🪜 Parallel Circuits — Multiple Paths for Current

In electrical engineering, a parallel circuit is defined by its ability to provide multiple paths for current to flow. Unlike series circuits, where there is only one route for electrons, parallel circuits allow the current to split and travel through different branches simultaneously. This configuration is the foundation of almost all residential and industrial electrical installations.

In a parallel circuit, all components are connected across the same two points (nodes). Because they share the same starting and ending nodes, the voltage across every branch in the circuit is identical. If you connect a 30V battery to three resistors in parallel, every single resistor will experience exactly 30V.

While the voltage remains constant, the current behaves differently. According to Kirchhoff’s Current Law (KCL), the total current entering the parallel network must equal the sum of the currents flowing through each individual branch. If one branch has very low resistance, it will draw a large amount of current, while a high-resistance branch will draw very little.

1. Total Resistance Behavior

بما أن هناك مسارات متعددة تفتح الأبواب لمرور التيار، فإن المقاومة الكلية لدارة التوازي تكون دائماً أصغر من قيمة أصغر مقاومة منفردة فيها.

• Water pipe analogy: If you have a water pipe with a diameter of 10 inches and you install another pipe with a diameter of 10 inches next to it, this will allow twice the amount of water to flow compared to the single pipe; this means that the total resistance of the two pipes together is less than the resistance of either one alone.
• Equal resistors case: If two equal resistors are connected in parallel, they will together carry twice the current that a single resistor would carry alone; thus, their total resistance becomes half the value of either one. For example, two resistors of 100 ohms in parallel give a total resistance of 50 ohms.

Case of unequal resistances:

The rule remains constant: the total resistance will always be less than the smallest resistance present.

• Water example: Imagine a pipe with a diameter of 10 inches that can carry 1,000 gallons per hour, and you installed a smaller pipe next to it with a diameter of 5 inches that can carry 250 gallons per hour. Together, the two pipes will carry 1,250 gallons per hour; this means that their combined resistance has decreased, allowing for a greater flow compared to the larger pipe alone (1,250 gallons versus 1,000 gallons).
• The electrical application: If a resistor of 5 ohms is connected in parallel with a resistor of 10 ohms, the total resistance of the circuit will definitely be less than 5 ohms.

2. Mathematical Methods

First: The "Product-over-Sum" method

This method is designed to calculate only two resistors in parallel, and is expressed by the following equation:

• Numerical application (5 ohms and 10 ohms):
  Note that the result ($3.33\ \Omega$) is indeed slightly less than the smallest individual resistance ($5\ \Omega$).
• Application of the method on three resistors: If you have three resistors (for example: two resistors of 100 ohms and one resistor of 50 ohms), apply the formula to any two resistors first, then take the result and apply the formula again with the third resistor:

1. We calculate the two resistors (100 ohms and 100 ohms): $(100 \times 100) / (100 + 100) = 50\ \Omega$.
2. We take the result (50 ohms) and calculate it in parallel with the remaining third resistor (50 ohms): $(50 \times 5) / (50 + 50) = 25\ \Omega$.
3. The final result is 25 ohms.

Secondly: "Reciprocal Method" (مقلوب المجموع)

This general method is valid for calculating any number of resistors connected in parallel:

$$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$$

Or it can be written directly to calculate the total value:

$$R_T = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots}$$

• Numerical application (10 ohms, 20 ohms, 30 ohms):

1. We take the reciprocal of each resistance (conductance):$$\frac{1}{10} = 0.1, \quad \frac{1}{20} = 0.05, \quad \frac{1}{30} = 0.0333$$
2. We gather these tools together:
   $$0.1 + 0.05 + 0.0333 = 0.183$$
3. We take the reciprocal of the resulting sum to obtain the total resistance:$$R_T = \frac{1}{0.183} = 5.46\ \Omega$$

🛠️ Practical Exercise (Test Yourself):

Calculate the total resistance for the following groups connected in parallel:

1. 5 ohms، 10 ohms= _________
2. 33 ohms، 47 ohms= _________
3. 5 ohms، 10 ohms، 15 ohms= _________
4. 33 ohms، 47 ohms، 100 ohms= _________
5. 50 ohms، 75 ohms، 80 ohms= _________
6. 200 ohms، 200 ohms، 100 ohms= _________

💡 Smart Tip for Assistance: Always remember that when two resistances in parallel have the same value, their total is simply half; apply this tip to the sixth and final problem: two resistors of 200 ohms in parallel give 100 ohms; now this total (100 ohms) is in parallel with the last resistor (100 ohms), resulting in a final total of 50 ohms simply and without complex calculations.