How to solve basic circuit problems, step by step

Circuit questions look busy but follow a pattern. Reduce the circuit to a single resistor, find the total current, then work back out to find what happens in each part. Here is the method.

Series: one path for the current

When components are in a line, the same current flows through each one. The resistances simply add: R_total = R1 + R2 + R3. The supply voltage is shared out between them in proportion to their resistance, so the biggest resistor takes the biggest share.

Parallel: several paths

When components sit side by side, each gets the full supply voltage and the current splits between the branches. The total resistance is always less than the smallest branch, because adding a path makes it easier for current to flow. For two resistors: R_total = (R1 × R2) ÷ (R1 + R2). In general, 1/R_total = 1/R1 + 1/R2 + 1/R3.

The method

1. Redraw the circuit if you need to, and label every resistor.
2. Combine series groups and parallel groups until only one resistor is left.
3. Find the total current with I = V ÷ R_total.
4. Work back out. For a series part, use V = IR for each resistor. For a parallel part, the voltage across each branch is the same, so I = V ÷ R for each branch.
5. Check. Branch currents should sum to the current that entered. Voltage drops around a loop should sum to the supply.

A worked example

A 12 V supply feeds a 4 ohm resistor in series with a pair of resistors, 6 ohm and 3 ohm, in parallel.

• The parallel pair: (6 × 3) ÷ (6 + 3) = 2 ohm.
• Total resistance: 4 + 2 = 6 ohm.
• Total current: 12 ÷ 6 = 2 A.
• Voltage across the 4 ohm resistor: 2 × 4 = 8 V. That leaves 4 V across the parallel pair.
• Branch currents: 4 ÷ 6 = 0.667 A and 4 ÷ 3 = 1.333 A, which add to 2 A as they should.

Common mistakes

Using the supply voltage across a single resistor in a series chain, forgetting to flip the reciprocal in the parallel formula, and mixing milliamps with amps. The last one is the sneakiest: always convert to amps before multiplying.