K323 / 2027
Current of electricity overview

Full chapter

Current of electricity

All 4 topics and the revision summary on one page.

01

Current, charge and electron flow

Current tells you how quickly charge passes a point in a circuit. A larger current means more charge passes each second.

Charge is measured in coulombs, C. Electrons carry negative charge. In a metal wire, mobile electrons can drift through the material while the positive ions remain in its structure.

Current = charge passing / time takenI = Q/t; Q = It; t = Q/I. For steady current, use Q in C, I in A and t in s.

Electric current is the rate of flow of charge. It is measured in amperes, A. One ampere means one coulomb passes a cross-section each second: 1 A = 1 C/s. Equivalently, 1 C = 1 A s: a coulomb is an ampere-second. Current is represented by I; charge by Q or q.

Imagine marking one cross-section of a wire and counting the total charge that passes it during an interval. A steady 0.30 A means 0.30 C passes each second. It does not mean the whole wire contains only 0.30 C of charge.

Follow charge through the external metal circuit

The cell is discharging into the resistor. Green arrows show conventional current; blue arrows show electron drift. Arrows give direction, not speed.

Conventional current and electron drift have opposite directionsA closed rectangular circuit contains a cell on the left and a resistor in the top wire. The cell's long upper plate is positive and its short lower plate is negative. In the external circuit, conventional current goes clockwise from the positive terminal, through the resistor, and back to the negative terminal. Electron drift in the metal wire goes the opposite way. One cross-section X is marked in the bottom wire. At a steady current of 0.30 ampere, 36 coulombs pass this cross-section in 120 seconds. No electron path through the cell's electrolyte is drawn.+-CellResistorConventionalcurrentElectrondriftCross-section X

Steady current: 0.30 A. Time: 2.0 min = 120 s.
Charge passing X: Q = It = 0.30 × 120 = 36 C.

The resistor transfers energy. It does not use up the charge passing through it.

In the external metal wire, conventional current and electron drift have opposite directions. The marked cross-section carries 0.30 C each second, so 36 C passes it in 120 s. The arrows show direction, not speed.

Conventional current and electron drift

Conventional current is defined in the direction in which positive charge would move. In the external circuit of a discharging cell, it goes from the positive terminal, through the components, towards the negative terminal.

Electrons are negative, so their drift in the metal wire is in the opposite direction: from the negative terminal towards the positive terminal through the external circuit. The cell symbol's long line marks its positive terminal and its short line the negative terminal.

A lamp transfers energy as charge passes through it; it does not use up that charge. Electrons are already present throughout the metal circuit. Do not picture the lamp waiting for one electron to leave the cell and complete a whole journey before it responds.

Charge passing a point

A current flows for two minutes

A steady current of 0.30 A flows for 2.0 min. First convert the time: 2.0 x 60 = 120 s.

Q = It = 0.30 x 120 = 36 C.

In a separate example, 18 C passes in 45 s. Its average current is I = Q/t = 18/45 = 0.40 A.

If the current is given as 300 mA, use 300/1000 = 0.300 A. The prefix milli means one thousandth.

Q = It applies directly when I is constant over the interval. For a changing current, use the average current for that interval, or find the charge in each stated constant-current part and add the amounts. An instantaneous reading does not establish the current throughout a long interval.

Measure current through a component

An ammeter, shown as a circle containing A, connects in series with the branch being measured. Open the circuit at the chosen position and include the meter in the path, so the branch's charge passes through it.

For a d.c. measurement, connect its positive terminal on the side from which conventional current enters. Choose a suitable range; if the expected value is uncertain, start with a sufficiently high range and reduce it when appropriate for a clearer reading. Keep the reading within the selected range.

An ammeter has very low resistance. Connecting it directly across a cell would provide a low-resistance path, rather than measure the current through the intended component. The resistance arrangement shows the correct current and voltage connections together.

An ammeter reads amperes, not coulombs. To determine charge for a steady current, also measure the time interval, for example with a stopwatch, then use Q = It.

Optional check A steady current of 0.30 A flows through a wire for 2.0 min. How much charge passes a marked cross-section?
A steady current of 0.30 A flows through a wire for 2.0 min. How much charge passes a marked cross-section?

02

E.m.f. and potential difference

Voltage describes energy transferred per coulomb of charge. The source and the component have different roles in that energy transfer.

Work done is an energy transfer, measured in joules, J. Charge is measured in coulombs, C. Current measures coulombs per second; voltage measures joules per coulomb.

Electromotive force: e.m.f.
The e.m.f. of a source is the work done per unit charge by the source in driving charge around a complete circuit. It is measured in volts. A 1.5 V source supplies 1.5 J per coulomb in this energy account.
Potential difference: p.d.
The p.d. across a component is the work done per unit charge in driving charge through the component. It is measured in volts. A p.d. of 3.0 V means 3.0 J is transferred in that component for each coulomb passing through it.
1 volt = 1 joule per coulomb1 V = 1 J/C. Potential difference = work done / charge; work done = potential difference x charge.

