K323 / 2027
D.C. circuits overview

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

Cell circuit symbolSupplies energy per charge. The longer plate is positive.+-

Supplies energy per charge. The longer plate is positive.

Battery

Battery circuit symbolTwo or more cells form the source shown here.+-

Two or more cells form the source shown here.

Switch

Switch circuit symbolOpens or closes a conducting path; shown open.

Opens or closes a conducting path; shown open.

Lamp

Lamp circuit symbolTransfers electrical energy into light and internal energy.

Transfers electrical energy into light and internal energy.

LED

LED circuit symbolEmits light in forward operation. The bar marks cathode K.K

Emits light in forward operation. The bar marks cathode K.

Fixed resistor

Fixed resistor circuit symbolProvides resistance in the conducting path.

Provides resistance in the conducting path.

Variable resistor

Variable resistor circuit symbolAdjusts resistance in a two-terminal path.

Adjusts resistance in a two-terminal path.

Fuse

Fuse circuit symbolIts link can melt and open the circuit if current is excessive.

Its link can melt and open the circuit if current is excessive.

Ammeter

Ammeter circuit symbolMeasures current through the path in which it is inserted.A

Measures current through the path in which it is inserted.

Voltmeter

Voltmeter circuit symbolMeasures p.d. between the two points to which it connects.V

Measures p.d. between the two points to which it connects.

D.c. supply

D.c. supply circuit symbolProvides a supply with the indicated fixed polarity.+-

Provides a supply with the indicated fixed polarity.

A.c. supply

A.c. supply circuit symbolProvides a supply whose polarity alternates.

Provides a supply whose polarity alternates.

Potentiometer

Potentiometer circuit symbolTwo track ends and a wiper provide three connections.

Two track ends and a wiper provide three connections.

Electric bell

Electric bell circuit symbolProduces sound when operated by the circuit.

Produces sound when operated by the circuit.

LDR

LDR circuit symbolResistance falls as more light reaches it; arrows point in.

Resistance falls as more light reaches it; arrows point in.

NTC thermistor

NTC thermistor circuit symbolResistance falls as its temperature rises.

Resistance falls as its temperature rises.

Each card shows a circuit symbol, its name and its role. A component's symbol identifies it; the wires attached to its terminals determine its place in the circuit.

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

One drawing of the networkR1 joins A to B. R2 joins B to C along the upper path and R3 joins the same B and C nodes along the lower path. The vertical wires at B and C connect the respective upper and lower junctions. There is no plain wire directly from B to C.ABCR1R2R3R1: A/B; R2 and R3: B/C

Move R3 around the outside

Move R3 around the outsideThe same network is redrawn. A joins the left terminal of R1. R1 ends at B. R2 connects directly from B to C. From B, a wire rises to the right terminal of R3; the left terminal of R3 connects around the left and bottom edges to C. The outer path contains R3 and is not a short circuit. The drawing has no unmarked wire crossing.ABCR1R2R3The named connections are unchanged
Both drawings keep R1 between A and B, and R2 and R3 each between B and C. R2 and R3 are therefore parallel. R1 is in series with that whole parallel combination, regardless of where the symbols sit on the page.
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?
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.

I = I1 = I2 = ...Same unbranched path: the same current passes every series component.

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.

Vs = V1 + V2 + ...Equal current does not require equal voltages across unequal resistances.

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.

Rseries = R1 + R2 + ...Adding a positive resistance in series increases the total resistance.

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

One path: same current through both resistorsAn ideal 6.0 V source drives a single closed path through R1: 2 ohm between A and B and R2: 4 ohm between B and C. The same 1.0 A passes through both. The p.d. across A/B is 2.0 V and across B/C is 4.0 V. The bottom switch is closed.+-6.0 VR1: 2 ohmR2: 4 ohm2.0 V4.0 VABCI = 1.0 ASwitch closed

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

Open the only pathA 6.0 V ideal source and two series resistors have their only path opened by the bottom switch. Steady current is zero throughout the circuit and each resistor has zero p.d. The source still has 6.0 V. In this ideal model, the right switch contact is 6.0 V above the left one, so the open gap has a 6.0 V p.d.+-6.0 VR1: 2 ohmR2: 4 ohm0 V0 VCurrent = 0 AOpen switch: 6.0 V across the gap

Zero steady current does not mean zero source voltage. The only path is broken.

