Mixed Circuit Analysis
Sydney Metro trains (NSW Transport, 2019): 1,500 V DC overhead supply. Each train has 4 traction motors wired in series (375 V each) to form one group; two such groups are then wired in parallel, giving a total of 8 motors with peak power 1.68 MW. The series-parallel topology lets engineers control startup current (series resistors) while maintaining redundancy (parallel groups).
A 4 Ω resistor is connected in series with a parallel combination of two 12 Ω resistors. The whole circuit is powered by a 12 V battery.
Predict 1: What is the equivalent resistance of the two 12 Ω resistors in parallel?
Predict 2: What is the total current from the battery?
When solving a mixed (series-parallel) circuit, which group should you simplify first?
Know
- A systematic 4-step strategy for mixed circuits
- How to identify series and parallel groups within a circuit
- Equivalent resistance reduces a group to a single value
Understand
- Why the order of simplification matters
- How to track voltages and currents through a multi-step reduction
- That all circuit rules (V, I, R) must be consistent at every step
Can Do
- Reduce a mixed circuit to find total resistance
- Find the current through and voltage across every component
- Verify answers using Kirchhoff's laws
Core Content
Look at a phone charger — it contains a transformer, then a bridge rectifier (four diodes), then a smoothing capacitor, then a voltage regulator. If you drew the circuit out, you would see some components side-by-side (parallel), others end-to-end (series), layered in groups. No single "series rule" or "parallel rule" solves the whole thing. You have to identify the innermost group, replace it with one equivalent resistor, then repeat — peeling the circuit like an onion until only one loop remains. Here is the four-step strategy.
- Identify all series groups and all parallel groups.
- Simplify innermost groups first — replace each group with its equivalent resistance.
- Repeat until one equivalent resistance remains.
- Work back — use $V = IR$ at each stage to find individual voltages and currents.
$R_1 = 4\ \Omega$ in series with ($R_2 = 12\ \Omega$ parallel $R_3 = 12\ \Omega$), $V_{\text{supply}} = 12\ \text{V}$.
- Parallel group: $R_{2-3} = (12 \times 12)/(12 + 12) = 6.0\ \Omega$
- Total: $R_{\text{total}} = R_1 + R_{2-3} = 4 + 6 = 10\ \Omega$
- Total current: $I = 12/10 = 1.2\ \text{A}$
- Voltage across series $R_1$: $V_1 = 1.2 \times 4 = 4.8\ \text{V}$
- Voltage across parallel group: $V_{2-3} = 12 - 4.8 = 7.2\ \text{V}$
- Branch currents: $I_2 = 7.2/12 = 0.60\ \text{A}$; $I_3 = 7.2/12 = 0.60\ \text{A}$ ✓ (sum = 1.2 A)
To solve a mixed circuit: identify series and parallel groups, reduce the innermost parallel group first (product-over-sum), repeat until one equivalent resistance remains, then work back using $V = IR$ at each stage to find individual voltages and currents.
Pause — write the highlighted strategy into your book before the check below.
In a mixed circuit, you must solve all resistors simultaneously.
After finding the total current, you use $V = IR$ to work back and find each component's voltage and current.
A 10 Ω resistor ($R_1$) is in series with a parallel combination of $R_2 = 20\ \Omega$ and $R_3 = 30\ \Omega$. The supply voltage is 25 V.
- Find the equivalent resistance of the parallel group ($R_2 \| R_3$).
- Find the total resistance of the circuit.
- Find the total current and the voltage across the parallel group.
- Find the current through $R_2$ and $R_3$.
In a circuit with $R_1 = 6\ \Omega$ in series with [$R_2 = 12\ \Omega$ parallel $R_3 = 6\ \Omega$], the equivalent resistance of the parallel section is:
$R_1 = 5\ \Omega$ is in series with a parallel combination of $R_2 = 10\ \Omega$ and $R_3 = 15\ \Omega$. Supply = 20 V.
- Find $R_{\text{total}}$ and $I_{\text{total}}$.
- Find the voltage across $R_1$ and the voltage across the parallel section.
- Find $I_2$ and $I_3$ and verify they sum to $I_{\text{total}}$.
- Verify total power using $P = IV$ for each component and for the supply.
Three of these statements about mixed circuits are correct. Pick the odd one out.
A 6 Ω resistor is in series with a parallel pair of 12 Ω resistors. The parallel pair has an equivalent resistance of _____ Ω, making the total circuit resistance _____ Ω. (Write the total resistance only.)
In the circuit from Activity 1 ($R_1 = 10\ \Omega$ series, $R_2 = 20\ \Omega \| R_3 = 30\ \Omega$, $V = 25\ \text{V}$), the voltage across the parallel group is approximately:
ApplyBand 3(3 marks) 3. $R_1 = 8\ \Omega$ is in series with $R_2 = 24\ \Omega \| R_3 = 12\ \Omega$, connected to a 16 V supply. Calculate (a) total resistance, (b) total current, and (c) voltage across the parallel section.
