Full chapter
D.C. circuits
All 6 topics and the revision summary on one page.
01
Circuit symbols and connections
A circuit diagram shows what is connected to what. Follow the wires and component terminals before deciding which calculation to use.
Current is charge passing per second, measured in amperes. Potential difference is energy transferred per charge between two points, measured in volts. Current is measured through a path; p.d. is measured across two points.
Draw the components
Use the standard symbols below, with lines for connecting wires. Direct current (d.c.) flows in one direction. The worked models here additionally use steady current, whose value remains constant during each measurement.
Recognise the component and its connections
Names and short roles accompany the actual circuit symbols. Light arrows on an LED point out; light arrows on an LDR point in.
Cell
Supplies energy per charge. The longer plate is positive.
Battery
Two or more cells form the source shown here.
Switch
Opens or closes a conducting path; shown open.
Lamp
Transfers electrical energy into light and internal energy.
LED
Emits light in forward operation. The bar marks cathode K.
Fixed resistor
Provides resistance in the conducting path.
Variable resistor
Adjusts resistance in a two-terminal path.
Fuse
Its link can melt and open the circuit if current is excessive.
Ammeter
Measures current through the path in which it is inserted.
Voltmeter
Measures p.d. between the two points to which it connects.
D.c. supply
Provides a supply with the indicated fixed polarity.
A.c. supply
Provides a supply whose polarity alternates.
Potentiometer
Two track ends and a wiper provide three connections.
Electric bell
Produces sound when operated by the circuit.
LDR
Resistance falls as more light reaches it; arrows point in.
NTC thermistor
Resistance falls as its temperature rises.
For a cell, the longer plate is positive and the shorter plate negative. A battery symbol shows repeated cell pairs. Mark the polarity when it matters to the current direction or meter connection.
A variable resistor changes the resistance included in a two-terminal path. Its adjustment arrow is part of the symbol, not a current-direction arrow. A switch opens or closes a path; a fuse is a protective link that melts and breaks the path if the current is sufficiently large for long enough.
An LED is a light-emitting diode. In forward operation, conventional current enters its anode and leaves its cathode, the side marked by the bar. The small arrows pointing outwards represent emitted light. Use an appropriate series current-limiting resistor in a practical LED circuit; different LEDs need different operating conditions.
An a.c. supply reverses its polarity; its symbol is distinct from a d.c. supply's. The resistive-network examples here use a steady d.c. source.
An LDR has light arrows pointing towards it. A thermistor responds to temperature. The sensor circuits page explains their resistance changes. The potentiometer has two track-end connections and a separate wiper connection; that third terminal is essential to its divider use.
Identify the connection points
A junction dot marks connected wires. Where wires cross without a connection, draw a clear bridge or gap so the crossing cannot be mistaken for a junction. Do not add a junction simply because two lines pass near each other.
A node is a set of points joined by ideal connecting wire with no component between them. These points have the same potential in the negligible-wire-resistance model. A wire can bend or take a longer route on the page without changing the node it represents.
Trace connections instead of judging position
Junction dots show connected wires. These two drawings show only the A/B/C resistor network; a source can be connected across its outer terminals A and C.
One drawing of the network
Move R3 around the outside
- Series connection
- Components share one path, so the current does not split between them. R1 carries the current entering the complete R2/R3 group; it does not necessarily carry the same current as either individual branch.
- Parallel connection
- Components or groups share both endpoints. R2 and R3 each connect B to C. Sharing just one point is not enough.
To redraw a circuit, first record each component's two endpoints. Keep those endpoints and junctions unchanged as you move the symbols. An extra plain wire joining B directly to C would change this circuit: it would bypass the resistors rather than provide another drawing of the same network.
Place meters for the quantity you need
An ammeter goes in series in the selected path. In a main lead it measures total current; inside one branch it measures that branch's current. Connecting it directly across the source would make a low-resistance bypass, not the intended current measurement.
A voltmeter connects across the two endpoints of the component or group. Across A/C it measures the whole network's p.d.; across B/C it measures the parallel group's p.d. A reading across a pair of series resistors is not automatically the voltage across either one.
We treat the ammeter's resistance as negligible and the voltmeter's current as negligible in the model calculations. Actual instruments approximate these conditions. The measurement examples show how to pair readings for a component or a whole network.