The closed 6.0 V circuit has one path through 2.0 Ω and 4.0 Ω. Its current is 1.0 A, with voltage drops of 2.0 V and 4.0 V. The separate open-switch view has zero steady current while the ideal source still maintains its voltage.

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.

  1. Total resistance = 2.0 + 4.0 = 6.0 Ω.
  2. Shared current = 6.0/6.0 = 1.0 A.
  3. Across 2.0 Ω: V = 1.0 x 2.0 = 2.0 V.
  4. 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?
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.

V1 = V2 = ... = VgroupEach branch spans the same two potentials. If the group is directly across the supply, Vgroup = Vs.

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.

Itotal = I1 + I2 + ...At equal branch p.d., a smaller resistance carries a larger current. The split need not be equal.

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

Reconnect the two resistors across the same two nodesAn ideal 6.0 V source connects to nodes A and C. A 2.0 ohm resistor and a 4.0 ohm resistor each connect from A to C, so each has 6.0 V. The upper branch carries 3.0 A and the lower branch 1.5 A, both from A towards C. The source current is 4.5 A. There is no wire directly bridging the two nodes.+-6.0 VAC2.0 ohm4.0 ohm4.5 A3.0 A; 6.0 V1.5 A; 6.0 VThe currents split and then recombine

The equivalent resistance is 4/3 ohm, about 1.33 ohm. This is a different connection from the 6.0 ohm series circuit.

The 2.0 Ω and 4.0 Ω resistors have been reconnected in parallel across 6.0 V. Each branch now has the full 6.0 V: their currents are 3.0 A and 1.5 A, giving 4.5 A in the main lead.

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:

1/Rparallel = 1/R1 + 1/R2 + ...Find the sum of reciprocals, then take its reciprocal to obtain resistance.

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?
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

Start with the complete networkA 12.0 V ideal source connects A to C through the complete external network. R1 is 4.0 ohm between A and B. R2 is 6.0 ohm between B and C, and R3 is 12.0 ohm between the same B and C nodes. R1 carries the source current, 1.50 A. R2 carries 1.0 A and R3 carries 0.50 A. Each series group has 6.0 V across it.+-12.0 VABCR1: 4.0 ohmR2: 6.0 ohmR3: 12.0 ohm1.50 A1.0 A0.50 AV_AB = 6.0 V; V_BC = 6.0 V

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

Replace the B/C group by 4.0 ohmThe original 6.0 ohm and 12.0 ohm parallel branches between B and C are replaced by one 4.0 ohm equivalent. R1 remains 4.0 ohm between A and B. A and C are still the external source terminals. The two groups are in series, giving 8.0 ohm in total and a source current of 1.50 A at 12.0 V.+-12.0 VR1: 4 ohmB/C: 4 ohm6.0 V6.0 VABCI = 1.50 ASwitch closed

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

Reduce the complete A/C network to 8.0 ohmThe complete network is replaced by an 8.0 ohm equivalent between A and C across the same 12.0 V ideal source. The source current is 1.50 A. B was an internal connection and is absent after this final replacement; no separate component voltage is implied by the single equivalent resistor.+-12.0 VAC8.0 ohm1.50 A

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

Open only the 12 ohm branchThe 12.0 V source still supplies R1, 4.0 ohm, between A and B. The 6.0 ohm branch between B and C remains connected. Only the 12.0 ohm branch has an open switch. Source current is now 1.20 A, all of which passes through the remaining 6.0 ohm branch. The open branch has no steady current. V_AB is 4.8 V and V_BC is 7.2 V.+-12.0 VABCR1: 4.0 ohmR2: 6.0 ohmR3: 12.0 ohm1.20 A1.20 A0 AV_AB = 4.8 V; V_BC = 7.2 V

The total current falls, but the current through the 6.0 ohm branch rises. Its shared series resistor changes the branch voltage.

R1 = 4.0 Ω connects A/B; the 6.0 Ω and 12.0 Ω branches each connect B/C. Replace only the B/C group first. The final comparison opens just the 12.0 Ω branch while leaving the 4.0 Ω and 6.0 Ω path intact.

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.