Small voltages may be given in millivolts: 1 mV = 0.001 V. Divide a reading in millivolts by 1000 to convert it to volts.

A cell transfers energy from its chemical store into the electrical pathway. In a resistor, energy is transferred to internal stores; in a motor, part can be transferred mechanically to a load. Charge carries on through the circuit while energy is transferred. The energy stores and pathways account helps keep these two ideas separate.

The name electromotive force does not mean a force measured in newtons. E.m.f. is an energy-per-charge quantity, just as p.d. is. Neither quantity measures how many coulombs pass each second.

Energy per coulomb

Find a component's potential difference

A component transfers 60 J when 12 C passes through it.

P.d. = 60/12 = 5.0 J/C = 5.0 V.

For a separate component with p.d. 3.0 V, passing 8.0 C transfers work of 3.0 x 8.0 = 24 J.

These are supplied energy and charge amounts. A temperature rise alone would not measure every energy transfer unless the heated mass, its properties and other transfers were also accounted for.

Measure voltage across two points

A voltmeter, shown as a circle containing V, connects in parallel across the component: one lead at each of its terminals. This measures the p.d. between those two points. It is not inserted into the main current path as an ammeter is.

For a positive d.c. reading, connect the meter's positive lead to the higher-potential side and its negative lead to the lower-potential side. Reversing the leads reverses the sign shown by a suitable digital meter; it does not mean the component has changed its resistance.

Select a voltage range that includes the expected reading. For example, a 0-5 V range is unsuitable for a 6.0 V p.d. Once a suitable range is established, finer resolution helps distinguish nearby readings. For an analogue scale, read at eye level to reduce parallax.

A voltmeter connected across a source with no other external circuit can give an estimate of its e.m.f. when the meter draws negligible current. For the ideal-source model, the source's terminal p.d. equals its e.m.f. The combined meter diagram instead measures the p.d. across a resistor.

Optional check A component transfers 60 J of energy when 12 C passes through it. What is the potential difference across it?
A component transfers 60 J of energy when 12 C passes through it. What is the potential difference across it?

Add sources with their polarities

For ideal sources connected in series, the total e.m.f. is their signed sum. Choose a direction through the connected sources. Crossing a cell from its negative terminal to its positive terminal adds its e.m.f.; crossing it from positive to negative subtracts it.

Add the voltage rises and falls in a stated direction

These are open arrangements of ideal cells. A long plate marks a positive terminal and a short plate a negative terminal. The separate arrow sets the calculation reference from A to B.

Three cells aiding one another

Three cells aiding one anotherFrom A to B, each cell is crossed from its negative short plate to its positive long plate. Each contributes a 1.5 V rise, so B is 4.5 V higher than A in this ideal open arrangement. The terminals A and B are open, with no return wire. The arrow below is a reference for adding source contributions, not an arrow of current in a closed circuit.1.5 V-+1.5 V-+1.5 V-+ABReference: A to B

1.5 + 1.5 + 1.5 = 4.5 V.

From A to B, each cell is crossed from its negative short plate to its positive long plate. Each contributes a 1.5 V rise, so B is 4.5 V higher than A in this ideal open arrangement.

The final cell is reversed

The final cell is reversedFrom A to B, the first two cells each give a 1.5 V rise. The last cell is crossed from positive to negative and gives a 1.5 V fall. B is therefore 1.5 V higher than A in this ideal open arrangement. The terminals A and B are open, with no return wire. The arrow below is a reference for adding source contributions, not an arrow of current in a closed circuit.1.5 V-+1.5 V-+1.5 V+-ABReference: A to B

1.5 + 1.5 - 1.5 = 1.5 V.

From A to B, the first two cells each give a 1.5 V rise. The last cell is crossed from positive to negative and gives a 1.5 V fall. B is therefore 1.5 V higher than A in this ideal open arrangement.

Use the same A-to-B direction in both source combinations. Three aiding 1.5 V cells give +4.5 V. Reversing the last cell changes its contribution to -1.5 V, giving a net +1.5 V. These are open source combinations, not complete circuits.

Keep the reference direction

One of three cells is reversed

Going from A to B through three 1.5 V cells:

  • All aiding: total e.m.f. = 1.5 + 1.5 + 1.5 = 4.5 V.
  • Last cell opposing: total e.m.f. = 1.5 + 1.5 - 1.5 = 1.5 V.

B is at higher potential than A in both illustrated combinations, but by different amounts. Counting the cells without checking their terminal signs would miss the difference.

Optional check Going from A to B through three ideal 1.5 V cells in series, the first two are crossed from negative to positive and the last from positive to negative. What is the net e.m.f. in the A-to-B direction?
Going from A to B through three ideal 1.5 V cells in series, the first two are crossed from negative to positive and the last from positive to negative. What is the net e.m.f. in the A-to-B direction?

03

Resistance and wire dimensions

Resistance relates the potential difference across a component to the current through it. At the same p.d., a greater resistance gives a smaller current.

