Series and parallel circuits and Kirchhoff's laws (QCE Physics Unit 1)

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.

Examples in context

Example 1. A Cairns suburban distribution board feeds six $40 \text{ W}$ LED downlights wired in parallel across $240 \text{ V}$. Each draws $I = P/V = 0.167 \text{ A}$ for a total current of $1.0 \text{ A}$ via Kirchhoff's junction rule, satisfying the $10 \text{ A}$ lighting circuit breaker. If one lamp fails open-circuit, the other five remain on (parallel branches independent). Wiring the same six lamps in series would push each lamp to one-sixth of $240 \text{ V}$ ($40 \text{ V}$), well under rating, and any single failure would dark the whole bank, which is why residential lighting in Queensland is universally parallel.

Example 2. A Bremer River bridge $24 \text{ V}$ navigation light is fed via a $35 \text{ m}$ submarine cable whose total resistance is $1.2 \text{ ohms}$. The cable resistance and lamp ($R = 12 \text{ ohms}$) are in series, so $I = 24/(12+1.2) = 1.82 \text{ A}$, with $V_{cable} = IR_{cable} = 2.18 \text{ V}$ dropped before the lamp. Engineers either upsize the cable or compensate with a higher source voltage to deliver the rated $24 \text{ V}$ at the lamp terminals, a worked Kirchhoff's-voltage-law calculation common in QCAA EA Unit 1 stems.

Try this

Q1. State Kirchhoff's current law and voltage law. [2 marks]

• Cue. $\sum I_{in} = \sum I_{out}$ at a junction; $\sum V = 0$ around a loop.

Q2. Three resistors of $6.0$, $3.0$ and $2.0 \text{ ohms}$ are connected in parallel across a $12 \text{ V}$ supply. Calculate $R_{eq}$, the total current, and the current in the $3.0 \text{ ohm}$ branch. [4 marks]

• Cue. $1/R_{eq} = 1/6 + 1/3 + 1/2 = 1$, so $R_{eq} = 1.0 \text{ ohms}$; $I_{tot} = 12 \text{ A}$; $I_{3} = 4 \text{ A}$.

Q3. A circuit has a $12 \text{ V}$ EMF with $4.0 \text{ ohms}$ in series with a parallel pair $6.0 \text{ ohms}$ and $3.0 \text{ ohms}$. (a) Determine the equivalent resistance. (b) Calculate the current from the EMF and the voltage across the parallel pair. (c) Verify Kirchhoff's voltage law around one loop. [3+3+2 marks; ISMG: Analysis and interpretation]

• Cue. (a) $R_{eq} = 4 + 2 = 6.0 \text{ ohms}$; (b) $I = 2.0 \text{ A}$, $V_{par} = 4.0 \text{ V}$; (c) $12 - 8 - 4 = 0$.