  1. 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.
  2. Find the whole resistance: 4.0 + 4.0 = 8.0 Ω across A/C.
  3. Find source current: 12.0/8.0 = 1.50 A. This is also the current through R1.
  4. Find group voltages: VAB = 1.50 x 4.0 = 6.0 V. Therefore VBC = 12.0 - 6.0 = 6.0 V.
  5. 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?
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

Measure the complete series networkThe source drives an ammeter in the main positive lead before node A. A 10.0 ohm resistor connects A to B and a 20.0 ohm resistor connects B to C. A voltmeter connects across A and C, the complete pair. With ideal meters its reading is 3.00 V and the series ammeter reads 0.100 A. The individual resistor drops are 1.00 V and 2.00 V. The source return contains a closed switch.+-3.00 VA0.100 A+-B10.0 ohm20.0 ohm1.00 V2.00 VVAC+-3.00 VSwitch closed

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

Measure the complete parallel networkThe source drives an ammeter in the main positive lead before node A. A 10.0 ohm resistor and a 20.0 ohm resistor are in separate branches from A to C. A voltmeter connects across A and C, the complete network. With ideal meters and 3.00 V across A/C, the branch currents are 0.300 A and 0.150 A and the main ammeter reads 0.450 A. The source return contains a closed switch.+-3.00 VA0.450 A+-10.0 ohm20.0 ohm0.300 A0.150 AVAC+-3.00 VSwitch closed

Whole-network resistance: 3.00/0.450 = 6.67 ohm. The main ammeter reads the sum of the two branch currents.

These separate measurement examples use 10.0 Ω and 20.0 Ω with a supplied terminal p.d. of 3.00 V. The main ammeter measures total current; the voltmeter across A/C measures the complete network's p.d. Component measurements need that component's own current and endpoints.

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.

  1. 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.
  2. 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.
  3. 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.
  4. 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:

I = Vs/(Rupper + Rlower)VBC = IRlower = Vs x Rlower/(Rupper + Rlower).

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

Wiper one quarter of the track from CA uniform 10.0 kilohm potentiometer track connects A at 6.0 V to C at the chosen zero-volt reference. Wiper B touches it one quarter of the resistive length from C. B is a separate third connection. An ideal voltmeter measures B relative to C and reads 1.50 V. The output takes negligible current; the whole track resistance remains 10.0 kilohm.+-6.0 VVA: 6.0 VBC: 0 V reference+-7.50 kohm2.50 kohm

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

Wiper three quarters of the track from CA uniform 10.0 kilohm potentiometer track connects A at 6.0 V to C at the chosen zero-volt reference. Wiper B touches it three quarters of the resistive length from C. B is a separate third connection. An ideal voltmeter measures B relative to C and reads 4.50 V. The output takes negligible current; the whole track resistance remains 10.0 kilohm.+-6.0 VVA: 6.0 VBC: 0 V reference+-2.50 kohm7.50 kohm

V_BC = 4.50 V. The lower portion's resistance is 7.50 kohm. C is a chosen reference, not an Earth connection.

A uniform 10.0 kΩ track spans A at 6.0 V to C at 0 V. With negligible output current, a wiper one quarter of the resistive length from C gives VBC = 1.50 V; three quarters from C gives 4.50 V.

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?
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

NTC thermistor below a fixed upper resistorA fixed 6.0 V source is across A and C. A fixed 3.0 kilohm resistor is the upper arm from A to B; an NTC thermistor is the lower arm from B to C. The thermistor has a diagonal temperature-dependent symbol without an adjustment arrowhead. The ideal output voltmeter is between B and C, across the NTC. When the supplied NTC resistance falls from 6.0 to 2.0 kilohm as temperature rises, V_BC falls from 4.0 to 2.4 V.+-6.0 VVA: 6.0 VBC: 0 V reference+-Fixed3.0 kohmNTC

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

LDR above a fixed lower resistorA fixed 6.0 V source is across A and C. An LDR is the upper arm from A to B; a fixed 4.0 kilohm resistor is the lower arm from B to C. Incident-light arrows point towards the LDR symbol. The ideal output voltmeter is between B and C, across the fixed resistor, not across the sensor. When the supplied LDR resistance falls from 12.0 to 2.0 kilohm, V_BC rises from 1.5 to 4.0 V.+-6.0 VVA: 6.0 VBC: 0 V reference+-LDRFixed4.0 kohm

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.

Both supplied circuits use a fixed 6.0 V source and negligible output current. The NTC is the lower arm in the first circuit; the LDR is the upper arm in the second. In both, VBC is measured across the lower arm.

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?
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

  1. Identify the nodes and a valid series or parallel group.
  2. Reduce the group while preserving its external terminals.
  3. Find total resistance and source current.
  4. Work back to group voltages and individual currents.
  5. 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.
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