By Selma Leathem
Updated Mar 24, 2022

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In complex electrical networks, reducing the layout to series‑parallel combinations simplifies the determination of key parameters such as resistance, voltage, and current. While series connections keep all current on a single path, parallel circuits split the current among multiple branches, ensuring that the lowest‑resistance path carries the most current. This behavior allows us to compute each resistor’s value and the overall equivalent resistance using straightforward formulas.

Step 1 – Measure Voltage and Current

Obtain the supply voltage and the current through each resistor. In a parallel network, the voltage drop is identical across every resistor, so measuring once is sufficient. However, each branch will carry a different current, so you must record the current I_j (j = 1…n) for all n resistors.

Step 2 – Compute Individual Resistances

Use Ohm’s law to calculate the resistance of each element: R_j = V / I_j. For example, with a 9‑V supply and currents I₁ = 3 A, I₂ = 6 A, and I₃ = 2 A, the resistances are R₁ = 3 Ω, R₂ = 1.5 Ω, and R₃ = 4.5 Ω.

Step 3 – Determine the Equivalent Resistance

Replacing the parallel network with a single resistor simplifies subsequent analysis. The equivalent resistance, R_eq, is found by summing the reciprocals of the individual resistances:

1 / R_eq = 1 / R₁ + 1 / R₂ + … + 1 / R_n

Because the parallel arrangement offers multiple conduction paths, R_eq is always smaller than any single R_j. In the example above, R_eq ≈ 0.82 Ω. This single resistor, under the same 9‑V supply, would carry the total current I_total = I₁ + I₂ + I₃ = 11 A.

TL;DR (Too Long; Didn’t Read)

For two resistors in parallel, the currents are inversely proportional to their resistances. The relation V = I₁ R₁ = I₂ R₂ can be rearranged to R₁ / R₂ = I₂ / I₁.