage (VC) lags the current by 90 degrees.

• The supply voltage (V) is the vector sum of VR and VC, not the simple arithmetic sum.

Capacitive Reactance (Xc)

The opposition offered by a capacitor to AC is called capacitive reactance, given by:

Xc = 1 / (2 × π × f × C)

where

• f = frequency in hertz (Hz)

• C = capacitance in farads (F)

Impedance of the Circuit (Z)

The total opposition to current is called impedance (Z), which combines both resistance (R) and capacitive reactance (Xc):

Z = √(R² + Xc²)

Current in the Circuit (I)

By Ohm’s law,

I = V / Z

where V is the supply voltage.

Voltage Across the Resistor (VR)

Since the resistor obeys Ohm’s law:

VR = I × R

Voltage Across the Capacitor (VC)

For a capacitor, the voltage is given by:

VC = I × Xc

Phase Angle (θ)

The phase angle between current and supply voltage is calculated as:

θ = arctan(Xc / R)

• If R >> Xc, the circuit is more resistive and θ is small.

• If Xc >> R, the circuit is more capacitive and θ approaches 90°.

Worked Example (Numerical Calculation)

Let us consider:

• Resistance, R = 200 Ω

• Capacitance, C = 10 µF = 10 × 10⁻⁶ F

• Frequency, f = 50 Hz

• Supply voltage, V = 50 V

Step 1: Calculate Xc
Xc = 1 / (2 × π × f × C)
= 1 / (2 × 3.1416 × 50 × 10 × 10⁻⁶)
= 1 / (0.0031416)
≈ 318.3 Ω

Step 2: Calculate Z
Z = √(R² + Xc²)
= √(200² + 318.3²)
= √(40000 + 101300)
= √141300
≈ 376 Ω

Step 3: Calculate Current I
I = V / Z
= 50 / 376
≈ 0.133 A (133 mA)

Step 4: Calculate VR
VR = I × R
= 0.133 × 200
≈ 26.6 V

Step 5: Calculate VC
VC = I × Xc
= 0.133 × 318.3
≈ 42.3 V

Step 6: Check Supply Voltage (Vector Addition)
V = √(VR² + VC²)
= √(26.6² + 42.3²)
= √(707 + 1789)
= √2496
≈ 50 V (matches supply voltage)

Step 7: Calculate Phase Angle θ
θ = arctan(Xc / R)
= arctan(318.3 / 200)
= arctan(1.5915)
≈ 58° (lagging)