Full chapter
Circuits
All 6 topics and the revision summary on one page.
01
Read connections and place meters
The connections determine the circuit. Moving a component on the page changes nothing if both of its terminals still connect to the same nodes.
A node is a set of points joined by ideal wire without an intervening component. They share one potential. A junction dot shows a connection; an unconnected crossing needs a clear gap or bridge. Do not decide whether two components are in series or parallel from their visual alignment alone.
Recognise the component and its terminals
Cell
The longer plate is the positive terminal.
Battery
Connected cells provide the indicated overall polarity.
D.c. supply
The two terminals have the labelled fixed polarity.
Switch
The open gap breaks this conducting path.
Fixed resistor
Resistance belongs to the component between its two terminals.
Variable resistor
The diagonal adjustment arrow denotes an adjustable two-terminal resistance.
Ammeter
Insert it in the current path being measured.
Voltmeter
Connect its terminals to the two points whose p.d. is required.
Filament lamp
Its settled resistance changes as the filament warms.
NTC thermistor
The marking identifies a thermistor; this NTC model has lower resistance when warmer.
LDR
The two arrows represent incoming light, not current.
Semiconductor diode
The bar identifies the cathode K; the other terminal is the anode.
Potentiometer
Two track ends and a separate wiper make three terminals.
A cell's long plate is positive and its short plate negative; a battery contains multiple cells. An open switch breaks its path. A semiconductor diode's bar marks the cathode, while the other terminal is the anode. Its symbol does not mean that current always flows: the voltage polarity and component behaviour matter.
A two-terminal variable-resistor connection changes the resistance in its path. A three-terminal potentiometer has two track ends and a sliding wiper; using both ends and the wiper gives a selectable potential-divider output. A thermistor changes resistance with temperature, while an LDR responds to illumination. Their symbols do not specify one universal resistance or switching threshold.
Redraw the same network
Let R1 join A to B. Let R2 and R3 each join B to C. The two latter resistors are parallel because they share both B and C. R1 carries the total current before it divides.
Read the terminal pairs
A node can be drawn in several places. Follow the uninterrupted wire and the named endpoints.
The unfamiliar drawing is equivalent
Changing the shape or position of a branch changes nothing if its terminal pair is preserved.
To check an unfamiliar drawing, trace each continuous wire and label its node. List the terminal pair of each component. A parallel pair needs equal p.d. because it shares both endpoints; a series pair needs an unbranched path between its components so they carry the same current.
Adding a wire directly from B to C would be a new connection, not a harmless redrawing. It would bypass both resistors and make VB - VC zero in the ideal-wire model.
Match the meter to the quantity
An ammeter goes in the selected current path. A voltmeter connects across the selected endpoints. To find a component's resistance, pair its own current with its own p.d. To find the equivalent resistance of the complete external network, use total entering current and VA - VC.
Ideal calculations assume negligible ammeter resistance and negligible voltmeter current. A real voltmeter can draw current and change a high-resistance circuit; a real ammeter can introduce a voltage drop. Check instrument range, resolution, polarity and loading against the intended measurement.
A source model can include an ideal e.m.f. between C and a private internal node D, followed by internal resistance from D to external terminal A. Its terminal voltmeter belongs across A/C. D/C spans only the ideal e.m.f. inside that model. The source calculation explains the difference.
Optional check R1 joins A to B, while R2 and R3 each join B to C. A source has external terminals A/C and a private internal node D before its internal resistance. Which redraw and terminal-voltage measurement preserve the intended circuit?
02
Resistance, material and geometry
Resistance belongs to a component at a stated operating condition. Resistivity describes its material at a stated temperature; length and cross-sectional area then determine a uniform wire's resistance.
For nonzero current, define resistance using the p.d. across the component and the current through it:
1 Ω = 1 V/A = kg m2 s-3 A-2
The rearrangement V = IR does not establish that R is constant. An ohmic component has proportional I and V under constant physical conditions. On an I-against-V graph its straight-line gradient is 1/R. For a curved non-ohmic characteristic, V/I at a point is not generally the reciprocal tangent gradient.
Use the conducting cross-section
For a uniform wire of electrical length l, cross-sectional area A and resistivity ρ:
ρ = RA/l
A = πd2/4 (circular wire of diameter d)
Resistivity has unit Ω m; the symbol ρ here does not mean mass density. A is the area perpendicular to current, not the curved outer surface of the wire. The length is the separation of the relevant electrical contacts. State uniform material and temperature, and assess lead/contact contributions before neglecting them.
