In a parallel circuit, the current has multiple paths to travel, just like water flowing through several pipes that join back together. The total resistance is always lower than any single resistor, which means the circuit is easier for the current to pass through.
Analogy 🚗: Imagine two roads (resistors) leading to the same destination. If you can choose either road, the overall “traffic resistance” is reduced because traffic can split. The more roads you add, the easier it is to get through.
The combined resistance \(R{\text{eq}}\) of two resistors \(R1\) and \(R_2\) in parallel is:
\( \displaystyle \frac{1}{R{\text{eq}}} = \frac{1}{R1} + \frac{1}{R_2} \)
or, solved for \(R_{\text{eq}}\):
\( \displaystyle R{\text{eq}} = \frac{R1 R2}{R1 + R_2} \)
\( R{\text{eq}} = \dfrac{R1 R2}{R1 + R_2} \)
Calculate the equivalent resistance of \(R1 = 8\,\Omega\) and \(R2 = 12\,\Omega\) in parallel.
| Step | Calculation |
|---|---|
| 1. Multiply | \(8 \times 12 = 96\) |
| 2. Add | \(8 + 12 = 20\) |
| 3. Divide | \(96 \div 20 = 4.8\) |
| Result | \(R_{\text{eq}} = 4.8\,\Omega\) |
Parallel circuits let current split, which reduces the overall resistance. Use the reciprocal formula or the product‑over‑sum shortcut to find \(R_{\text{eq}}\).