# Parallel Circuit – en

A parallel circuit is an electrical configuration in which components are connected in multiple paths, allowing current to flow through more than one path simultaneously. In parallel circuits, the voltage across each component remains constant and equal to the source voltage, while the current divides among the components inversely proportional to their resistances or reactances. Parallel circuits are commonly used in various electronic devices and systems, such as power distribution networks and household wiring, due to their ability to maintain consistent voltage across components and provide redundancy. One advantage of parallel circuits is that if one component fails or is disconnected, the remaining components continue to function, as the current can still flow through the other paths. However, parallel circuits can be more challenging to analyze than series circuits, especially when dealing with complex impedances in AC circuits.

## Parallel Circuits

Characteristics:

1. Components are connected in multiple paths, with each path having a common voltage.
2. The total resistance is determined by the reciprocal of the sum of the reciprocals of the individual resistances.
3. The source current is distributed among the components based on their resistance values.
4. The voltage across each component is equal to the source voltage.

Analysis:

1. Determine the total resistance: 1/R_total = 1/R1 + 1/R2 + … + 1/Rn
2. Calculate the source current using Ohm’s Law: I_total = V_total / R_total
3. Determine the current flowing through each component: I_component = V_total / R_component
4. Calculate power dissipation for each component: P_component = V_total^2 / R_component

Applications:

1. Power distribution: Parallel circuits are commonly used in power distribution systems, as they allow for equal voltage across multiple loads and facilitate the addition or removal of loads without affecting the overall system.
2. Redundancy and fault tolerance: Parallel connections can provide redundancy and fault tolerance in systems, as the failure of one component does not necessarily lead to the failure of the entire circuit.
3. Current sharing: Parallel circuits can be used to share the current among multiple components, such as batteries or power supplies, reducing the burden on individual components.

Understanding the characteristics, analysis methods, and applications of series and parallel circuits is essential for anyone working with electrical or electronic systems. These fundamental circuit configurations form the basis for more complex circuits and systems, and mastering their principles is crucial for designing, troubleshooting, and optimizing various devices and applications.

## Example – Calculation of DC Parallel Circuit

Let’s consider a simple DC parallel circuit with a voltage source (V) and three resistors (R1, R2, and R3) connected in parallel. The goal is to calculate the total current (I_total) supplied by the voltage source and the current through each resistor.

Given values:

1. V = 12 V (DC)
2. R1 = 4 Ω
3. R2 = 6 Ω
4. R3 = 3 Ω

Step 1: Determine the total resistance (R_total) of the parallel circuit:

1 / R_total = 1 / R1 + 1 / R2 + 1 / R3 = 1 / 4 + 1 / 6 + 1 / 3 1 / R_total = 1 / 2 R_total = 2 Ω

Step 2: Calculate the total current (I_total) supplied by the voltage source:

We can use Ohm’s Law to find the total current:

I_total = V / R_total = 12 V / 2 Ω = 6 A

Step 3: Calculate the current through each resistor:

We can use Ohm’s Law again to find the current through each resistor:

I_R1 = V / R1 = 12 V / 4 Ω = 3 A I_R2 = V / R2 = 12 V / 6 Ω = 2 A I_R3 = V / R3 = 12 V / 3 Ω = 4 A

In conclusion, the total current (I_total) supplied by the voltage source is 6 A, and the current through R1 (I_R1), R2 (I_R2), and R3 (I_R3) are 3 A, 2 A, and 4 A, respectively. Note that the sum of the individual currents equals the total current:

I_total = I_R1 + I_R2 + I_R3 = 3 A + 2 A + 4 A = 6 A

## Example – Calculation of AC Parallel Circuit

Let’s consider a simple AC parallel circuit with an AC voltage source (V) and two components: a resistor (R) and a capacitor (C) connected in parallel. The goal is to calculate the total current (I_total) supplied by the voltage source and the current through each component.

Given values:

1. V = 120 Vrms at a frequency of 60 Hz
2. R = 20 Ω
3. C = 50 μF

Step 1: Calculate the angular frequency (ω):

ω = 2 * π * f = 2 * π * 60 ≈ 377 rad/s

Step 2: Calculate the capacitive reactance (X_C):

X_C = 1 / (ω * C) = 1 / (377 * 50 * 10^(-6)) ≈ 53.1 Ω

Step 3: Calculate the impedance of each component:

The resistor has an impedance of Z_R = R = 20 Ω The capacitor has an impedance of Z_C = -j * X_C = -j * 53.1 Ω

Step 4: Calculate the current through each component using Ohm’s Law:

I_R = V / Z_R = (120 V) / (20 Ω) = 6 A

I_C = V / Z_C = (120 V) / (-j * 53.1 Ω) ≈ -j * 2.26 A

Step 5: Calculate the total current (I_total) supplied by the voltage source:

Since the resistor and capacitor currents are out of phase, we need to sum them as complex numbers:

I_total = I_R + I_C = 6 A + (-j * 2.26 A) ≈ 6 – j * 2.26 A

Step 6: Calculate the magnitude of the total current:

|I_total| = √(Real(I_total)² + Imag(I_total)²) = √(6² + (-2.26)²) ≈ 6.34 A

In conclusion, the total current (I_total) supplied by the voltage source is approximately 6.34 A, the current through the resistor (I_R) is 6 A, and the magnitude of the current through the capacitor (I_C) is approximately 2.26 A. Note that the sum of the magnitudes of the individual currents does not equal the total current magnitude due to the phase differences between the currents through the resistor and the capacitor.

The primary purpose of this project is to help the public to learn some exciting and important information about electricity and magnetism.