A rough p.d. of 0.4 V and current of 0.1 A suggest a few ohms. A diameter of order 4 × 10-4 m gives area of order 10-7 m2. For a roughly metre-long wire, resistivity is then of order 10-7 to 10-6 Ω m. This estimate helps catch powers-of-ten errors before exact calculation.
Worked wire and selected readings
The source voltage is not the wire voltage
A supplied uniform wire has l = 0.800 m and d = 0.400 mm. In this worked model, a voltmeter reads 0.400 V across contacts P/Q and an ammeter reads 0.125 A through the wire. The measured source-terminal supply is 6.0 V; a series control takes the remaining voltage.
Use the wire readings, not the source reading
The control and wire share 0.125 A. The wire has 0.400 V across P/Q; using the source's 6.00 V would include the control resistance.
Double length and diameter at fixed resistivity
Length and diameter use different display scales, each shared by both wires. Length doubles but area quadruples; at unchanged resistivity the resistance is 1.60 ohm, half the original 3.20 ohm.
Resistance falls as the reciprocal of area
Doubling area halves resistance. Doubling diameter is different: it gives four times the area and one quarter of the resistance if length stays fixed.
d = 0.400 × 10-3 m
A = π(0.400 × 10-3)2/4
= 1.2566 × 10-7 m2
ρ = (3.20)(1.2566 × 10-7)/0.800
= 5.03 × 10-7 Ω m
Convert diameter to metres before squaring.
Rcontrol = 5.60/0.125 = 44.8 Ω
Neglecting meter and lead effects, the control and wire drops add to the 6.0 V terminal supply. Using 6.0/0.125 would give 48 Ω for the combined external path, not the wire's 3.20 Ω.
A second wire of the same material and temperature has twice the length and twice the diameter. Its area is four times as large:
Rnew = 1.60 Ω
Recognise the reciprocal relationship
Hold ρ and l fixed, and call the original area A0 and resistance R0. Then:
Area ratios 0.5, 1, 2 and 4 give resistance ratios 2, 1, 0.5 and 0.25. The positive-domain graph is a falling reciprocal curve, not a straight line. R against 1/A would instead be linear. R against diameter follows 1/d2, a different horizontal variable.
Determine resistivity from a wire
- Use a uniform wire and measure the contact separation l. Connect the voltmeter across those contacts and the ammeter in the wire path, with a suitable low-voltage source and series current control.
- Estimate the expected V and I, then choose actual available meter ranges and resolution. A 0-0.1 A range cannot contain the supplied 0.125 A current. Use modest current and check that heating does not appreciably change the wire's resistance.
- Use a suitable micrometer to measure diameter at several positions and orientations. Check its zero, use its contact mechanism consistently and retain the raw readings and instrument resolution.
- Vary the electrical length, recording matched V/I values to calculate R at each length. Plot R vertically against l; a gradient gives ρ/A, so multiply the gradient by measured A to infer ρ.
- Inspect any intercept and departure from a line. A roughly constant contact/lead contribution can produce a nonzero intercept; forcing the line through zero can conceal it. Changing temperature or diameter along the wire challenges the uniform model.
Because area contains d2, a small 1% diameter error produces about a 2% area error and, with R/l fixed, about a 2% resistivity error. Repetition can assess scatter; it cannot correct a persistent micrometer zero, using diameter as radius, or measuring the wrong contact separation.
Optional check A uniform wire has resistance 3.20 ohm. A second wire of the same material at the same temperature has twice the length and twice the diameter. What is its resistance?
03
I-V curves and temperature
A component characteristic connects its current and p.d. under specified conditions. Temperature, settling and polarity can change the relationship, so a graph needs more than the component's name.
Here current is vertical and p.d. horizontal. On a straight line through the origin, gradient I/V is 1/R. On a curved graph, calculate the operating resistance as V/I at that point. A local tangent describes a small change in current per change in voltage, which is a different ratio. If the axes are reversed, reconsider the gradient interpretation.
Ohmic conductor at fixed temperature
A constant gradient here means constant resistance: 20 ohm. Temperature and the other physical conditions are fixed.
A settled filament lamp
The filament warms as current increases. Read V/I at a point for its resistance; this curved graph has no single constant resistance.
A self-heating NTC thermistor model
These readings include self-heating. A low-current test that holds its temperature fixed is a different comparison. Notice the milliampere axis.
Diode: a qualitative characteristic
The axes have no numerical scale. The forward behaviour depends on the diode and its conditions; do not infer a universal turn-on voltage.
