Parallel Circuits: Current, Voltage, and Resistance

In a parallel circuit the components are connected across the same two points, so each branch has the same potential difference, the total current is the sum of the branch currents (I = I1 + I2 + …), and the total resistance is found from 1/R = 1/R1 + 1/R2 + …, which is always smaller than the smallest branch resistance. In brief: adding a branch in parallel lowers the total resistance and increases the total current drawn from the supply. If one branch fails, the others continue to work, which is why household appliances are wired in parallel. For two resistors, R = R1R2 ÷ (R1 + R2).

The sockets in a house are connected in parallel, so each appliance receives the full mains voltage and works independently. This guide states the rules for parallel circuits, explains why the total resistance decreases, compares series and parallel circuits in a table, works through numerical examples and corrects common mistakes.

What are the rules for parallel circuits?

Why does the resistance decrease?

Each additional branch provides another path for the current, like adding extra lanes to a road. The conductance G = 1/R adds: Gtotal = G1 + G2 + …. Equivalently, the effective cross-sectional area of the conductor is increased, so the resistance falls. The sum of the currents at a junction equals the current entering (conservation of charge, Kirchhoff’s current law). The p.d. is equal across the branches because they connect the same two points, and the energy per coulomb is the same whichever path the charge takes. The total resistance is smaller than the smallest resistance in the group: for example, 10 Ω in parallel with 1,000 Ω is 9.9 Ω.

Two resistors in parallel across a 12 volt battery with branch currents 12 V total 1.0 A 20 Ω 0.60 A 30 Ω 0.40 A 12 V across each branch
The worked example: 20 Ω and 30 Ω resistors in parallel across 12 V. Each branch has the full 12 V. The current splits as 0.60 A and 0.40 A, which add up to the 1.0 A drawn from the battery, and the smaller resistor carries the larger current.

Series and parallel circuits compared

Series and parallel connections
Quantity Series Parallel
CurrentSame in allSplits among branches
VoltageDivided among componentsSame across all branches
Total resistanceR1 + R2 + …1/R = 1/R1 + 1/R2 + …
Effect of adding a componentResistance up, current downResistance down, current up
One component failsAll stopOthers continue

Worked example: two resistors in parallel

A 20 Ω resistor and a 30 Ω resistor are connected in parallel to a 12 V supply.

  1. Total resistance: R = (20 × 30) ÷ (20 + 30) = 600 ÷ 50 = 12 Ω. Check: 1/20 + 1/30 = 0.050 + 0.0333 = 0.0833 = 1/12 ✓.
  2. Voltage: 12 V across each.
  3. Branch currents: I1 = 12 ÷ 20 = 0.60 A; I2 = 12 ÷ 30 = 0.40 A.
  4. Total current: 0.60 + 0.40 = 1.0 A, which agrees with 12 ÷ 12 = 1.0 A ✓.
  5. Powers: P1 = 12 × 0.60 = 7.2 W; P2 = 12 × 0.40 = 4.8 W. Total = 12 W.
  6. Adding a third branch of 60 Ω: 1/R = 0.050 + 0.0333 + 0.0167 = 0.100, so R = 10 Ω, and the total current rises to 1.2 A. The first two branches are unchanged (0.60 A and 0.40 A).

How were parallel circuits used?

Thomas Edison’s electric lighting system in the 1880s used parallel wiring so that each lamp could be switched on and off independently, in contrast with the early arc lamp systems in series. Parallel wiring is the standard for home circuits worldwide. Kirchhoff’s current law (1845) gives the rules for the analysis of any network.

Which mistakes are common?

Mistake 1: Adding the resistances directly

For parallel resistors, add the reciprocals.

Mistake 2: Forgetting to invert at the end

After adding 1/R values, take the reciprocal to get R.

Mistake 3: Dividing the voltage among branches

The voltage is the same across each branch. It is the current that is divided.

Mistake 4: Expecting the total resistance to be greater than each branch

It is always less than the smallest branch resistance.