DC circuits (Ohm's law, series & parallel)
The idea
A circuit is a closed loop that lets charge flow, and Ohm's law is its central relation: V = IR, where resistance R (in ohms, Ω) measures how strongly a component opposes current. Given any two of voltage, current, and resistance, you can find the third. Real circuits combine resistors two basic ways: in series, single file along one path, or in parallel, side by side across the same two points.
Each arrangement has its own logic. Series resistors share one current, their voltages add, and the total resistance is the sum: R = R₁ + R₂. Parallel resistors share one voltage, their currents add, and the combination resists LESS than any branch alone: 1/R = 1/R₁ + 1/R₂, because extra branches are extra lanes for traffic. Two conservation rules anchor every analysis: voltage drops around any loop sum to the battery's voltage, and current into a junction equals current out.
The misconception that causes the most wrong answers is assuming the battery's full voltage appears across every component. In a series chain the voltage divides among the resistors in proportion to their resistance; only parallel branches connected directly across the battery see its full voltage. Always ask which two points a component actually spans.
Worked example
A 12 V battery drives a circuit with a 4.0 Ω resistor in series with a parallel pair: 6.0 Ω and 3.0 Ω. Find the total current from the battery and the current through each parallel resistor.
- Collapse the parallel pair first: 1/R = 1/6.0 + 1/3.0 = 1/2.0, so the pair behaves like a single 2.0 Ω resistor — smaller than either branch, as parallel combinations must be.
- Add the series resistance: total R = 4.0 + 2.0 = 6.0 Ω across the 12 V battery.
- Apply Ohm's law to the whole circuit: I = V/R = 12/6.0 = 2.0 A flowing from the battery and through the 4.0 Ω resistor.
- Find how the voltage divides: the 4.0 Ω resistor drops V = IR = 2.0 × 4.0 = 8.0 V, leaving 12 − 8.0 = 4.0 V across the parallel section.
- Apply Ohm's law inside each branch, which shares that 4.0 V: the 6.0 Ω branch carries 4.0/6.0 ≈ 0.67 A and the 3.0 Ω branch carries 4.0/3.0 ≈ 1.33 A.
- Verify with the junction rule: 0.67 + 1.33 = 2.0 A, matching the current that arrived — charge is conserved, and the smaller resistance correctly took the larger share.
Answer. The battery supplies 2.0 A; the parallel section splits it into about 0.67 A through the 6.0 Ω resistor and 1.33 A through the 3.0 Ω resistor.
Check your understanding
- Why does adding a resistor in parallel decrease the total resistance when adding one in series increases it?
- How do the loop and junction rules express conservation of energy and conservation of charge in a circuit?
- What systematic order of moves would you teach a friend for reducing any mixed series-parallel network?
- Why are the lights in a house wired in parallel, and what would daily life look like if they were wired in series?
Build the foundations first
DC circuits (Ohm's law, series & parallel) builds on these concepts. If any feel shaky, start there.