Circuits

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

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.

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 V_B - V_C 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 V_A - V_C.

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.

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:

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:

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.

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

1. 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.
2. 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.
3. 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.
4. 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.
5. 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.

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 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:

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:

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.

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. E_s is energy supplied by the source per unit charge. The terminal p.d. V_terminal 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:

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.

Take E_s = 6.0 V and r = 1.0 Ω, held constant. Substituting each external load gives:

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:

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, P_load = E_s²R/(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 E_s 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 E_s 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.

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:

In parallel, each branch shares the same endpoints and therefore the same V. Total current is the sum of branch currents:

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.

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:

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.

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 R_upper connect A to B and R_lower connect B to C. Apply a terminal supply V_s across A/C and take the output V_BC. With negligible output current, both resistors carry the same current:

The numerator contains the resistance across the chosen output, not automatically the sensor. If source internal resistance matters, calculate the actual terminal V_s first. E.m.f. is not necessarily the p.d. across A/C.

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:

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:

Warming lowers the NTC resistance and total resistance, so total current rises. Nevertheless, the NTC's share of the supply falls and V_BC 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:

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 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.

07

Capacitor combinations

Capacitor combinations follow charge and voltage constraints. Identify which plates share a node and whether charge can enter an intermediate node before choosing a rule.

For a capacitor with equal and opposite plate charges, C = Q/V, where Q is the magnitude on one plate and V is the p.d. magnitude. The net charge of the two-plate device can be zero while it stores energy. Capacitance has unit farad: 1 F = 1 C/V, with 1 µF = 10^-6 F and 1 µC = 10^-6 C. The capacitance explanation develops these quantities.

Parallel: the same p.d.

Capacitors in parallel share both endpoints and therefore V. The total charge supplied to the positive plates is the sum:

For 3.0 µF and 6.0 µF in parallel across 12 V, C_equivalent = 9.0 µF. The plate-charge magnitudes are 36 and 72 µC, giving total supplied charge 108 µC. Do not double this by adding both positive and negative plate-charge magnitudes.

Series: the middle-node charge constraint

Start with uncharged capacitors and keep their intermediate node isolated from any other conducting connection. No net charge can enter that node, so its two connected plates acquire equal and opposite charges. The series capacitors therefore have equal plate-charge magnitudes Q, while their voltages add:

For 3.0 µF on A/B and 6.0 µF on B/C across 12 V:

With C at 0 V and A at 12 V, B is at 4 V. Its connected plates carry -24 and +24 µC, so B has zero net charge, not zero potential. The smaller capacitance has the greater p.d. under this equal-charge condition.

Initially charged capacitors can leave a nonzero net charge on an isolated middle node. Then conserve that given node charge rather than assuming equal magnitudes automatically. The combination rules above use the stated initial and connection conditions.

A mixed arrangement and an energy check

Add a 4.0 µF capacitor directly across A/C, in parallel with the 2.0 µF series equivalent. The complete capacitance is 6.0 µF. At 12 V the added branch has 48 µC, while the series branch still has 24 µC, giving total supplied charge 72 µC.

For an ideal capacitor, stored energy is U = ½CV². Add each component's energy using its own voltage:

Each total also equals ½C_equivalentV_supply². These are final stored energies, not automatically all the energy supplied during charging. An ideal capacitor has no steady conduction through its dielectric; the charging process determines the transient currents and other energy transfers.

08

RC charging and discharging

Trace the switch connection first. Then keep one sign convention as the capacitor charges or discharges, so the current graph agrees with the change in its plate charge.

Use a resistor R between switch terminal S and capacitor plate B. The other plate connects to common return node C. The source positive terminal is A, at E_s above C. Here the letter C labels a node in the drawing; the italic quantity C in an equation is capacitance.

Define Q as the charge on plate B and V_C = V_B - V_{node C} = Q/C. Define positive I through the resistor from S towards B. Then I = dQ/dt: positive current increases Q and negative current reduces it. In the examples Q stays positive while falling during discharge.

A switch that simply disconnects every conducting path does not discharge an ideal capacitor. The discharge state shown provides a closed resistor-capacitor loop without connecting the source across a short circuit.

Charging from zero

Take an initially uncharged capacitor, a constant ideal source e.m.f. E_s, fixed R and C, and negligible leakage and meter loading. At each instant, E_s = IR + Q/C. Initially the capacitor p.d. is zero, so current is E_s/R. As Q and its p.d. grow, less voltage remains across R and current decreases.

Charge and capacitor p.d. rise towards their final values; charging current decreases towards zero. The source voltage stays constant in this model. A finite resistor prevents capacitor voltage from jumping instantaneously, though the current can change abruptly when the switch changes a connection.

Discharging from a specified initial voltage

Disconnect the source and connect R across a capacitor initially at V₀, with Q₀ = CV₀. The stored charge now drives current through R. Keep the same positive current reference:

The current is negative and approaches zero from below. Its magnitude decreases. Calling it negative does not mean that the capacitor has reversed plate polarity. If a question chooses the opposite current reference, reverse the current sign while keeping the physical behaviour unchanged.