AnalyseBand 4(3 marks) 4. In the circuit from Q3, calculate (a) the current through $R_2$ and $R_3$, (b) verify that branch currents add to the total, and (c) calculate the power dissipated by $R_1$.
EvaluateBand 6(4 marks) 5. A student proposes that because the total voltage is conserved in any closed loop, you can simply add all resistances in a mixed circuit regardless of their configuration. Evaluate this claim, using the circuit $R_1 = 10\ \Omega$ series with $R_2 = 20\ \Omega \| R_3 = 20\ \Omega$ at 30 V as a specific counterexample.
Show all answers
Activity 1 — Model Answers
- $R_{2\|3} = (20 \times 30)/(20 + 30) = 600/50 = 12\ \Omega$
- $R_{\text{total}} = 10 + 12 = 22\ \Omega$
- $I = 25/22 \approx 1.14\ \text{A}$; $V_{\text{parallel}} = 1.14 \times 12 \approx 13.6\ \text{V}$; (or exact: $I = 25/22$, $V_p = 300/22 \approx 13.6\ \text{V}$)
- $I_2 = 13.6/20 \approx 0.68\ \text{A}$; $I_3 = 13.6/30 \approx 0.45\ \text{A}$. Sum $\approx 1.13\ \text{A} \approx 1.14\ \text{A}$ ✓ (rounding)
Activity 2 — Model Answers
- $R_{2\|3} = (10 \times 15)/(10+15) = 150/25 = 6.0\ \Omega$; $R_{\text{total}} = 5 + 6 = 11\ \Omega$; $I = 20/11 \approx 1.82\ \text{A}$
- $V_1 = 1.82 \times 5 = 9.09\ \text{V}$; $V_p = 20 - 9.09 = 10.9\ \text{V}$
- $I_2 = 10.9/10 = 1.09\ \text{A}$; $I_3 = 10.9/15 = 0.727\ \text{A}$; sum = 1.82 A ✓
- $P_1 = 1.82 \times 9.09 \approx 16.5\ \text{W}$; $P_2 = 1.09 \times 10.9 \approx 11.9\ \text{W}$; $P_3 = 0.727 \times 10.9 \approx 7.9\ \text{W}$; total $\approx 36.3\ \text{W}$. Supply: $P = 20 \times 1.82 = 36.4\ \text{W}$ ✓
Short Answer — Model Answers
Q3 (3 marks): (a) $R_{2\|3} = (24 \times 12)/(24+12) = 288/36 = 8.0\ \Omega$; $R_{\text{total}} = 8 + 8 = 16\ \Omega$. (b) $I = 16/16 = 1.0\ \text{A}$. (c) $V_p = 1.0 \times 8 = 8.0\ \text{V}$.
Q4 (3 marks): (a) $I_2 = 8.0/24 = 0.333\ \text{A}$; $I_3 = 8.0/12 = 0.667\ \text{A}$. (b) $0.333 + 0.667 = 1.0\ \text{A}$ ✓. (c) $V_1 = 1.0 \times 8 = 8.0\ \text{V}$; $P_1 = 1.0 \times 8.0 = 8.0\ \text{W}$ (or $I^2 R = 8.0\ \text{W}$).
Q5 (4 marks): The claim is incorrect because it ignores circuit configuration. You cannot add resistances in parallel using simple addition. Counterexample: naively adding gives $R_{\text{total}} = 10 + 20 + 20 = 50\ \Omega$, $I = 30/50 = 0.60\ \text{A}$. Correct reduction: $R_{2\|3} = (20 \times 20)/(20+20) = 10\ \Omega$; $R_{\text{total}} = 10 + 10 = 20\ \Omega$; $I = 30/20 = 1.5\ \text{A}$. The results differ by a factor of 2.5. Kirchhoff's voltage law requires that the voltage across each parallel branch is equal, which is only satisfied by the reduction method. Simply adding resistances treats all components as if they were in series, which would force different voltages on parallel branches — violating conservation of energy.
The 2019 Sydney Metro trains use exactly the structure from this lesson: 4 traction motors wired in series (375 V each, 1,500 V total per group), then two groups wired in parallel. The series wiring shares voltage; the parallel wiring shares the load — combining both topologies achieves 1.68 MW peak power with built-in redundancy. The analysis method is identical to what you just practised: reduce the innermost group first, then work back with V = IR.
Now check your Think First answers: the parallel pair of 12 Ω resistors gives R₂₋₃ = (12×12)÷(12+12) = 6 Ω. Total resistance = 4 + 6 = 10 Ω. Total current = 12 ÷ 10 = 1.2 A.