Circuits & Ohm's Law — Print · Grade 12 Science
Chapter: Electricity, Magnetism & CircuitsCircuits & Ohm's Law
Learning goals
- I can apply Ohm's law (V = IR) to find voltage, current, or resistance in a circuit.
- I can find the total resistance of resistors wired in series and in parallel.
- I can calculate electrical power with P = IV and relate it to energy used.
Readings for this unit
Before this unit
Ohm’s law and circuit continuity.
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Start with something to explain
Why can adding a parallel branch increase battery current?
Illustrative evidence case. A second resistor connects across the same two ideal source terminals.
Notice and wonder
Record two observations using the stated quantities or relationships. Keep observations separate from possible causes. Write one question that a new measurement could help answer.
Why can adding a parallel branch increase battery current?
Your first model
Label shared nodes and trace a complete return path.
Build the language from the idea
- Ohm's law
- The relationship V = IR linking voltage, current, and resistance for a resistor at constant temperature.
- series circuit
- A circuit with a single path, so the same current flows through every component and their resistances add.
- parallel circuit
- A circuit with two or more branches, so each branch has the full voltage and the total resistance is less than any one branch.
- equivalent resistance
- The single resistance that could replace a whole combination of resistors and draw the same current.
- electrical power
- The rate at which a device converts electrical energy, P = IV, measured in watts (W).
The three quantities from the last lesson lock together in the most useful equation in all of electronics, Ohm's law: V = I × R. Voltage equals current times resistance, and the same relationship rearranges to I = V ÷ R and R = V ÷ I, so knowing any two of the three always gives you the third. Read it as a story: for a fixed battery voltage, piling on more resistance chokes the current down, while lowering the resistance lets more current surge through. A resistor that obeys this straight-line relationship at constant temperature is called 'ohmic,' and it is the assumption behind almost every circuit calculation you will do.
How the components are wired matters just as much as their values. A series circuit offers just one single loop for charge to travel, so the very same current flows through every component in turn. Because the charge must pass through each resistor one after another, their resistances simply add up: R_total = R₁ + R₂ + R₃ + …, and the total is always larger than any single resistor. The voltage from the battery divides among them, with the biggest resistor taking the biggest share. The weakness of a series circuit is its single path — break it anywhere and the current stops everywhere, which is why one dead bulb once darkened an entire old string of holiday lights.
A parallel circuit gives charge several branches to choose from at once. Here every branch is connected across the same two points, so each one receives the full battery voltage, and the branch currents add up to the total. Counterintuitively, adding more parallel branches *lowers* the overall resistance, because you are opening more lanes for charge to flow: 1 ÷ R_total = 1 ÷ R₁ + 1 ÷ R₂ + …, giving a total that is always smaller than the smallest branch. For just two resistors this tidies up to R = R₁R₂ ÷ (R₁ + R₂). This independence is why your home is wired in parallel — you can switch off one lamp, or have one appliance fail, without cutting power to everything else.
Many resistor networks can be simplified through series and parallel groups, but a bridge network may require general node and loop equations instead. Series/parallel formulas assume ideal wires and the stated component behavior. Real devices can contain capacitors, inductors and nonlinear elements, so a constant-R calculation does not predict every transient or arbitrary appliance. Trace topology, identify conditions and then choose an appropriate method.
Circuits exist to *do* something, and that means converting electrical energy into other forms — light, heat, motion, sound. The rate of that conversion is electrical power, P = I × V, measured in watts, where one watt is one joule of energy converted per second. Combine it with Ohm's law and it takes two other handy forms, P = I²R and P = V² ÷ R, letting you find power from whichever two quantities you know. A 60-watt bulb converts 60 joules of electrical energy every second; multiply a device's power by the time it runs and you get the total energy it uses, which is exactly what your utility meters and bills.
Together, Ohm's law, series and parallel rules, and the power equation let you predict a circuit completely — how much current each part draws, how the voltage splits up, and how much energy is consumed and where. That predictive power is not academic: it is how a fuse or circuit breaker is sized to trip before too much current overheats the wiring, how a phone charger is designed to deliver exactly the right current, and how engineers keep the enormous circuit of the electrical grid balanced. Build a real device and these are the rules you will actually reach for.
