Real circuits are rarely all-series or all-parallel, they mix the two. The trick is to never panic: break the network into series and parallel chunks, collapse each chunk to a single resistance one step at a time, and the whole thing becomes solvable. Then you will measure such a circuit in the lab and see how close real readings come to your calculations.
A circuit has two resistors in parallel, and that combination is joined in series with a third resistor and a battery.
Before reading: to find the total resistance, which part should you simplify first, the parallel pair or the series resistor? Write your prediction and a one-sentence reason.
Collapse parallel groups, then add what is in series
A combination circuit is solved by replacing the trickiest cluster, usually a parallel group, with one equivalent resistor, until the whole network is a single simple loop.
The strategy is to simplify from the inside out. Find any group of resistors that is purely series or purely parallel, replace it with its single equivalent resistance, redraw, and repeat. Eventually the network reduces to one resistance $R_T$, from which you find the supply current $I = V/R_T$. Then work backwards: the supply current is the series current through the main path, and you use $V = IR$ and the parallel rules to find the voltage and current of each individual component.
Solve a combination circuit by reduction: replace each pure series or pure parallel group with a single equivalent resistance until one $R_T$ remains, find the supply current $I = V/R_T$, then work backwards with $V = IR$ to get each component's voltage and current.
Pause, copy the highlighted method into your book before moving on.
Did you get this? True or false: in a combination circuit, you should replace each series or parallel group with a single equivalent resistance before finding the total.
A worked reduction from start to finish
We just saw the reduction strategy. That raises a question: how does it play out on a real network? This card answers it → collapse the parallel pair, add the series resistor, then back-substitute for each component.
Working a single example end to end shows how each rule slots in: reciprocals for the parallel part, addition for the series part, and $V = IR$ to share the values back out.
Consider a $12\ \text{V}$ battery in series with a $4.0\ \Omega$ resistor, which then feeds a parallel pair of $6.0\ \Omega$ and $3.0\ \Omega$.
A $12\ \text{V}$ battery is in series with a $4.0\ \Omega$ resistor ($R_1$), which is in series with a parallel pair $R_2 = 6.0\ \Omega$ and $R_3 = 3.0\ \Omega$. Find the total resistance, the supply current, and the voltage across the parallel section.
Reduce the parallel section first ($R_p = 2.0\ \Omega$), add the series resistor for $R_T$, find the supply current $I = V/R_T$, then back-substitute: the series resistor takes $V_1 = IR_1$ and the parallel section gets the rest of the supply voltage.
Add the highlighted worked route to your notes before the check below.
Quick check: Two $6.0\ \Omega$ resistors are in parallel, and that combination is in series with a $5.0\ \Omega$ resistor. What is the total resistance?
Pattern C, Predict, measure, account for difference
Build a fixed-resistance combination circuit (e.g. a $10\ \Omega$ resistor in series with a parallel pair of $20\ \Omega$ and $20\ \Omega$) from a low-voltage supply. In your book: (1) calculate the total resistance, supply current, and the voltage across each section; (2) measure the current at different points (ammeter in series) and the voltage across each component (voltmeter in parallel); (3) tabulate measured vs calculated values; (4) write two sentences accounting for any variation, refer to the resistance of connecting wires, the loading effect of real meters, and the manufacturing tolerance of the resistors.
Two truths, one lie: three of these steps are correct when SOLVING a combination circuit. Which one is the lie?
Dot point: Solve problems involving voltage, current and resistance, in series, parallel and combination circuits
Accounting for variation in the practical
We just learned to solve and measure a combination circuit. That raises a question: why don't the meter readings exactly match the calculations? This card answers it → real components and meters are not ideal, so small, explainable differences appear.
Calculations assume perfect resistors and ideal meters; real equipment differs slightly, so measured values land close to, but not exactly on, the predictions.
Several real-world effects cause variation: connecting wires and switches have a small resistance not included in the calculation; a real ammeter has a little resistance and a real voltmeter draws a little current, slightly changing the circuit (meter loading); and resistors are made to a tolerance (often $\pm 5\%$), so the actual values differ from the marked ones. Temperature can also change resistance. None of these mean the experiment "failed", a good report calculates the percentage difference and identifies which of these effects best explains it.
Measured values differ from calculated ones because of connecting-wire resistance, meter loading (real ammeters have some resistance, real voltmeters draw some current), and resistor tolerance. These are systematic effects, so small differences are expected and explainable.
Pause, copy the highlighted list of causes into your book before the check below.
Last check: Decide if the statement is true or false.
Resistor tolerance, connecting-wire resistance and meter loading can all cause measured circuit values to differ slightly from calculated ones.
Fill the blanks: Complete the account of an experimental result.
In a combination-circuit experiment, measured voltages are usually slightly than the calculated values, because the connecting leads and the supply have their own that the calculation ignored.
