IP Physics Notes (Upper Secondary, Year 3-4): 12) Direct Current Circuits

Study guide30 Sep 2025, 00:00 Z

Combine series/parallel rules, potential dividers, and sensor behaviour to solve IP DC circuit problems.

Download PDF Join our Telegram study group

Q: What does IP Physics Notes (Upper Secondary, Year 3-4): 12) Direct Current Circuits cover?
A: Combine series/parallel rules, potential dividers, and sensor behaviour to solve IP DC circuit problems.

Quick recap -- DC circuit analysis hinges on current and voltage sharing rules. Once you master how resistances combine and how potential dividers behave, thermistor/LDR sensor circuits become straightforward.

The core idea is simple: Series shares current; parallel shares voltage.

Use it as a working check: Simplify resistance first, then find total current, voltage drops, and branch currents. Potential dividers turn resistance changes into voltage changes.

Then go one layer deeper: Work through the mixed network and sensor notes to practise drawing the simplified circuit, applying the correct sharing rule, and checking current and voltage totals.

Keep your practice loop tight via our Sec 3 IP Physics tuition hub. It links each topic here to quizzes, diagnostics, and WA-style problem sets.

New to the Integrated Programme? Start with What is IP? | Browse all free IP notes.

These notes align with SEAB GCE O-Level Physics (6091) content used in IP programmes (exams from 2026).

Status: SEAB O-Level Physics 6091 syllabus (exams from 2026) checked 2025-11-30 - scope unchanged; remains the reference for these notes.

Series vs Parallel Essentials

Series
- Current is identical everywhere in the loop.
- Potential differences add: $V = V_1 + V_2 + \cdots$

Reviewed by

Chee Wei Jie·Academic Advisor (Physics)

What you know for one component	Fastest expression	Why this is safer	Common trap
Current through the component and its resistance	$P = I^2R$	Series components share current, so this often avoids finding each voltage first.	Using the supply voltage for every series resistor.
Voltage across the component and its resistance	$P = \dfrac{V^2}{R}$	Parallel branches share voltage, so this often avoids finding branch current first.	Using total current in every parallel branch.
Component voltage and component current	$P = VI$	This is the direct definition once both quantities are already known.	Multiplying supply voltage by a branch current unless the branch is across the full supply.

Measurement wanted	Where the meter goes	Ideal resistance reason	Common trap
Current through one component	Ammeter in series with that component	Very low resistance so it does not significantly reduce the current	Connecting the ammeter across the component and creating a near short circuit.
Potential difference across one component	Voltmeter in parallel across that component	Very high resistance so it takes negligible current	Putting the voltmeter in series and almost opening the circuit.
Total current from the supply	Ammeter in the main loop before any branch split or after branches recombine	The same total current passes through that main section	Measuring only one branch current and calling it total current.
Voltage across a parallel block	Voltmeter across the two junctions of the whole block	Every branch shares the same potential difference	Placing the voltmeter across just one resistor outside the block.

Circuit clue	First move	Rule to apply	Common trap
Components share one path with no junction between them	Treat them as a series group.	Add their resistances directly.	Splitting the same current into different values inside a series path.
Components connect across the same two junctions	Treat them as a parallel group.	Add inverse resistances, then invert the result.	Adding parallel resistances directly.
One resistor is outside a parallel block	Replace the parallel block first, then combine with the outside resistor.	Parallel first, then series.	Using the supply voltage across each resistor before simplifying.
Asked for a branch current	Find the voltage across the whole parallel block first.	Use $I = V/R$ for that branch.	Using the total circuit current as every branch current.

Circuit change	First comparison	Brightness result	Common trap
Add another identical bulb in series	Total resistance increases, so total current decreases	Every series bulb becomes dimmer than a single bulb on the same supply	Saying the first bulb keeps the original current because it is nearest the battery.
Add another identical bulb in parallel	Each branch still gets the supply voltage	Each identical parallel bulb is about as bright as one bulb alone, while total current from the supply increases	Saying each bulb gets half the voltage just because there are two bulbs.
One bulb is removed from a series circuit	The circuit opens if there is no bypass path	All bulbs in that series path go off	Thinking the remaining bulbs get brighter automatically.
One branch is removed from a parallel circuit	Other branches still connect across the supply	Bulbs in the remaining branches stay lit with about the same brightness	Treating a parallel circuit as one broken loop.

Sensor position	Sensor resistance increases	Sensor resistance decreases
Output across the sensor	$V_\text{out}$ increases because the sensor takes a larger share of the supply.	$V_\text{out}$ decreases because the sensor takes a smaller share of the supply.
Output across the fixed resistor	$V_\text{out}$ decreases because the fixed resistor takes a smaller share of the supply.	$V_\text{out}$ increases because the fixed resistor takes a larger share of the supply.

IP Physics Notes (Upper Secondary, Year 3-4): 12) Direct Current Circuits

Series vs Parallel Essentials

Power formula choice checkpoint

Meter-placement checkpoint

Mixed-circuit simplification checkpoint

Bulb brightness checkpoint

Worked Example: Mixed Network

Potential Divider Principle

Divider direction map

Worked Example: Thermistor Divider Direction

Sensor Behaviours

Tips for Divider Problems

Threshold decision checkpoint

Practice Quiz

Key Takeaways

Related Posts

Question clue	First decision	Then compare	Common trap
"Switch turns on when $V_\text{out} \gt \pu{3.0 V}$ "	Does the condition raise or lower $V_\text{out}$ ?	Turn on only when the output is above the threshold.	Saying the switch turns on just because the sensor resistance changed.
"Output is taken across the fixed resistor"	If the sensor resistance falls, the fixed resistor takes a larger voltage share.	Compare the new fixed-resistor voltage with the threshold.	Using the sensor's voltage when the output terminals are not across it.
"Output is taken across the sensor"	If the sensor resistance falls, the sensor takes a smaller voltage share.	Compare the new sensor voltage with the threshold.	Assuming an LDR or thermistor always increases output when the stimulus increases.
"Choose a sensor position"	Decide whether the output should rise or fall when the stimulus rises.	Place the sensor across the output if you want its voltage share to control the signal directly.	Drawing a divider before deciding the required output direction.