Model check: A series circuit has equal current; a parallel circuit has equal branch voltage. 1/R=1/6+1/6=1/3, soR=3ohm; lower than either branch.
Build your understanding
Use the detailed parent plan: two core 50–60-minute sessions per focused unit, with a third assigned for selected calculation/modeling topics. First study the core explanation and Parts 1–2 with their figures/checks; next study Parts 3–4, the whole-unit application, worked cases, varied practice and exit check. Attempt reasoning before opening guidance. Extend time when actual understanding requires it; investigations and chapter assessments have separate sessions.
Part 1 · Recognize connections from shared nodes
A node joins conductors that share one ideal potential. Components are in series when the same unbranched current path passes through them sequentially. In parallel, branches connect to the same pair of nodes and share voltage. Drawing components side by side does not establish either relation. Trace wires and label nodes before using a series/parallel formula; the complete closed loop here places 2 Ω and 4 Ω in series across 6 V.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What steady current does the ideal series loop carry?
Visual reasoning — after your attempt
Rtotal=2+4=6 Ω, so I=6/6=1 A through each resistor.
Series components share a single path, giving the same current in a steady ideal circuit. Parallel branches connect across the same two nodes, giving the same potential difference. Visual placement on a page does not establish a connection; trace the wires and label nodes.
Two resistors drawn side by side may still be series if wired end to end. A schematic is an interaction map, not a literal room layout.
Pause and explain
What establishes that two components are parallel?
Check your reasoning after answering
Both ends connect to the same pair of circuit nodes, so they share the potential difference between those nodes.
Part 2 · Apply charge and energy accounting at nodes
Kirchhoff’s current rule expresses steady charge conservation at a node: incoming current equals outgoing current. The loop rule accounts for energy per charge: potential rises and drops sum to zero around an ideal steady loop. Both parallel branches here have 6 V, giving 1 A through 6 Ω and 2 A through 3 Ω. The source supplies 3 A, which divides and later recombines rather than being consumed in the branches.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What current returns to the source in this ideal steady circuit?
Visual reasoning — after your attempt
3 A, the sum of the 1 A and 2 A branch currents.
For a steady node without accumulating charge, incoming current equals outgoing current. Around a compatible ideal circuit loop, signed potential changes sum to zero. These accounts are related to charge and energy conservation; specify loop direction and source orientation before assigning voltage signs.
A parallel branch can carry a smaller current than another branch even though both have the same voltage.
Pause and explain
Why do branch currents not need to be equal?
Check your reasoning after answering
The branch resistances or component behaviors can differ; their common voltage does not require equal current.
Part 3 · Use equivalent resistance with its assumptions
An equivalent resistance gives the same voltage–total-current relation at a chosen pair of terminals as the ideal resistor network. Series resistances add; parallel conductances 1/R add. A 2 Ω equivalent for 6 Ω parallel with 3 Ω is smaller than either branch resistance because it offers more conducting paths. This does not mean either actual resistor changed its own resistance or that branch currents equal the total current.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
Is a parallel equivalent larger or smaller than its smallest positive branch resistance?
Visual reasoning — after your attempt
Smaller for more than one finite positive branch in this ideal model.
Series resistances add. For parallel ohmic resistors, reciprocal resistances add, and the equivalent is smaller than the smallest positive branch resistance. Equivalent resistance simplifies source behavior but does not remove the need to calculate individual branch quantities.
Two equal 10 Ω resistors in parallel give 5 Ω, while series gives 20 Ω. Check which physical connection the formula represents.
Pause and explain
Why should a parallel equivalent exceed no individual branch resistance?
Check your reasoning after answering
An added positive conducting branch increases total conductance, reducing the resistance seen by the ideal source.
Part 4 · Distinguish an ideal source from actual behavior
An ideal voltage source holds its specified voltage regardless of current within the model. A simple real-source model includes internal resistance r, giving Vterminal=ε−Ir while delivering current. Here terminal voltage falls as load current rises. Chemical state, temperature and protection limits can introduce further behavior. Do not assume adding a parallel load preserves exactly the same voltage or guarantees unlimited current from a real battery.
