Series and Parallel Resistance
Series (Ω)
200
Parallel (Ω)
50
How it works
Resistors in series add directly: R_total = R₁ + R₂ + R₃... Resistors in parallel combine as the reciprocal of the sum of reciprocals: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃... For two parallel resistors, the product-over-sum shortcut applies: R = (R₁ × R₂) / (R₁ + R₂).
**Why parallel resistance is always less than the smallest resistor** Adding a parallel path always provides more routes for current to flow, reducing total resistance. Two 10Ω resistors in parallel give 5Ω. A 10Ω and 1,000Ω in parallel give 9.9Ω — the small resistor dominates. This asymmetry has practical implications: a single bad solder joint (high resistance path) parallel to a good one barely affects the circuit; a good component parallel to a short circuit fails the entire branch.
**Series-parallel combinations** Real circuits combine both topologies. Solve by simplifying: replace parallel groups with their equivalent resistance, then add series elements. Work from the innermost nested combination outward. For complex networks (ladder networks, Wheatstone bridges), apply Kirchhoff's laws instead.
**Current sharing in parallel branches** Parallel resistors share voltage equally; current splits inversely proportional to resistance. A 1Ω and 2Ω in parallel across 6V: the 1Ω draws 6A, the 2Ω draws 3A, total 9A. Current through each branch is V/R for that branch.
**Tolerance effects** Real resistors have tolerances (±1%, ±5%). Series chains accumulate tolerance. If a precise ratio matters (voltage divider, precision measurement), use matched resistors or resistor networks with correlated tolerances.
Frequently Asked Questions
- Work from the innermost combinations outward. Identify groups of pure series or pure parallel resistors, replace each with its equivalent, then repeat. Example: R1 series with (R2 parallel R3). Step 1: R_parallel = (R2 × R3)/(R2 + R3). Step 2: R_total = R1 + R_parallel. Draw the circuit, redraw after each simplification. For circuits with bridges or loops (Wheatstone bridge), simplification fails — use Kirchhoff's voltage and current laws or mesh/nodal analysis.
- Each parallel branch provides an additional current path. Total current increases for the same voltage, meaning the combined resistance (V/I_total) must be lower. Mathematically: 1/R_total = 1/R1 + 1/R2 — adding any positive term to 1/R1 makes the sum larger, so R_total must be smaller than R1. In the extreme: parallel a resistor with a short circuit (R=0) and total resistance is zero — all current flows through the short regardless of the resistor's value.
- Two resistors in series form a voltage divider: V_out = V_in × R2/(R1+R2), where V_out is measured across R2. A 10kΩ and 20kΩ divider on 12V: V_out = 12 × 20/(10+20) = 8V. Voltage dividers are used for level shifting (reducing 5V logic to 3.3V), biasing transistors, and ADC input scaling. Important limitation: voltage dividers are only accurate when the load resistance is much larger than R2 (typically 10× or more) — a heavy load changes the effective R2 and shifts the output voltage.
- For a 5% tolerance divider (two 5% resistors), worst case the ratio R2/(R1+R2) can vary by roughly ±7%. A divider set for 3.3V from 5V could output anywhere from 3.07V to 3.53V — potentially outside an input's acceptable range. For precision analog signals, use 0.1% or 1% resistors. For digital level shifting (high or low threshold detection), 5% is usually acceptable. Matching resistors from the same production batch reduces effective tolerance even if individual tolerance is wide.