Ohmic resistor: fixed physical conditions
A straight line through the origin represents proportional I and V. A 20.0 Ω model gives 0.050, 0.100 and 0.150 A at 1.00, 2.00 and 3.00 V. Each V/I is 20.0 Ω. Significant heating would violate the fixed-temperature comparison.
Filament lamp: settled heating changes resistance
As voltage magnitude increases, current heats the metal filament. Its resistivity rises, so current increases less rapidly and the I-V curve becomes less steep. The supplied settled model gives:
At 2.00 V, I = 0.160 A: R = 12.5 Ω
At 3.00 V, I = 0.200 A: R = 15.0 Ω
A cold filament's initial response is not the same as its settled operating characteristic. Do not predict an exact lamp current by assuming it keeps one fixed resistance while heating.
Semiconductor diode: polarity matters
The diode conducts strongly in its forward direction once the applied voltage produces appreciable current. Reverse current is small over the shown range before breakdown. Do not mirror its forward branch to obtain the reverse branch or assign every diode one universal forward threshold.
The real characteristic is different from an ideal zero-drop conducting state and perfectly blocked reverse state. An ideal model is useful only when that approximation is stated.
NTC thermistor: identify the experiment
For a negative temperature coefficient thermistor, increasing temperature reduces resistance. In the supplied settled voltage sweep, self-heating increases as voltage magnitude rises, so current can rise progressively more steeply. At 1.00, 2.00 and 3.00 V, the model currents are 5.0, 14.0 and 30.0 mA, where 1 mA = 10-3 A:
R = 2.00/0.0140 ≈ 143 Ω
R = 3.00/0.0300 = 100 Ω
If temperature is instead externally controlled and self-heating is kept small, an approximately linear low-field I-V characteristic can be observed at that fixed temperature. These are different measurement conditions, not contradictory properties.
Why typical metals and NTC semiconductors differ
The carrier model I = nAvq separates carrier number density from mean drift speed. Compare the same geometry and applied electric field when discussing temperature:
- Typical metal: greater lattice vibration increases carrier scattering. Carrier number density stays roughly unchanged, but mean drift speed decreases at the same field. Current decreases at the same p.d., so resistance and resistivity increase. Faster random thermal motion does not imply faster directed drift.
- NTC semiconductor: warming makes more charge carriers available. Increased number density is the main explanation for its falling resistivity in this regime; it is not simply that the same carriers move faster. This is not a claim that every semiconductor resistance decreases under every condition.
If current is fixed instead of field, warming a typical metal requires a larger field and p.d. to maintain that current. State the controlled quantity before comparing the microscopic explanation with a circuit reading.
Obtain and interpret a characteristic
Use an adjustable low-voltage source or a suitable series control, an ammeter in the component path and a voltmeter across its terminals. Record actual I/V pairs with ranges, resolution and polarity. Choose suitable ratings and a limiting resistance for a diode. Reverse polarity deliberately for the reverse branch rather than changing graph signs without changing the circuit.
For a lamp characteristic, allow its thermal state to settle at each point. For a separate resistance-versus-temperature study, measure temperature, allow the sensor response to settle and keep unwanted electrical self-heating small. Record which condition was controlled; otherwise a temperature effect can be confused with a different voltage sweep.
Optional check Compare higher temperatures at the same applied electric field in a typical metal and in the stated NTC semiconductor. Which explanation is appropriate?
04
Sources with internal resistance
A source can transfer energy internally as well as to its external load. Its e.m.f. and its terminal p.d. therefore need not be equal while it supplies current.
The e.m.f. Es is energy supplied by the source per unit charge. The terminal p.d. Vterminal is energy transferred to the external circuit per unit charge. Both are measured in volts, or J/C. The subscript s distinguishes source e.m.f. from other uses of E; see the source-energy explanation.
Model the source as an ideal e.m.f. from C to D, followed by a resistance r from D to A. Current leaves external terminal A, passes through the load and returns to C. The internal drop is Ir:
Vterminal = Es - Ir
I = Es/(R + r) for external load R
These signs describe a discharging source supplying the load. Do not apply them unchanged when an external supply drives current back into a rechargeable source. Here r is an internal resistance in ohms, not a geometrical radius.
For example, a source e.m.f. of 1.60 V and terminal p.d. of 1.50 V while supplying 0.400 A imply r = (1.60 - 1.50)/0.400 = 0.250 Ω. The terminal reading alone would not reveal the e.m.f. without the current and source model.