Read the visual model
Identify the labeled structures and units, then trace the arrows or compare values under the caption’s stated conditions. Separate the displayed evidence from a further inference. Write your reasoning before opening the guidance.
What terminal voltage does this model give at 1.5 A?
Visual reasoning — after your attempt
4.5 V; internal power dissipation is I²r=2.25 W in the model.
An ideal voltage source holds its stated voltage despite load. Real source internal resistance and capacity can alter terminal voltage and available current. Opening a parallel branch and opening a shared series path have different effects. Avoid inferring every fault from one ideal equivalent-resistance calculation.
A larger load may reduce a real battery’s terminal voltage. The ideal model is useful but should be checked against actual source specifications.
Pause and explain
What observation would challenge an ideal constant-voltage source model?
Check your reasoning after answering
A reproducible terminal-voltage change under changing load beyond measurement uncertainty would indicate source behavior requiring a revised model.
Apply the whole unit · Analyze unequal parallel branches
Illustrative ideal 12.0 V source connects across parallel 6.00 Ω and 12.0 Ω ohmic resistors.
Your task
Find branch currents, source current, equivalent resistance, and branch powers. Predict what happens if the 12 Ω branch opens and distinguish ideal-source from real-source behavior.
What your response should show
- Define the system, quantities, units, and relevant conditions before selecting a model.
- Show intermediate calculations or a labeled evidence account, including signs or denominators.
- Explain how the supplied evidence supports the conclusion rather than reporting only an answer.
- Identify a meaningful uncertainty, model limit, or measurement needed to test the conclusion.
Response guidance — after your attempt
Branch currents are 12/6=2.00 A and 12/12=1.00 A.
Total current=3.00 A; equivalent R=12/3=4.00 Ω.
Branch powers are 24.0 W and 12.0 W; opening the 12 Ω branch leaves 2.00 A in the ideal remaining branch.
The ideal-source voltage remains 12 V and the open branch carries no sustained current. A real source may change terminal voltage as its load changes, altering remaining-branch current. The total 36 W equals source voltage times total current under the ideal assumptions.
Scientific references and source-model foundations
Parent / teacher teaching notes
Require node labels and a power sum as an independent check. Keep branch-current equality separate from the actual equal-voltage property of parallel connections. Ask the learner to explain each symbol, axis, unit and arrow before calculating. Calculated examples, supplied illustrative records and actual historical observations are labeled separately. Selected public OpenSciEd middle-school source figures are foundation examples, not complete high-school editions or an endorsement; all advanced explanations are supplied locally.
A simple circuit diagram with a battery, switch, and resistor or bulb, showing current flowing around the loop when the switch is closed.
The key idea
Ohm's law V = IR ties voltage, current, and resistance together; series resistances add while parallel resistances give a smaller total (1/R = 1/R₁ + 1/R₂ + …); electrical power is P = IV.
Ohm's Law
V = IR
V is voltage in volts (V); I is current in amps (A); R is resistance in ohms (Ω). Rearranges to I = V ÷ R and R = V ÷ I.
Series Resistance
R = R₁ + R₂ + R₃ + …
resistances in series simply add; the total is always larger than any single resistor.
Parallel Resistance
1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ + …
for resistors in parallel; the total is always smaller than the smallest single resistor. For two, R = R₁R₂ ÷ (R₁ + R₂).
Electrical Power
P = IV
P is power in watts (W); I is current in amps (A); V is voltage in volts (V). Can also be written P = I²R or P = V² ÷ R.
Worked examples
Example 1
- Rearrange Ohm's law V = IR to solve for current: I = V ÷ R.
- I = 12 ÷ 4.
- = 3.
Example 2 · Worked application
- Series resistance=4+8=12 Ω.
- Current I=12/12=1 A through both.
- Drops are 1×4=4 V and 1×8=8 V, totaling 12 V.
Kirchhoff's loop balance provides a check.
Example 3
- In series, resistances add: R_total = 3 + 6 = 9 Ω.
- Find the current with Ohm's law: I = V ÷ R = 18 ÷ 9 = 2 A (the same current flows through both).
- Voltage across the 6 Ω resistor: V = IR = 2 × 6 = 12 V. (The 3 Ω resistor takes the other 6 V, and 12 + 6 = 18 V checks out.)
