Analyse series and parallel resistor combinations using Kirchhoff's current and voltage laws, including problems with mixed series and parallel branches

What this dot point is asking

QCAA wants you to analyse DC circuits made of resistors in series and parallel combinations. Two foundational laws apply: Kirchhoff's current law (charge conservation at junctions) and Kirchhoff's voltage law (energy conservation around loops). These let you derive every other circuit relationship.

Kirchhoff's current law (KCL)

The sum of currents into a junction equals the sum of currents out:

This is conservation of charge. Charge does not pile up at a node.

Kirchhoff's voltage law (KVL)

The sum of potential differences around any closed loop is zero:

This is conservation of energy. The total energy gained from sources equals the total energy dissipated in resistors around the loop.

Series circuits

Resistors are in series when they form a single line; the same current flows through each.

• Equivalent resistance: $R_{\text{series}} = R_1 + R_2 + R_3 + \cdots$
• Same current through every resistor: $I_1 = I_2 = I_3 = \cdots$
• Voltage divides in proportion to resistance: $V_i = I R_i$.
• Total voltage: $V_{\text{total}} = V_1 + V_2 + V_3 + \cdots$ (by KVL).

A single break in a series circuit (one bulb out) stops current in the whole circuit; this is why old Christmas-tree lights wired in series all went dark when one bulb blew.

Parallel circuits

Resistors are in parallel when they share two common nodes; the same voltage is across each.

• Equivalent resistance: $\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$
• Same voltage across every branch: $V_1 = V_2 = V_3 = \cdots$
• Current divides in inverse proportion to resistance: $I_i = V / R_i$.
• Total current: $I_{\text{total}} = I_1 + I_2 + I_3 + \cdots$ (by KCL).

The parallel resistance is less than the smallest of the individual resistances. Adding more parallel paths drops the equivalent resistance and increases the total current at fixed voltage.

For two parallel resistors specifically, $R_p = R_1 R_2 / (R_1 + R_2)$ (the product over the sum).

Mixed circuits

Most real circuits combine series and parallel sections. Reduce them step by step:

1. Identify a parallel block and replace with $R_p$.
2. Combine series resistors into $R_s$.
3. Repeat until one equivalent resistance remains.
4. Use $V = IR$ on the equivalent to get total current.
5. Work backward, applying $V_i = I R_i$ for series sections and $I_i = V / R_i$ for parallel branches.

Worked example

Three identical $6\,\Omega$ resistors are connected to a $12$ V battery, all in parallel.

Parallel total: $1/R_p = 3 \times (1/6) = 1/2$, so $R_p = 2.0\,\Omega$.

Total current: $I = V/R_p = 12 / 2.0 = 6.0$ A.

Current through each branch: $I_i = V / R_i = 12 / 6 = 2.0$ A (matches the KCL check $3 \times 2.0 = 6.0$ A).

Common traps

Adding parallel resistors directly. $1/R$ values add, not $R$ values. Two $6\,\Omega$ resistors in parallel give $3\,\Omega$, not $12\,\Omega$.

Forgetting that parallel branches have the same voltage. This is the most useful single fact: at any parallel node, $V$ is common.

Treating a junction as having different currents on the same wire. Within a single segment between junctions, the current is the same at every point.

Applying KVL with the wrong signs. Going around a loop, voltage drops across resistors in the direction of current flow are negative; battery EMFs going from negative to positive terminal are positive. Pick a direction and stick with it.

In one sentence

Series resistors share the same current and add directly ($R_s = R_1 + R_2 + \cdots$), parallel resistors share the same voltage and combine reciprocally ($1/R_p = 1/R_1 + 1/R_2 + \cdots$), and Kirchhoff's current law (junction rule) and voltage law (loop rule) let you solve any DC network by working from the equivalent resistance back to the individual branches.