Optional check R1 connects A to B. R2 and R3 each connect B to C, with no plain wire directly joining B to C. Which description remains correct when the drawing is rearranged?
02
Series circuits
Series components carry the same current. Their individual potential differences add to the p.d. across the complete path.
For each resistor, V = IR uses the voltage across that resistor and the current through it. Here Vs means the voltage across the complete external network.
Why the current is the same
In a steady unbranched circuit, charge does not continually pile up at one component. The same amount of charge per second passes each position. A resistor transfers energy; it does not consume part of the current before the next resistor.
Each coulomb undergoes successive energy transfers as it passes the components. The work per coulomb across the whole path is the sum of the work per coulomb in its parts, so the potential differences add.
Add the series resistances
For two resistors carrying the same current, V1 = IR1 and V2 = IR2. Adding gives Vs = I(R1 + R2). The single equivalent resistance must therefore be their sum.
One path has the same current; component voltage drops add
These supplied calculation models use fixed resistances, negligible wire resistance and an ideal source. Current arrows indicate direction, not a current scale.
One path: same current through both resistors
The drops 2.0 V and 4.0 V add to 6.0 V. Current is 1.0 A in the whole unbranched path.
Open the only path
Zero steady current does not mean zero source voltage. The only path is broken.
One current, two voltage drops
2.0 Ω and 4.0 Ω across 6.0 V
Assume fixed resistances, negligible wire resistance and an ideal source whose terminal p.d. stays at 6.0 V.
- Total resistance = 2.0 + 4.0 = 6.0 Ω.
- Shared current = 6.0/6.0 = 1.0 A.
- Across 2.0 Ω: V = 1.0 x 2.0 = 2.0 V.
- Across 4.0 Ω: V = 1.0 x 4.0 = 4.0 V.
The drops add to 6.0 V. One coulomb transfers 2.0 J in the first resistor and 4.0 J in the second. The charge passing them has not decreased.
At fixed supply voltage, a larger total resistance means a smaller current. These example values describe a model; practical components must also have suitable ratings and remain close to the assumed temperature.
What an open switch changes
Opening the only path stops the steady current throughout this simple circuit. Each fixed resistor then has V = IR = 0 across it, but the source does not lose its e.m.f. In the ideal open-switch arrangement, the source voltage appears across the gap.
Zero current does not always mean zero voltage. An open gap can have a p.d. across it even though there is no conducting path through it.
Optional check Fixed 3.0 ohm and 6.0 ohm resistors are in series across an ideal 9.0 V source. What is the potential difference across the 6.0 ohm resistor?
03
Parallel circuits
Parallel branches share the same two endpoints, so they have the same potential difference. Their currents add to the total current entering the group.
Follow the connection points, not the shape of the drawing. Two branches are parallel only when each begins and ends at the same pair of nodes.
The entering current divides at a junction and recombines at the other. Charge is conserved, so in steady conditions the total current entering equals the sum of the branch currents.
Each branch shares the same two endpoints
The source holds 6.0 V across the whole parallel network. Both finite-resistance branches carry current. Arrow lengths do not compare current magnitudes.
Reconnect the two resistors across the same two nodes
The equivalent resistance is 4/3 ohm, about 1.33 ohm. This is a different connection from the 6.0 ohm series circuit.
A changed circuit
The same two resistors, connected in parallel
With an ideal fixed 6.0 V source:
- Current in 2.0 Ω = 6.0/2.0 = 3.0 A.
- Current in 4.0 Ω = 6.0/4.0 = 1.5 A.
- Total current = 3.0 + 1.5 = 4.5 A.
Both branches carry current. The equivalent resistance is Vs/Itotal = 6.0/4.5 = 4/3 Ω, approximately 1.33 Ω.
Find the effective resistance
Each branch current is the common p.d. divided by its own resistance. Adding them gives Itotal = V/R1 + V/R2 + ... . Since Itotal = V/Rparallel, dividing by the common nonzero p.d. gives:
For exactly two resistors, this can be rearranged to Rparallel = R1R2/(R1 + R2). The general reciprocal relationship also works for three or more branches.
Three parallel branches
6 Ω, 12 Ω and 4 Ω
1/R = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 1/2, so R = 2 Ω.