Example 4 · Find and fix the mistake
- Test the claim against the stated conditions: Kirchhoff's loop balance provides a check.
- They share current; voltage drops depend on resistance.
Kirchhoff's loop balance provides a check.
Example 5 · Reason from the visual model
- An equivalent resistance gives the same voltage–total-current relation at a chosen pair of terminals as the ideal resistor network. Series resistances add; parallel conductances 1/R add. A 2 Ω equivalent for 6 Ω parallel with 3 Ω is smaller than either branch resistance because it offers more conducting paths. This does not mean either actual resistor changed its own resistance or that branch currents equal the total current.
- Smaller for more than one finite positive branch in this ideal model.
Original exact equivalent-resistance comparison for two positive ohmic resistors.
Avoid these mistakes
- A series circuit has equal current; a parallel circuit has equal branch voltage.
Check your understanding
Two 6ohm resistors in parallel have what equivalent R?
Discuss the answer
1/R=1/6+1/6=1/3, soR=3ohm; lower than either branch.
Return to the phenomenon
Return to the initial case. Why can adding a parallel branch increase battery current? Revise your drawing or explanation using two named ideas and one stated limitation.
Compare explanations after your attempt
Parallel paths reduce equivalent resistance; branch currents add while each shares terminal voltage in the ideal-source model.
What should we investigate next?
Which comparison would test a remaining uncertainty? State what you would change, keep comparable and measure.
Investigate, calculate and revise
Keep a first model and an evidence-supported revision. Attempt each task before opening guidance; all required explanations and supplied-data routes are available locally.
Activity 1 · Return to the opening case
Use the opening phenomenon, unit definitions and explained Parts 1–4.
Your task
Return to the initial case. Why can adding a parallel branch increase battery current? Revise your drawing or explanation using two named ideas and one stated limitation.
Success criteria
- Use the stated observations and appropriate units.
- Connect a labeled model with the physical/environmental mechanism.
- State assumptions and an evidence limitation.
- Explain a substantive revision or next test.
Hint
Keep object/material identity, material transfer and energy account distinct.
Response guidance — discuss after your attempt
Parallel paths reduce equivalent resistance; branch currents add while each shares terminal voltage in the ideal-source model.
Activity 2 · Transfer to a changed case
Cover the worked answer and use the explained models, relationships and notebook.
Your task
Two resistors 4 Ω and 12 Ω are in parallel across an ideal 12 V source. Find each current, total current, equivalent resistance and total power.
Success criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.
Hint
Name known quantities and conditions, show intermediate steps and check conservation.
Response guidance — discuss after your attempt
Currents are 3 A and 1 A; total 4 A. Req=12/4=3 Ω and P=12×4=48 W. Branch powers 36 W and 12 W sum consistently.
Words to know
- Ohm's law
- The relationship V = IR linking voltage, current, and resistance for a resistor at constant temperature.
- series circuit
- A circuit with a single path, so the same current flows through every component and their resistances add.
- parallel circuit
- A circuit with two or more branches, so each branch has the full voltage and the total resistance is less than any one branch.
- equivalent resistance
- The single resistance that could replace a whole combination of resistors and draw the same current.
- electrical power
- The rate at which a device converts electrical energy, P = IV, measured in watts (W).
Georg Ohm was ridiculed before he was revered
When German schoolteacher Georg Simon Ohm published his law relating voltage, current, and resistance in 1827, the reception was chilly — some colleagues dismissed the work as unscientific fantasy, and it cost him professionally for years. Time proved him spectacularly right: V = IR became one of the most-used equations in all of engineering, and the SI unit of resistance, the ohm (Ω), now carries his name. It's a useful reminder that even a relationship that looks obvious in hindsight can take a generation to be accepted, and that careful measurement eventually wins the argument.
Ohm's 'law' isn't actually a law of nature
It's easy to assume V = IR always holds, but it's really a property that only some materials happen to obey, and only under steady conditions. A device where current stays proportional to voltage is called ohmic — a plain resistor at constant temperature is the classic example. Plenty of everyday components are non-ohmic: a light-bulb filament's resistance climbs sharply as it heats up, so doubling the voltage does not double the current, and a diode conducts freely in one direction but blocks the other, giving a wildly curved voltage-current graph. Ohm's law is an excellent model within its limits, not an ironclad rule of the universe.