Read voltage and power as the load changes
A few volts across a total resistance of a few ohms suggest a current of order 1 A and powers of a few watts. A very small external resistance need not produce a large useful output: the internal heating can dominate.
Separate the source from its external terminals
D belongs to the internal model. The external terminal reading is across A/C, after the internal voltage drop.
Terminal voltage falls as current rises
The intercept gives the e.m.f.; the negative gradient gives the internal resistance. The 6 A endpoint is a model limit, not an instruction to short the source.
Load power has a maximum
Increasing current eventually increases the internal loss enough to reduce the power delivered to the load. The maximum is 9 W at 3 A.
Take Es = 6.0 V and r = 1.0 Ω, held constant. Substituting each external load gives:
| Load R / Ω | I / A | Vterminal / V |
|---|---|---|
| 5.0 | 1.0 | 5.0 |
| 1.0 | 3.0 | 3.0 |
| 0.25 | 4.8 | 1.2 |
The V-against-I graph has e.m.f. as its vertical intercept and gradient -r. Here it passes through (0 A, 6 V), (1 A, 5 V), (3 A, 3 V) and the mathematical limit (6 A, 0 V).
Use power as the rate of energy transfer to account for the two destinations:
Pinternal = I2r
Psource = EsI = Pload + Pinternal
| R / Ω | Load power / W | Internal power / W |
|---|---|---|
| 5.0 | 5.0 | 1.0 |
| 1.0 | 9.0 | 9.0 |
| 0.25 | 5.76 | 23.04 |
For this source, load power reaches 9.0 W at 3.0 A, when R = r. Reducing the load further increases current but reduces terminal p.d. enough that load power falls. Equivalently, Pload = Es2R/(R + r)2.
Maximum load power is not maximum efficiency. At R = r, equal powers go to the load and internal heating, giving 50% of source power to the load. A larger R can give a greater fraction to the load while transferring less power.
With an open circuit, I = 0, so the ideal voltmeter reads Es and both load and internal powers are zero. At the mathematical short-circuit limit R = 0, this model gives I = 6 A, terminal p.d. zero and 36 W of internal heating. This limiting calculation is not a measurement procedure.
Infer source parameters from measurements
Vary a suitable external load while measuring total current and source-terminal p.d. across A/C. Choose ranges that include the expected values, retain instrument resolution and avoid source heating or depletion that changes Es or r during the record. A high-resistance voltmeter makes its own loading small.
A generated example with I = 0, 0.1, 0.2, 0.3, 0.4 A and V = 6.0, 5.9, 5.8, 5.7, 5.6 V gives intercept 6.0 V and gradient -1.0 V/A, hence r = 1.0 Ω. For actual readings, inspect the spread and possible drift before adopting one straight-line source model; do not force the intercept to an assumed e.m.f.
Optional check A discharging source has e.m.f. 6.0 V and internal resistance 1.0 ohm. It supplies a 1.0 ohm external load. What is the power account?
05
Reduce a network and work back
Find the equivalent resistance to obtain total current, then return to the original branches. A single equivalent resistor does not preserve every branch current or voltage.
Derive the two combination rules
In a series path, steady charge flow gives the same I through every resistor. The total p.d. is the sum of the individual drops:
Rseries = R1 + R2 + …
In parallel, each branch shares the same endpoints and therefore the same V. Total current is the sum of branch currents:
1/Rparallel = 1/R1 + 1/R2 + …
For exactly two parallel resistors, this rearranges to R1R2/(R1 + R2). Use the reciprocal sum for two or more: 3 Ω, 6 Ω and 6 Ω in parallel give 1/R = 1/3 + 1/6 + 1/6, hence R = 1.5 Ω.
Current conservation at a junction expresses conservation of charge; voltage drops along a path describe energy transferred per charge. Resistors transfer energy without consuming charge. A parallel equivalent must be smaller than the smallest positive branch resistance, while a series equivalent exceeds every individual positive resistance.
Worked network
Keep internal resistance in the complete path
A source has e.m.f. 9.0 V and internal resistance 1.0 Ω. The external circuit has R1 = 2.0 Ω on A/B, with R2 = 6.0 Ω and R3 = 3.0 Ω both on B/C.
Account for the whole circuit
The branch currents add to 1.8 A. Both parallel branches have 3.6 V across the same B/C endpoints.
Open only the three-ohm branch
The surviving branch receives 6.0 V and carries 1.0 A. Opening its parallel neighbour changes the whole circuit, including the internal drop.