Across 6 V, the branch currents are 1.0 A, 0.50 A and 1.5 A. Their sum is 3.0 A, which agrees with 6 V / 2 Ω.
For positive finite branch resistances, the parallel equivalent is smaller than the smallest branch resistance. The added path lets more total current flow at the same p.d.; it does not require all the current to choose only one path.
Open one branch
If these branches are connected directly across an ideal fixed-voltage source, opening one branch leaves the same p.d. across the remaining branches. Their currents stay the same, while total source current falls.
This prediction depends on the connection and supply condition. If another resistor is shared in series with the branches, their p.d. can change when a branch opens. The complete-circuit example shows why.
Optional check Fixed 3.0 ohm and 6.0 ohm resistors are connected directly in parallel across an ideal 6.0 V source. What current does the source supply?
04
Solving a complete circuit
Reduce a valid group to find the source current, then work back through the original connections to find each component's voltage and current.
Series groups share a path current. Parallel branches share endpoint voltage. R = V/I must refer to the same component or the same complete group.
Keep the same external endpoints
An equivalent resistance replaces a group at its two external terminals. It gives the same total current for the same p.d. across those terminals. It does not mean every original component had that current or the whole group's voltage.
Reduce the network while preserving terminal pairs
R1 is between A and B. The two parallel branches are between B and C. Keep those names when replacing a group or recovering its voltage.
Start with the complete network
First replace only the B/C parallel group. Its two branches do not each receive the full 12.0 V.
Replace the B/C group by 4.0 ohm
The drops 6.0 V and 6.0 V add to 12.0 V. Current is 1.50 A in the whole unbranched path.
Reduce the complete A/C network to 8.0 ohm
Work back from the source current to the voltage across each original group, then to the individual branch currents.
Open only the 12 ohm branch
The total current falls, but the current through the 6.0 ohm branch rises. Its shared series resistor changes the branch voltage.
Reduce, then work back
12.0 V across 4.0 Ω plus a parallel group
Assume an ideal source, negligible wire resistance and fixed component resistances.
- Reduce B/C: the 6.0 Ω and 12.0 Ω parallel equivalent is (6.0 x 12.0)/(6.0 + 12.0) = 4.0 Ω.
- Find the whole resistance: 4.0 + 4.0 = 8.0 Ω across A/C.
- Find source current: 12.0/8.0 = 1.50 A. This is also the current through R1.
- Find group voltages: VAB = 1.50 x 4.0 = 6.0 V. Therefore VBC = 12.0 - 6.0 = 6.0 V.
- Recover branch currents: 6.0/6.0 = 1.0 A through the 6.0 Ω resistor; 6.0/12.0 = 0.50 A through the 12.0 Ω resistor.
Check the result: 1.0 + 0.50 = 1.50 A at B, and 6.0 + 6.0 = 12.0 V along either complete path.
The parallel group has only 6.0 V across it here. Using 12.0 V for either branch would ignore the voltage across R1. Dividing the group's 6.0 V by total current 1.50 A finds the group's 4.0 Ω equivalent, not the resistance of either individual branch.
A branch current can rise while total current falls
Now open only the 12.0 Ω branch. The remaining path is 4.0 Ω in series with 6.0 Ω, so its total resistance is 10.0 Ω.
The new source current is 12.0/10.0 = 1.20 A. All of it passes through the remaining 6.0 Ω branch. That branch current has risen from 1.0 A to 1.20 A, even though source current has fallen from 1.50 A to 1.20 A.
R1 now has a smaller drop, 1.20 x 4.0 = 4.8 V. This leaves 7.2 V across the 6.0 Ω resistor. The changed voltage share explains its larger current. It was never connected directly across the ideal source.
Optional check An ideal 12.0 V source supplies a 4.0 ohm resistor in series with parallel 6.0 ohm and 12.0 ohm branches. Initially the source current is 1.50 A and the 6.0 ohm branch current is 1.0 A. Only the 12.0 ohm branch is then opened. What happens?
Test the rules with paired measurements
Use a suitable low-voltage cell or battery, switch and appropriately rated resistors. Disconnect the source while changing connections. Use components whose heating is small enough for the intended constant-resistance comparison.
Match a voltage reading to the correct current reading
These are expected readings for the supplied low-voltage models, not experimental records. Meters are ideal: negligible ammeter resistance and negligible voltmeter current.