Kirchhoff's rules and why your house is wired in parallel
For circuits too tangled for a single series-or-parallel shortcut, Gustav Kirchhoff gave two bookkeeping rules in 1845. His current law says all the current flowing into any junction must equal the current flowing out (charge can't pile up). His voltage law says the voltage changes around any complete loop must sum to zero (energy per charge returns to where it started). These rules explain why homes are wired in parallel: every outlet connects across the same two supply lines, so each receives the full 120 V independently, and switching one device off doesn't starve the rest — a series house would go completely dark the moment any single bulb failed.
The kilowatt-hour and where your electric bill comes from
Electrical power is P = IV, measured in watts, but your utility bills you for energy, which is power multiplied by time. The everyday unit is the kilowatt-hour (kWh) — the energy used by 1,000 watts running for one hour, equal to 3.6 million joules. A 1,500-watt space heater running for 4 hours uses 1.5 kW × 4 h = 6 kWh; at, say, 15 cents per kWh, that's about 90 cents. This is also why high-power appliances like ovens and dryers get their own dedicated, thick-wire circuits: P = IV means high power at household voltage demands a large current, and only a thick wire can carry that current without its own resistance overheating it.
In the real world
Your home is wired in parallel so each outlet gets the full 120 V, and a circuit breaker watches the total current — using P = IV, an electrician can predict when too many high-power appliances on one circuit will trip it.
Exit check
Two 6ohm resistors in parallel have what equivalent R?
Check your explanation after answering
1/R=1/6+1/6=1/3, soR=3ohm; lower than either branch.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Answer and explanation
10 V
V = IR = 2 × 5 = 10 V.
Problem 2
Answer and explanation
8 Ω
R = V ÷ I = 24 ÷ 3 = 8 Ω.
Problem 3
Answer and explanation
20 Ω
In series, resistances add: 4 + 6 + 10 = 20 Ω.
Problem 4
- the same through all of them
- larger in the bigger resistors
- smaller in the bigger resistors
- zero
Answer and explanation
the same through all of them
A series circuit has one single path, so the identical current flows through every component.
Problem 5
Answer and explanation
6 Ω
1/R = 1/12 + 1/12 = 2/12 = 1/6, so R = 6 Ω. (For two equal resistors, the parallel total is always half of one.)
Problem 6
Answer and explanation
0.5 A
Rearranging P = IV gives I = P ÷ V = 60 ÷ 120 = 0.5 A.
Problem 7
Hints
Kirchhoff's loop balance provides a check.
Answer and explanation
R_eq=3 Ω; I_total=12/3=4 A, with 2 A per branch.
Problem 8
Hints
State what the supplied observations or conditions establish, and separate that from any assumption.
Answer and explanation
Each branch spans the same two electrical nodes, so it has the same potential difference.
Problem 9
Answer and explanation
Smaller for more than one finite positive branch in this ideal model.
An equivalent resistance gives the same voltage–total-current relation at a chosen pair of terminals as the ideal resistor network. Series resistances add; parallel conductances 1/R add. A 2 Ω equivalent for 6 Ω parallel with 3 Ω is smaller than either branch resistance because it offers more conducting paths. This does not mean either actual resistor changed its own resistance or that branch currents equal the total current.
Smaller for more than one finite positive branch in this ideal model.
Response criteria
- Interpret labels, structures, axes, units and directions correctly.
- Explain the physical or environmental relationship or mechanism.
- Respect stated assumptions and distinguish observation from inference.
Problem 10
Answer and explanation
Currents are 3 A and 1 A; total 4 A. Req=12/4=3 Ω and P=12×4=48 W. Branch powers 36 W and 12 W sum consistently.
Currents are 3 A and 1 A; total 4 A.
Req=12/4=3 Ω and P=12×4=48 W.
Branch powers 36 W and 12 W sum consistently.
Response criteria
- Use the stated measurements or conditions, with correct units.
- Connect the result to the physical or environmental mechanism or evidence relationship.
- Identify the assumption, model limit or further evidence requested.