In the general rising form x = x₀(1 - e^-t/τ), x₀ is the final limit when the initial value is zero. In the decreasing form x = x₀e^-t/τ, x₀ is the initial value. Check which quantity the symbol describes.

Compare matched charge and discharge models

Let E_s = V₀ = 10 V, R = 200 kΩ and C = 10 µF. These are two separate runs, each with its own t = 0: charging starts uncharged, while discharging starts fully at 10 V.

Extra digits show the model comparison; final experimental precision would depend on the actual component values and readings. Use micro = 10^-6 consistently for both µF and µA.

What the time constant does and does not mean

At t = τ, a decaying quantity is e^-1 = 0.368 of its initial value, while a rising quantity is 1 - e^-1 = 0.632 of its final value. RC has units of time:

The time constant is not a half-life and is not the time for mathematically complete charging. At 5τ, charging reaches about 99.33% of its final value. To find a different fraction, solve the exponential:

Doubling R alone makes τ = 4.0 s and halves the initial charging current to 25 µA, while final charge stays 100 µC. Doubling C alone also makes τ = 4.0 s, but now final charge is 200 µC and initial current remains 50 µA. An equal change in time constant does not imply identical changes in every quantity.

Energy and the actual resistor path

For charging from zero in this ideal constant-voltage model, the source supplies energy E_sQ_final = CE_s². The capacitor stores ½CE_s²; the rest is transferred to internal energy in the resistance. Here the source supplies 1.00 mJ, the capacitor stores 0.500 mJ and the resistor receives 0.500 mJ. In a complete discharge through R, the stored energy is transferred through that resistance.

Choose R from the path present in the relevant state. Source internal resistance can add to a charging path while being absent from the separate discharge loop. A finite-resistance voltmeter adds a parallel path; leakage can also matter. An arbitrary product of a whole network's named resistance and capacitance is not automatically its time constant.

09

Infer a time constant from data

A discharge record can reveal a time constant through its exponential shape or a straight-line logarithmic plot. The area under its current-time graph gives charge transferred, with the sign set by the chosen current reference.

For a capacitor discharging through fixed R, let V₀ be its initial p.d. and τ = RC. With positive current defined towards the initially positive plate, V = V₀e^-t/τ and I = -V/R. The capacitor voltage falls while current is negative and approaches zero. The switching diagrams identify the corresponding paths.

Make the exponential relationship linear

Divide by V₀ before taking the natural logarithm:

The logarithm's input is a positive, dimensionless ratio. Plot ln(V/V₀) vertically against time t horizontally. The ideal straight line has gradient -1/τ, with unit s^-1, and zero intercept. Its gradient gives τ; a known discharge resistance then gives C = τ/R.

A raw ln(V) label would conceal the required reference unit. A zero or negative ratio cannot be processed by LN. In an actual record, preserve such readings and investigate zero offset, noise, polarity and the usable range; do not silently delete them or invent positive replacements.

Turn a real record into a defensible time constant

Record the actual switch event and time zero, sample times, resistor value, initial capacitor voltage and capacitor information. Connect the voltage sensor across the capacitor's B/C terminals. Check its zero, polarity, range, resolution and response time; sample often enough to resolve the change over the expected time constant. A range ending below the initial 10 V would not capture this supplied example.

Use the real circuit's discharge path, check component ratings and capacitor polarity where applicable, and repeat the timing where useful. Retain the original record and document any excluded interval or correction. The measurement-record procedure explains how to enter or import actual readings while keeping them distinct from generated examples.

The simple exponential assumes negligible loading. A finite-resistance voltmeter across the capacitor is also a discharge path. For a 10 MΩ meter across the supplied 200 kΩ resistor, using 1 MΩ = 10⁶ Ω:

This is about 1.96% shorter than 2.0 s. Dividing that measured time constant by 200 kΩ while ignoring the meter would underestimate capacitance. In a charging circuit, the same finite parallel meter can also lower the final capacitor voltage. Check the actual connections before choosing the ideal formula.

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

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

These source signs describe current delivered to the load. A terminal-V-against-I line has intercept E_s and gradient -r. Maximum load power occurs at R_external = 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.

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.

Capacitors and switching

Parallel capacitors share voltage. In the stated initially uncharged series arrangement, an isolated neutral middle node enforces equal charge magnitudes; its potential need not be zero. Keep any supplied initial node charge in a more general arrangement.

For the simple RC path, τ = RC. Take Q on the initially positive plate, V = Q/C and positive current towards that plate:

At one time constant, a rising quantity reaches 63.2% of its final value and a decaying quantity retains 36.8% of its initial value. The half-value time is τ ln 2, while 90% charging takes -τ ln(0.10).

Read discharge data

Log ratios and exponential arguments are dimensionless. Fit ln(V/V₀) against numeric time; the gradient has unit s^-1. A coarse trapezium sum approximates a curved area, and rounded model values can move a fitted result slightly. Neither is evidence of measured scatter. Meter loading and the actual switched resistor path matter when interpreting a real record.

Quantity and unit reference

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 or the capacitance symbol 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.