Rexternal = 2.0 + 2.0 = 4.0 Ω
I = 9.0/(4.0 + 1.0) = 1.8 A
Now work back through the original network:
VAB = (1.8)(2.0) = 3.6 V
VBC = 7.2 - 3.6 = 3.6 V
I6 = 3.6/6.0 = 0.60 A
I3 = 3.6/3.0 = 1.20 A
The branch currents add to 1.8 A. Alternatively, the simultaneous relationships (6.0 Ω)I6 = (3.0 Ω)I3 and I6 + I3 = 1.8 A give the same two values.
With C chosen as 0 V, the model potentials are D = 9.0 V, A = 7.2 V and B = 3.6 V. The ideal source raises potential from C to D; internal resistance lowers it from D to A. This is why applying 9.0 V directly across the external network would be wrong.
Pinternal = (1.8)2(1.0) = 3.24 W
Psource = (9.0)(1.8) = 16.20 W
The two destinations add to the source power, providing an independent check on the calculation.
Opening a branch can increase another branch's current
Open only the 3.0 Ω branch. The external path is now 2.0 + 6.0 = 8.0 Ω, giving:
VAC = 9.0 - 1.0 = 8.0 V
VAB = (1.0)(2.0) = 2.0 V
VBC = 8.0 - 2.0 = 6.0 V
Total current falls from 1.8 to 1.0 A, yet current in the 6.0 Ω branch rises from 0.60 to 1.0 A. Its p.d. has increased from 3.6 to 6.0 V. The original branch voltage was not fixed independently of the rest of this circuit.
For another network, first establish what remains connected and which source model is supplied. Keep total, branch and component readings separate; a statement that one current falls does not determine every other current.
Optional check A 9.0 V source with 1.0 ohm internal resistance feeds a shared 2.0 ohm resistor, then parallel 6.0 and 3.0 ohm branches. Initially total current is 1.80 A and the 6.0 ohm branch carries 0.600 A. What happens when only the 3.0 ohm branch is opened?
06
Potential dividers and sensors
A divider output depends on both resistances and the selected endpoints. Identify the output arm before predicting whether its voltage rises or falls.
Let Rupper connect A to B and Rlower connect B to C. Apply a terminal supply Vs across A/C and take the output VBC. With negligible output current, both resistors carry the same current:
VBC = IRlower
= VsRlower/(Rupper + Rlower)
The numerator contains the resistance across the chosen output, not automatically the sensor. If source internal resistance matters, calculate the actual terminal Vs first. E.m.f. is not necessarily the p.d. across A/C.
A wiper selects a fraction of a uniform track
The lower 35% of the track gives 35% of 8.0 V: 2.80 V across B/C. The voltmeter draws negligible current.
NTC below the output junction
Warmer NTC: lower resistance falls from 12 to 4 kilohm, so output falls from 6.0 to 4.0 V.
LDR above the output junction
Brighter LDR: upper resistance falls from 14 to 2 kilohm, so output rises from 1.0 to 4.0 V.
Uniform potentiometer: convert position to resistance
A 20 kΩ track is connected across 8.0 V. Its wiper is 35% of the track length from C. With a uniform track and an unloaded output:
RAB = 13.0 kΩ
VBC = 8.0 × 7.0/20.0 = 2.80 V
I = 8.0/20000 = 0.400 mA
Here 1 kΩ = 103 Ω and 1 mA = 10-3 A. Moving the wiper changes the selected fraction, while the unloaded end-to-end resistance stays 20 kΩ. A nonuniform track would require its resistance distribution rather than position alone.
NTC in the lower arm
Keep the supply at 8.0 V and upper resistor at 4.0 kΩ. The lower NTC changes from 12 kΩ to 4.0 kΩ when warmed:
VBC = 8.0 × 12/16 = 6.0 V
Warmer: I = 8.0/8000 = 1.00 mA
VBC = 8.0 × 4/8 = 4.0 V
Warming lowers the NTC resistance and total resistance, so total current rises. Nevertheless, the NTC's share of the supply falls and VBC decreases. The complementary upper-arm p.d. rises from 2.0 to 4.0 V.
LDR in the upper arm
Now use an upper LDR and a fixed lower 2.0 kΩ resistor, again at 8.0 V. Increasing illumination changes the supplied LDR resistance from 14 kΩ to 2.0 kΩ:
VBC = 8.0 × 2/16 = 1.0 V
Brighter: I = 8.0/4000 = 2.00 mA
VBC = 8.0 × 2/4 = 4.0 V
Lower LDR resistance increases the current through the unchanged lower resistor, so its output rises. Swapping which arm contains the sensor changes this conclusion. Neither example supplies a universal resistance-temperature or resistance-illumination calibration.