Dot point: conduct a laboratory experiment to measure the voltage differences across components and current at different points in a combination circuit of fixed resistance, and account for variation from the calculated values
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A $10\ \text{V}$ battery is in series with a $3.0\ \Omega$ resistor, which feeds a parallel pair of $4.0\ \Omega$ and $4.0\ \Omega$. Calculate the total resistance and the supply current, then state one reason why a measured value of current in a real version of this circuit might differ from your calculated value.
1 mark: parallel pair $R_p = 2.0\ \Omega$ · 1 mark: $R_T = 5.0\ \Omega$ · 1 mark: $I = 2.0\ \text{A}$ · 1 mark: a valid reason for variation (wire resistance, meter loading or resistor tolerance)
ApplyBand 5(4 marks) 2. In a combination circuit a $6.0\ \Omega$ resistor is connected in parallel with a $3.0\ \Omega$ resistor, and this pair is in series with an unknown resistor $R$. The total resistance of the circuit is $8.0\ \Omega$. Determine $R$, and calculate the current through $R$ when the circuit is connected to a $16\ \text{V}$ supply.
1 mark: parallel pair $R_p = 2.0\ \Omega$ · 1 mark: $R = R_T - R_p$ · 1 mark: $R = 6.0\ \Omega$ · 1 mark: $I = V/R_T = 2.0\ \text{A}$, and this is the current through $R$ because it is in series
Dot point: Solve problems involving voltage, current and resistance, in series, parallel and combination circuits
AnalyseBand 5(3 marks) 3. In a laboratory investigation a student measures the voltages across components in a combination circuit and finds they are consistently a little lower than the values calculated from the nominal resistances. Suggest two reasons for this systematic difference and state how the measurements could be improved.
1 mark: resistors have manufacturing tolerance, so actual values differ from nominal · 1 mark: connecting wires and the supply have their own small resistance, and meters are not ideal, which lowers the measured values · 1 mark: an improvement, e.g. measure each resistance directly with an ohmmeter and use meters of appropriate resistance, and repeat readings
Dot point: conduct a laboratory experiment to measure the voltage differences across components and current at different points in a combination circuit of fixed resistance, and account for variation from the calculated values
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): First reduce the parallel pair: $1/R_p = 1/4.0 + 1/4.0 = 2/4.0$, so $R_p = 2.0\ \Omega$. This is in series with the $3.0\ \Omega$ resistor, giving total resistance $R_T = 3.0 + 2.0 = 5.0\ \Omega$. The supply current is $I = V/R_T = 10/5.0 = 2.0\ \text{A}$. A measured current in a real circuit might differ from this because the connecting wires and switch add a small extra resistance that was not included, because a real ammeter has its own small resistance that slightly reduces the current, or because the resistors have a manufacturing tolerance (e.g. $\pm 5\%$) so their true values differ from the marked ones.
Q2 (4 marks): First reduce the parallel pair: $1/R_p = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0$, so $R_p = 2.0\ \Omega$. This pair is in series with $R$, and series resistances add, so $R_T = R + R_p$. Rearranging, $R = R_T - R_p = 8.0 - 2.0 = 6.0\ \Omega$. To find the current through $R$, note that $R$ lies on the single path carrying all the current before it divides between the branches, so the current through it is the total circuit current. That is $I = V/R_T = 16/8.0 = 2.0\ \text{A}$. As a check, the voltage across $R$ is $2.0 \times 6.0 = 12\ \text{V}$ and across the pair is $2.0 \times 2.0 = 4.0\ \text{V}$, which sum to the $16\ \text{V}$ supply.
Q3 (3 marks): One reason is that real resistors are manufactured to a stated tolerance, often five or ten per cent, so the actual resistance of each component differs from the nominal value printed on it, and the calculation used the nominal figure. A second reason is that the circuit contains resistance the calculation ignored: the connecting leads, the switch contacts and the internal resistance of the supply all add a small resistance in series, which reduces the current and therefore the voltage available across the components being measured. The voltmeter itself also draws a small current, slightly lowering the voltage it reads. The measurements could be improved by measuring the actual resistance of each resistor directly with an ohmmeter and using those values in the calculation, by using a voltmeter of very high resistance so it draws negligible current, by keeping leads short and connections clean, and by repeating each reading and averaging to reduce random error.
You can now reduce a combination circuit, solve for any component, measure it, and account for the difference between measurement and theory. That completes the electric-circuits strand, next we turn to magnetic fields.
↩ Back to focus-area hubAt the start you predicted which part to simplify first. The answer: simplify the parallel pair first into one equivalent resistor, because then the whole circuit becomes a simple series loop you can add up. If you chose the series resistor first, revisit Card 2, you reduce the inner cluster before combining the outer series path.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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