Measure the complete series network
Whole-network resistance: 3.00/0.100 = 30.0 ohm. The voltmeter spans both resistors, so its 3.00 V is not either single resistor drop.
Measure the complete parallel network
Whole-network resistance: 3.00/0.450 = 6.67 ohm. The main ammeter reads the sum of the two branch currents.
For the supplied 3.00 V model:
- Series: total current is 0.100 A. The 10.0 Ω resistor has 1.00 V across it and the 20.0 Ω resistor 2.00 V. Whole-network resistance is 3.00/0.100 = 30.0 Ω.
- Parallel: currents are 0.300 A through 10.0 Ω and 0.150 A through 20.0 Ω. Total current is 0.450 A, giving whole-network resistance 3.00/0.450 = 6.67 Ω.
These are expected values for the stated model. Record the actual current and terminal p.d. in an investigation. A real battery need not maintain exactly the same terminal voltage when the circuit changes.
- Choose the measurement. Put an ammeter in a main lead for total current, or in one selected branch for its current. Put the voltmeter across the exact component or group being investigated.
- Select the ranges and polarity. Use a suitable d.c. range. Conventional current should enter the ammeter's positive terminal; the voltmeter's positive lead goes to the higher-potential point. Check the zero and read an analogue scale without parallax. A nominal 3 V battery can exceed a 3 V range; establish a suitable range before choosing a finer one.
- Record positions with readings. Label the component or node pair and include units. Compare main current with the sum of branch currents, or complete p.d. with the series drops.
- Determine resistance from a matched pair. For one component, divide its p.d. by its current. For the whole network, divide the p.d. across A/C by total current entering it.
When moving one meter between positions, readings are taken at different times. Source drift and resistor heating can affect their comparison. Check the supply and allow conditions to settle; do not assume every discrepancy proves a failure of the circuit rule.
Loose contacts or contact resistance can change readings. Meter resistance, finite resolution and changes in temperature are separate limitations. Repeat readings to assess scatter, but correct a wrong connection or range directly. Finer resolution does not by itself guarantee better accuracy.
Brightness is not a calibrated current reading. A lamp's resistance changes as its filament heats. Do not treat a fixed-resistor calculation as an exact prediction for any lamp.
05
Potential dividers and the potentiometer
A potential divider provides an output voltage from part of a series resistance. Name the output terminals before deciding which resistance sets the output.
Series resistances carry the same current when no current branches off between them. Their voltage drops add to the supply p.d. Here C is labelled 0 V as a reference; this does not require a physical connection to Earth.
Derive the voltage share
Connect an upper resistance Rupper from A to B and a lower resistance Rlower from B to C. The supply is across A/C. Take the output VBC across the lower resistance, with B measured relative to C.
If the output draws negligible current, the same current passes through both resistances:
The numerator is the resistance across the chosen output. If output were measured across the upper resistance instead, its resistance would go in the numerator. The two drops add to Vs.
Move a potentiometer's wiper
A potentiometer has a resistive track with two end connections and a movable contact called the wiper. Connect the track ends A and C across the supply. The wiper is B, and output is taken between B and C.
Take the output between the wiper B and reference C
The uniform 10.0 kohm track is connected across 6.0 V at A/C. The output meter takes negligible current. The arrow touching the track is a connected wiper, not current flow.
Wiper one quarter of the track from C
V_BC = 1.50 V. The lower portion's resistance is 2.50 kohm. C is a chosen reference, not an Earth connection.
Wiper three quarters of the track from C
V_BC = 4.50 V. The lower portion's resistance is 7.50 kohm. C is a chosen reference, not an Earth connection.
Moving the wiper changes how much of the track lies between B and C. The complete A/C track resistance stays the same. In the ideal unloaded model, output ranges from 0 V when B is at C to the full supply voltage when B is at A.
Use the named reference end
One quarter of a uniform track from 0 V
The track is 10.0 kΩ across 6.0 V. Here 1 kΩ = 1000 Ω.
One quarter from C gives Rlower = 2.50 kΩ and Rupper = 7.50 kΩ. Thus VBC = 6.0 x 2.50/(7.50 + 2.50) = 1.50 V.
