Series and parallel DC circuits (VCE Physics Unit 1)

What this dot point is asking

The Unit 1 key knowledge asks you to model resistance in series and parallel circuits, work with equivalent resistance and voltage dividers, and calculate power using $P = VI$ (quantitative analysis is restricted to $I = V/R$ and $P = VI$). Kirchhoff's laws are a useful way of thinking about current and voltage, but they are an extension beyond the study design, so do not expect to be asked to name or apply them as laws.

Kirchhoff's current law (KCL)

Sum of currents into a junction equals sum out. Charge conservation at every node.

Kirchhoff's voltage law (KVL)

Sum of potential differences around any closed loop is zero. Energy conservation around any loop.

Resistors in a single line; same current through each.

• $R_{\rm series} = R_1 + R_2 + R_3 + \cdots$
• Same $I$ in all resistors.
• $V_i = I R_i$. Total voltage: $V = V_1 + V_2 + V_3 + \cdots$.

Parallel circuits

Resistors share two common nodes; same voltage across each.

• $\dfrac{1}{R_{\rm parallel}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \cdots$
• Same $V$ across all branches.
• $I_i = V/R_i$. Total current: $I = I_1 + I_2 + I_3 + \cdots$.

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

The parallel resistance is always less than the smallest individual resistance.

Mixed circuits

Reduce step by step:

1. Find parallel blocks; replace with $R_p$.
2. Combine series resistors.
3. Repeat until one $R_{\rm eq}$ remains.
4. Use $V = IR$ on $R_{\rm eq}$ to get total current.
5. Work backward: in series sections use $V_i = I R_i$; in parallel sections use $I_i = V/R_i$.

Electrical power and energy

For a resistor:

SI unit: watt (W).

Electrical energy: $W = Pt$ (in joules). For household billing: $1$ kWh $= 3.6 \times 10^6$ J.

VCAA exam style

VCE Year 11 SAC tasks typically include:

• Compute $R_{\rm eq}$ for a circuit with $3$-$5$ resistors mixed series and parallel.
• Find current through and voltage across each component.
• Compute power dissipated in each resistor and verify the total matches the battery's $P = VI$.

Common traps

Adding parallel resistors directly

$1/R$ values add, not $R$ values.

Forgetting same-voltage rule in parallel

At any parallel node, $V$ is common.

Mixing series and parallel in mixed circuits

Identify the topology first. Two resistors in series across a battery are not in parallel with each other.

Confusing power and energy

A $100$ W device left on for an hour uses $360\,000$ J, not $100$ J.

In one sentence

Series resistors share the same current and add directly ($R_s = R_1 + R_2$); parallel resistors share the same voltage and combine reciprocally ($1/R_p = 1/R_1 + 1/R_2$); Kirchhoff's current and voltage laws ensure charge conservation at junctions and energy conservation around loops, and power dissipation in any resistor is $P = VI = I^2 R = V^2 /R$.

Examples in context

Example 1. Melbourne tram catenary feeder. Yarra Trams runs $600$ V DC traction through overhead catenary fed from rectifier substations. Consider a $200$ m run of copper feeder with resistance $0.05$ $\Omega$ supplying a tram drawing $400$ A on acceleration. The voltage drop along the feeder is $V = IR = 400 \times 0.05 = 20$ V, so the tram sees $580$ V at its pantograph. Power dissipated in the feeder is $P = I^2 R = 400^2 \times 0.05 = 8000$ W. Engineers run two parallel feeders so the effective resistance halves to $0.025$ $\Omega$, dropping losses to $4000$ W and ensuring trams climbing the Bourke Street rise still get adequate voltage.

Example 2. Grid battery string configuration (simplified model). Large grid batteries such as the Hornsdale Power Reserve (owned by Neoen, rated at $150$ MW) combine cells in series to reach a high DC bus voltage, then place the strings in parallel to deliver the required power. The figures below are a made-up model, not published specifications: a bus voltage of $1500$ V. A single LFP cell at $3.2$ V needs approximately $469$ cells in series for $1500$ V, then thousands of such strings in parallel to share the current. If a string of resistance $0.4$ $\Omega$ delivers $130$ A, the series voltage drop is $52$ V, so balance circuits trim the parallel currents. Because the strings are in parallel, every string sees the same bus voltage despite slightly different internal resistances.

Try this

Q1. Two resistors of $4$ $\Omega$ and $12$ $\Omega$ are connected in parallel. Calculate the equivalent resistance. [2 marks]

• Cue. $1/R = 1/4 + 1/12 = 4/12$, so $R = 3$ $\Omega$.

Q2. A $24$ V battery is connected in series with a $2$ $\Omega$ resistor and a parallel combination of $6$ $\Omega$ and $3$ $\Omega$. Calculate the total current and the power dissipated by the $3$ $\Omega$ resistor. [5 marks]

• Cue. Parallel pair is $2$ $\Omega$; total $4$ $\Omega$; $I = 6$ A; parallel pair voltage $12$ V; $P_{3} = 12^2/3 = 48$ W.

Q3. A tram feeder model has a $600$ V supply, $0.05$ $\Omega$ feeder, and a tram drawing $400$ A. (a) Determine the voltage at the tram. (b) Calculate the power lost in the feeder. (c) Explain why running two feeders in parallel reduces losses. [2+2+2 marks]

• Cue. (a) $V = 600 - 20 = 580$ V. (b) $I^2 R = 8000$ W. (c) Parallel halves resistance, halving $I^2 R$ losses at fixed current.