State that the combined resistance of resistors in parallel is less than that of any single resistor in that circuit

4.3.2 Series and Parallel Circuits ⚡️

Objective 🎯

State that the combined resistance of resistors in parallel is less than that of any single resistor in that circuit.

Key Concepts 🔌

• Series circuits: resistors connected end‑to‑end. Total resistance: \$R{\text{total}} = R1 + R2 + \dots + Rn\$.
• Parallel circuits: resistors connected across the same two points. Total resistance: \$\displaystyle \frac{1}{R{\text{total}}} = \frac{1}{R1} + \frac{1}{R2} + \dots + \frac{1}{Rn}\$.
• In parallel, \$R{\text{total}} < \min(Ri)\$.

Why Parallel Lowers Resistance 🚀

Think of water flowing through pipes. If you have one pipe (one resistor), the water can only flow through that path. Add another pipe side by side, and the water has more routes, so it flows more easily and the overall resistance to flow drops. The same idea applies to electric current in parallel circuits.

Example Calculation 🔢

1. Two resistors: \$R1 = 10\,\Omega\$, \$R2 = 10\,\Omega\$ in parallel.
2. Compute: \$\displaystyle \frac{1}{R_{\text{total}}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}\$.
3. Thus \$R_{\text{total}} = 5\,\Omega\$.
4. Notice \$5\,\Omega < 10\,\Omega\$.

Quick Check 🧪

Use the formula: \$R{\text{total}} = \frac{R1 R2}{R1 + R2}\$ for two resistors. If \$R1 = 15\,\Omega\$ and \$R2 = 30\,\Omega\$, then \$R{\text{total}} = \frac{15 \times 30}{15 + 30} = \frac{450}{45} = 10\,\Omega\$.

Table of Parallel Resistances 📊

Takeaway 📌

In a parallel circuit, the total resistance is always less than the smallest individual resistor. This is because the current has multiple paths to travel, reducing the overall opposition to flow.