At three quarters from C, the lower share is 7.50/10.0, giving 4.50 V. The track current is 6.0/10 000 = 0.000600 A = 0.600 mA at either position in this unloaded model.
The position-to-voltage relationship is linear here because the track has uniform resistance per unit length. Do not assume that every type of potentiometer has the same relationship.
A two-terminal variable resistor instead includes a changing amount of resistance in one series path. A potentiometer used as a divider needs both track ends and the wiper connection; it changes the voltage share rather than the resistance of the whole connected track.
Keep the output-current condition
A high-resistance voltmeter or suitable input circuit draws negligible current, so it approximately preserves the simple divider ratio. A significant load connected across B/C becomes another parallel path and changes the circuit.
For example, two 2 kΩ arms across 6 V give an unloaded output of 3 V. Adding a 2 kΩ load across the lower arm makes the lower parallel equivalent 1 kΩ. The new output is 6 x 1/(2 + 1) = 2 V. Apply the whole-circuit method when the loading cannot be ignored.
Optional check A uniform 10 kilohm potentiometer track is connected across 6.0 V, from A at 6.0 V to C at 0 V. Its wiper B is 80% of the resistive length from C towards A. What is the output V_BC when the output draws negligible current?
06
Thermistor and LDR circuits
A sensor changes resistance when its surroundings change. In a powered potential divider, that resistance change produces a changing output voltage.
For an unloaded divider, the output is the supply voltage multiplied by the fraction of the total resistance across the chosen output. Check which terminals are used before predicting whether the voltage rises or falls.
- NTC thermistor
- Its resistance decreases as temperature increases. NTC means negative temperature coefficient. This behaviour differs from the rising resistance of a hotter ordinary metallic conductor.
- Light-dependent resistor: LDR
- Its resistance decreases as the light falling on it increases. Its symbol's incoming arrows represent illumination.
These trends do not establish a straight-line or inverse-proportional relationship. Use supplied resistance values or a given characteristic for a calculation.
As an input transducer, the sensor lets a physical input such as temperature or light affect an electrical signal. The sequence is: surroundings change, sensor resistance changes, voltage sharing changes, output voltage changes. The source supplies the circuit's energy; the sensor does not provide a universal on/off voltage by itself.
A sensor changes the voltage share across a named output
Both supplied divider circuits keep the source at 6.0 V and take negligible output current. Read the output between B and C; the sensor is in a different arm in each case.
NTC thermistor below a fixed upper resistor
Supplied 20 deg C condition: NTC 6.0 kohm, output 4.0 V. At 40 deg C: NTC 2.0 kohm, output 2.4 V. The series current changes between conditions.
LDR above a fixed lower resistor
Supplied dim condition: LDR 12.0 kohm, output 1.5 V. Brighter condition: LDR 2.0 kohm, output 4.0 V. Output is across the fixed lower resistor.
An NTC in the lower arm
A fixed 3.0 kΩ resistor connects A/B. The NTC connects B/C, and output is across the NTC. Its supplied resistance is 6.0 kΩ at 20 °C and 2.0 kΩ at 40 °C.
Temperature to output voltage
Warm the NTC from 20 °C to 40 °C
At 20 °C: total resistance is 3.0 + 6.0 = 9.0 kΩ. Current is 6.0/9000 = 0.000667 A, or 0.667 mA. Output VBC = 6.0 x 6.0/(3.0 + 6.0) = 4.0 V.
At 40 °C: total resistance is 3.0 + 2.0 = 5.0 kΩ. Current is 6.0/5000 = 0.00120 A, or 1.20 mA. Output VBC = 6.0 x 2.0/(3.0 + 2.0) = 2.4 V.
Warming reduces the lower arm's resistance and its share of the supply voltage. The output falls even though the series current rises. The fixed upper resistor's p.d. rises from 2.0 V to 3.6 V, keeping the total at 6.0 V.
This changing voltage could be an input to a temperature monitor or alarm. The controller's required input and switching condition would need to be specified; the thermistor alone does not determine them.
An LDR in the upper arm
The LDR now connects A/B, with a fixed 4.0 kΩ resistor at B/C. Output is across the fixed lower resistor. The supplied LDR resistance is 12.0 kΩ in a dim condition and 2.0 kΩ in a brighter condition.
