6. Resistor Connections to AC Sources: Impedance, Power Dissipation, and Voltage Division

Introduction: Resistors connected to AC sources play a crucial role in electronic circuits, allowing for impedance control, power dissipation, and voltage division. Understanding how resistors interact with AC sources is essential for circuit design and analysis. In this blog post, we will delve into resistor connections to AC sources, discussing impedance, power dissipation, voltage division, and providing practical examples to solidify your understanding.

1. Impedance in AC Circuits: In AC circuits, resistors exhibit impedance, which is the opposition to the flow of alternating current. The impedance of a resistor is purely resistive and is equal to its resistance (R).
2. Voltage Division in AC Circuits: Similar to DC circuits, resistors connected in series to an AC source exhibit voltage division. The voltage across a resistor depends on its resistance and the total impedance in the circuit.
3. Power Dissipation in AC Circuits: When a resistor is connected to an AC source, power is dissipated in the form of heat. The power dissipated by a resistor in an AC circuit can be calculated using the same formula as in DC circuits.

P = (Vᵣ)² / R

Where: P is the power dissipated by the resistor, Vᵣ is the voltage across the resistor, R is the resistance of the resistor.

4. Practical Examples: Let's consider a few practical examples to illustrate resistor connections to AC sources:

Example 1: A 100 Ω resistor is connected in series to a 120 Vrms AC source with a frequency of 60 Hz.

Impedance of the resistor (Z) = R = 100 Ω

The voltage across the resistor (Vᵣ) can be calculated using Ohm's Law: Vᵣ = I * Z

Since the current (I) is the same throughout the series circuit, we can calculate it using: I = Vrms / Z

Vᵣ = (Vrms / Z) * Z Vᵣ = Vrms

Therefore, the voltage across the resistor is 120 Vrms.

Power dissipation in the resistor: P = (Vᵣ)² / R P = (120 Vrms)² / 100 Ω P ≈ 144 W

Example 2: A circuit consists of two resistors in series, R₁ = 50 Ω and R₂ = 75 Ω, connected to a 220 Vrms AC source with a frequency of 50 Hz.

Impedance of R₁ (Z₁) = R₁ = 50 Ω Impedance of R₂ (Z₂) = R₂ = 75 Ω

Voltage division across the resistors: Vᵣ₁ = (Z₁ / (Z₁ + Z₂)) * Vrms Vᵣ₁ = (50 Ω / (50 Ω + 75 Ω)) * 220 Vrms Vᵣ₁ ≈ 73.33 Vrms

Vᵣ₂ = (Z₂ / (Z₁ + Z₂)) * Vrms Vᵣ₂ = (75 Ω / (50 Ω + 75 Ω)) * 220 Vrms Vᵣ₂ ≈ 146.67 Vrms

Power dissipation in the resistors: P₁ = (Vᵣ₁)² / R₁ P₁ = (73.33 Vrms)² / 50 Ω P₁ ≈ 107.06 W

P₂ = (Vᵣ₂)² / R₂ P₂ = (146.67 Vrms)² / 75 Ω P₂ ≈ 289.39 W

Conclusion: Resistor connections to AC sources involve impedance, voltage division, and power dissipation. Understanding the concept of impedance allows you to determine the voltage across resistors in AC circuits, while power dissipation calculations help assess the heat generated by the resistors. By exploring practical examples, you can gain a deeper understanding of how resistors behave in AC circuits.