Full chapter
D.C. circuits
All 4 topics and the revision summary on one page.
01
Circuit symbols and connections
Make a guess. You are not marked.
Two wires cross on a circuit diagram, with no dot where they cross. What does that mean?
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They are not connected; the current in each stays on its own wire.
A junction is shown by a dot. Wires that cross without a dot pass over each other, so no current flows from one to the other.
In short
- Use the standard symbols for cells, lamps, resistors, switches and meters.
- A dot shows a junction, where wires join.
- Ammeters go in series; voltmeters go in parallel.
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.
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.
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.
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.
Check your understanding
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?
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R2 and R3 are in parallel with each other, and R1 is in series with their combination.
R2 and R3 share both endpoints, B and C. R1 carries the current entering or leaving that whole parallel group. Moving symbols on the page does not change these connections.
02
Series circuits
Make a guess. You are not marked.
Two identical lamps are connected in series with a cell and both glow. One lamp is unscrewed from its holder. What happens to the other lamp?
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It goes out.
A series circuit has one path. Removing a lamp breaks that path, so no current flows anywhere in the circuit and the other lamp goes out.
In short
- In series, the current is the same everywhere.
- The voltages across the components add up to the supply voltage.
- Total resistance = R1 + R2 + ...
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.
Try it step by step
Work it out in steps: Total resistance in ohm, then Current in A, then p.d. across the first resistor in V.
Show the answer
Problem 1.
Total resistance in ohm: R = 3.0 + 9.0 = 12 ohm
Current in A: I = V / R = 6.0 / 12 = 0.50 A
p.d. across the first resistor in V: V = IR = 0.50 x 3.0 = 1.5 V (2 sf)
Problem 2.
Total resistance in ohm: R = 5.0 + 15 = 20 ohm
Current in A: I = 10 / 20 = 0.50 A
p.d. across the first resistor in V: V = 0.50 x 5.0 = 2.5 V (2 sf)
Problem 3.
Total resistance in ohm: R = 8.0 + 4.0 = 12 ohm
Current in A: I = 9.0 / 12 = 0.75 A
p.d. across the first resistor in V: V = 0.75 x 8.0 = 6.0 V (2 sf)
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.
Check your understanding
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?
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6.0 V
Total resistance is 3.0 + 6.0 = 9.0 ohm, so the shared current is 9.0/9.0 = 1.0 A. The 6.0 ohm resistor has V = IR = 1.0 x 6.0 = 6.0 V; the other has 3.0 V.
03
Parallel circuits
Make a guess. You are not marked.
One lamp is connected across a cell. A second identical lamp is now connected in parallel with the first. What happens to the current drawn from the cell?
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It doubles.
Each lamp still has the full cell p.d. across it, so each draws the same current as before. The cell now supplies both currents, twice as much in total.
In short
- In parallel, each branch has the full supply voltage.
- The branch currents add up to the main current.
- Adding a branch lowers the total resistance.
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.
Try it step by step
Work it out in steps: Current in the smaller resistor in A, then Current in the larger resistor in A, then Total current in A, then Combined resistance in ohm.
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Problem 1.
Current in the smaller resistor in A: I1 = 6.0 / 3.0 = 2.0 A
Current in the larger resistor in A: I2 = 6.0 / 6.0 = 1.0 A
Total current in A: I = 2.0 + 1.0 = 3.0 A
Combined resistance in ohm: R = V / I = 6.0 / 3.0 = 2.0 ohm (check: 1/3 + 1/6 = 1/2)
Problem 2.
Current in the smaller resistor in A: I1 = 12 / 4.0 = 3.0 A
Current in the larger resistor in A: I2 = 12 / 12 = 1.0 A
Total current in A: I = 3.0 + 1.0 = 4.0 A
Combined resistance in ohm: R = 12 / 4.0 = 3.0 ohm (2 sf)
Problem 3.
Current in the smaller resistor in A: I1 = 10 / 5.0 = 2.0 A
Current in the larger resistor in A: I2 = 10 / 20 = 0.50 A
Total current in A: I = 2.0 + 0.50 = 2.5 A
Combined resistance in ohm: R = 10 / 2.5 = 4.0 ohm (2 sf)
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.
Check your understanding
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?
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3.0 A
Each branch has 6.0 V. Their currents are 6.0/3.0 = 2.0 A and 6.0/6.0 = 1.0 A, so the source supplies 2.0 + 1.0 = 3.0 A.
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.
In short
- Replace a parallel group by its combined resistance. Add the series resistances to get the total.
- Source current = supply voltage / total resistance.
- Work back: find the voltage across each part with V = IR, then the current in each branch.
- Check: branch currents add up to the main current, and voltages around each loop add up to the supply.
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. Inside the group, each component still has its own current and its own voltage, so work back to find them.
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.
Try it step by step
Work it out in steps: Parallel group resistance in ohm, then Total resistance in ohm, then Source current in A, then p.d. across the series resistor in V.
Show the answer
Problem 1.
Parallel group resistance in ohm: Group: 1/R = 1/3.0 + 1/6.0 = 1/2.0, so R = 2.0 ohm
Total resistance in ohm: Total = 2.0 + 2.0 = 4.0 ohm
Source current in A: I = 8.0 / 4.0 = 2.0 A
p.d. across the series resistor in V: V = IR = 2.0 x 2.0 = 4.0 V (the group takes the other 4.0 V)
Problem 2.
Parallel group resistance in ohm: Group: (4.0 x 12) / (4.0 + 12) = 3.0 ohm
Total resistance in ohm: Total = 3.0 + 3.0 = 6.0 ohm
Source current in A: I = 9.0 / 6.0 = 1.5 A
p.d. across the series resistor in V: V = 1.5 x 3.0 = 4.5 V (2 sf)
Problem 3.
Parallel group resistance in ohm: Group: two equal 10 ohm resistors in parallel give 5.0 ohm
Total resistance in ohm: Total = 6.0 + 5.0 = 11 ohm
Source current in A: I = 11 / 11 = 1.0 A
p.d. across the series resistor in V: V = 1.0 x 6.0 = 6.0 V (2 sf)
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.
Check your understanding
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?
Show the answer
Source current falls to 1.20 A, while the 6.0 ohm branch current rises to 1.20 A.
The remaining path has 4.0 + 6.0 = 10.0 ohm, so its current is 12.0/10.0 = 1.20 A everywhere. The 4.0 ohm drop falls to 4.8 V, leaving 7.2 V across the 6.0 ohm resistor. Total current falls even though this branch current rises.
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.
More detail (optional)
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.
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.
Back to circuit symbols and connectionsPractise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
Two resistors are connected in parallel to a cell. State how the currents in the resistors are related to the current from the cell, and how the potential differences across them compare. [2]
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Exam-style question
A 12 V battery of negligible internal resistance is connected to a 2.0 Ω resistor in series with a parallel pair of resistors of 6.0 Ω and 3.0 Ω. Calculate the current from the battery and the current in the 6.0 Ω resistor. [4]
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Exam-style question
A 4.0 Ω resistor and an 8.0 Ω resistor are connected in series to a 6.0 V battery of negligible internal resistance. Calculate the current in the circuit and the potential difference across the 8.0 Ω resistor. [3]
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Quick recall
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?