Light to output voltage
Increase the illumination
Dim: total resistance is 12.0 + 4.0 = 16.0 kΩ. Current is 6.0/16 000 = 0.000375 A = 0.375 mA. Output VBC = 6.0 x 4.0/(12.0 + 4.0) = 1.5 V.
Brighter: total resistance is 2.0 + 4.0 = 6.0 kΩ. Current is 6.0/6000 = 0.0010 A = 1.0 mA. Output VBC = 6.0 x 4.0/(2.0 + 4.0) = 4.0 V.
More light lowers the upper resistance, giving the fixed lower resistor a larger share of the supply. The output rises. The LDR's own p.d. instead falls from 4.5 V to 2.0 V.
A light-level controller can use this voltage as its input. Moving the sensor to the other arm, or measuring across the other component, can reverse the output trend. Name the resistance that changes and the terminals being measured.
Compare readings under controlled conditions
Use supplied data or a suitable low-voltage arrangement with the output meter across the named terminals. Keep the source voltage fixed. Allow the sensor's response to settle before recording the input condition and output voltage.
For an NTC comparison, measure temperature with a suitable thermometer and keep other conditions consistent. Electrical current can also warm the thermistor, so do not attribute every change to the intended external temperature alone. A voltage response needs calibration before it can be read as a temperature scale.
For an LDR comparison, change the illumination while limiting unwanted heating and keeping other light sources consistent. These supplied resistance values describe the example; they are not a calibration for every LDR or thermistor.
The current can change between conditions. Both arms carry the same current within one unloaded-divider state, but changing a sensor's resistance changes the total resistance and therefore the current. Recalculate it or use the correctly referenced voltage ratio.
Optional check A 6.0 V divider has an NTC thermistor as its upper arm and a fixed 3.0 kilohm resistor as its lower arm. Output is across the fixed lower resistor. Warming reduces the NTC resistance from 6.0 to 2.0 kilohm. What happens to the unloaded output?
Revision summary
Start with connections
Use the component symbols and trace their endpoints. A junction connects wires. Series components share an unsplit path current; parallel branches share both endpoint potentials. A rearranged drawing keeps the same circuit only if all connections remain unchanged.
- Series
- I = I1 = I2 = ...; Vs = V1 + V2 + ...; Rseries = R1 + R2 + ... . Equal current does not mean equal voltage drops.
- Parallel
- V1 = V2 = ... = Vgroup; Itotal = I1 + I2 + ...; 1/Rparallel = 1/R1 + 1/R2 + ... . Take the reciprocal after summing. For two resistors only, Rparallel = R1R2/(R1 + R2).
- One component or one equivalent group
- V = IR, using its own p.d. and current. In these fixed-resistance examples, take wires and source internal resistance as negligible. Vs is the p.d. across the complete external network.
Solve and measure a complete circuit
- Identify the nodes and a valid series or parallel group.
- Reduce the group while preserving its external terminals.
- Find total resistance and source current.
- Work back to group voltages and individual currents.
- Check junction current sums and voltage drops along a full path.
An ammeter connects in the measured path; a voltmeter connects across the measured endpoints. Whole-network resistance needs the whole p.d. and total current. A component needs its own pair of readings. Ideal meters have negligible ammeter resistance and negligible voltmeter current.
Opening a branch directly across an ideal fixed-voltage source leaves the other branch currents unchanged. Opening a branch inside a larger network can change the voltage across the remaining branch. Recalculate the connections before predicting the effect.
Record actual readings with units, suitable ranges and named positions. Account for contact resistance, source drift and heating. 1 mA = 0.001 A; 1 mV = 0.001 V; 1 kΩ = 1000 Ω.
Potential divider
With negligible output current, Vout = Vs x resistance across output / total series resistance. For output B/C across the lower arm, use Rlower in the numerator.
A potentiometer connects both track ends across the supply and takes an output from the wiper relative to one end. A uniform unloaded track gives a voltage proportional to the resistive length from that reference end. A significant load changes the circuit.
Sensors as input transducers
- NTC: higher temperature gives lower resistance.
- LDR: more incident light gives lower resistance.
- Output: find which arm changes and which terminals are measured. Across a sensor its voltage share can fall while the complementary fixed-resistor share rises.
- Calculation: use supplied resistance values and the fixed supply. Current changes when total resistance changes; the sensor does not supply the circuit's energy.