In a real sensor investigation, record the actual supply and output, control the other environmental conditions, and allow thermal response to settle. Keep electrical self-heating small when measuring ambient temperature. A threshold for a connected device requires that device's specified input behaviour as well as the divider calculation.
A connected load changes the divider
Two 4.0 kΩ resistors across 8.0 V give an unloaded 4.0 V output across the lower resistor. Connect a 4.0 kΩ load across B/C: it is parallel with the lower resistor, not in series with the divider.
The load is another B/C branch
The parallel load changes the lower equivalent resistance. Output is 2.67 V; each lower branch carries about 0.667 mA.
= 2.0 kΩ
VBC = 8.0 × 2.0/(4.0 + 2.0)
= 2.67 V
The source current is 8.0/6000 = 1.333 mA, splitting equally into about 0.667 mA per lower branch. The output is no longer 4.0 V. A voltmeter is also an output load; calling it negligible means its resistance is sufficiently large for the required accuracy.
Optional check Two 4.0 kilohm divider arms connect across an ideal fixed 8.0 V supply. A 4.0 kilohm load is then connected across the lower arm B/C. What is V_B - V_C?
Revision
Circuits: revision summary
Identify the nodes, the component and the stated conditions. Then choose a relationship that uses the correct current, voltage and source model.
Connections and measurements
- Series components share current along an unbranched path; parallel components share both endpoints and therefore p.d.
- An ammeter measures current in its own path. A voltmeter measures p.d. between its own terminals. Pair component V with component I before using R = V/I.
- Check range, resolution, zero, polarity, loading and thermal conditions. Repeated readings assess scatter but do not remove a consistent wrong-endpoint or zero error.
Resistance and material behaviour
A = πd2/4
Rseries = R1 + R2 + …
1/Rparallel = 1/R1 + 1/R2 + …
Convert diameter before squaring it. At fixed material, temperature and length, resistance is proportional to 1/A; doubling diameter quarters resistance. On an I-against-V plot, only a straight line through the origin has gradient 1/R at all points.
- Ohmic resistor: proportional I and V under fixed physical conditions.
- Filament lamp: settled heating raises resistance, so the I-V curve becomes less steep at greater voltage magnitude.
- Semiconductor diode: forward conduction and small reverse current before breakdown; no universal forward threshold.
- NTC thermistor: warming lowers resistance. A self-heated voltage sweep differs from a fixed-temperature, low-field measurement.
For a typical metal at the same electric field, greater scattering lowers mean drift speed while carrier density changes little. For the stated NTC regime, increased carrier number density is the main explanation for falling resistivity. Random thermal motion is not the directed drift.
Sources, networks and outputs
I = Es/(Rexternal + r)
Pload = IVterminal
Pinternal = I2r
EsI = Pload + Pinternal
These source signs describe current delivered to the load. A terminal-V-against-I line has intercept Es and gradient -r. Maximum load power occurs at Rexternal = r for this fixed source model, where load and internal powers are equal; this is not maximum efficiency.
Reduce a network to find total current, then work back to its branch voltages and currents. After a connection changes, recalculate the complete network: another branch's voltage may change even if its resistance does not.
Vout = VsRlower/(Rupper + Rlower)
Use the actual supply-terminal p.d. The selected output determines the numerator. Work through sensor resistance, total current and output rather than memorising a direction. An output load is parallel with its selected arm and changes the effective divider.
Quantity and unit reference
| Quantity | Symbol | SI unit |
|---|---|---|
| Current | I | A |
| Potential difference | V | V = J/C |
| Source e.m.f. | Es, E | V |
| Resistance; internal resistance | R; r | Ω = V/A |
| Resistivity | ρ | Ω m |
| Length; diameter | l; d | m |
| Cross-sectional area | A | m2 |
| Power | P | W = J/s |
| Charge | Q | C = A s |
| Time | t | s |
Read symbols in context: A as area differs from A as the ampere unit, and C as charge's unit differs from a node labelled C. Useful prefixes are micro (µ, 10-6), milli (m, 10-3), kilo (k, 103) and mega (M, 106). Squared lengths and areas require squared conversion factors.
For a calculation, draw or trace the nodes, choose reference directions and endpoints, state the component conditions, solve with units, then check current and energy accounts. For a measurement, retain the actual readings and instrument information as well as the inferred result.