Analyse DC circuits containing resistors in series and parallel using Kirchhoff's current and voltage laws, including problems combining series and parallel branches and including electrical power and energy ($P = VI$, $W = Pt$)

What this dot point is asking

VCAA wants you to analyse DC circuits with resistors in series and parallel using Kirchhoff's laws and to compute power dissipation in each component.

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.

Series circuits

Resistors in a single line; same current through each.

• IMATH_1
• Same $I$ in all resistors.
• IMATH_3 . Total voltage: $V = V_1 + V_2 + V_3 + \cdots$.

Parallel circuits

Resistors share two common nodes; same voltage across each.

• IMATH_5
• Same $V$ across all branches.
• IMATH_7 . 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 $P = VI = I^2 R = V^2 /R$.