Current I is measured through a component in amperes. Potential difference V is measured across its terminals in volts. Use readings for the same component when finding its resistance.

R = V/IResistance = potential difference / current. Rearrangements: V = IR and I = V/R. The unit is the ohm, Ω; 1 Ω = 1 V/A.

This defines resistance at the stated operating point: the voltage and current under the conditions of that reading. It does not by itself establish that a component's resistance stays the same when conditions change.

Measure current through and p.d. across the same resistor

A is an ammeter in series. V is a voltmeter connected to the resistor's two terminals. The marked meter polarities suit this d.c. source.

An ammeter in series and a voltmeter in parallel measure the resistorThe cell has its positive plate above its negative plate on the left. The top branch runs through ammeter A and then a resistor before returning to the source along the right and bottom wires. The ammeter's left terminal is positive. Two separate connections descend from the exact left and right resistor terminals to voltmeter V; its left terminal is positive. The supplied ammeter reading is 0.20 ampere and the p.d. across the resistor is 6.0 volts. With ideal meters, the ammeter gives the resistor current, so its resistance is 30 ohm. No wire bypasses the resistor or either meter symbol.+-SourceA+-0.20 AMeasured resistorV+-6.0 VBoth readings concern this resistor

R = V/I = 6.0/0.20 = 30 Ω.

This arrangement uses the ideal meter model: the voltmeter takes negligible current and the ammeter causes negligible p.d. The ammeter therefore measures the resistor current.

The ammeter measures the current in the resistor's path, and the voltmeter connects across its two terminals. With ideal meters, the voltmeter draws negligible current and the ammeter adds negligible resistance. The supplied readings are 6.0 V and 0.20 A.

Use corresponding readings

Determine resistance, then predict current

The p.d. across a resistor is 6.0 V and the current through it is 0.20 A.

R = V/I = 6.0/0.20 = 30 Ω.

If the same resistor remains at 30 Ω when the p.d. becomes 9.0 V, then I = V/R = 9.0/30 = 0.30 A.

The prediction uses the stated constant-resistance assumption. The first measurement alone cannot show that heating or other changes will leave the resistance unchanged.

Optional check A voltmeter reads 6.0 V across a resistor while an ammeter measures 0.20 A through that resistor. What is its resistance at this operating point?
A voltmeter reads 6.0 V across a resistor while an ammeter measures 0.20 A through that resistor. What is its resistance at this operating point?

How wire dimensions affect resistance

Compare wires of the same material at the same temperature. Changing the material or temperature can change the relationship between their dimensions and resistance.

Greater length gives greater resistance
Resistance is proportional to length, R ∝ L. Doubling the length while keeping the cross-section unchanged doubles the resistance. Charge must travel through a longer section of the material.
Greater cross-sectional area gives smaller resistance
Resistance is inversely proportional to cross-sectional area, R ∝ 1/A. Doubling that area while keeping length unchanged halves the resistance. The wider cross-section provides more conducting material alongside the path.

The cross-sectional area is the area you see when looking straight at a cut end. It is not the curved outer surface of the wire. For a circular wire, A = πd2/4, so area changes with the square of the diameter.

Double the length and double the diameter

Compare the same material at the same temperature. Both drawings use the same scale; the labels give ratios, not dimensions to measure from your screen. The area is the circular end section.

Original wire: 8.0 ohm

Original wire: 8.0 ohmThe original wire has length L, circular diameter d and cross-sectional area A. Its supplied resistance is 8.0 ohm. A side view and an end-on circular section show the dimensions. The measured axial length is between the two end faces. Diameter brackets span the wire and the end-on circle. The side-view end faces are drawn as ellipses to suggest their orientation, while the end-on section is a true circle.Side viewLdEnd-on cross-sectionAd

Changed wire: 4.0 ohm

Changed wire: 4.0 ohmThe changed wire is twice as long and twice the diameter. At the same drawing scale, its circular section has twice the radius and four times the area. It is the same material at the same temperature, so its resistance is two divided by four, or one half, of 8.0 ohm: 4.0 ohm. The measured axial length is between the two end faces. Diameter brackets span the wire and the end-on circle. The side-view end faces are drawn as ellipses to suggest their orientation, while the end-on section is a true circle.Side view2L2dEnd-on cross-section4A2d

Twice the diameter gives four times the cross-sectional area. The resistance changes by 2/4 = 1/2, so 8.0 Ω becomes 4.0 Ω.

The second wire has twice the length and twice the diameter. Its end-on area is four times the original area. Material and temperature are kept the same, so the two changes together halve the resistance.

Compare one change at a time

Twice the length and twice the diameter

The original wire has resistance 8.0 Ω.

  1. Doubling its length gives a resistance factor of 2.
  2. Doubling its diameter gives an area factor of 22 = 4, so the resistance factor from area is 1/4.
  3. Combine the factors: new R = 8.0 x 2/4 = 4.0 Ω.

For any comparison at the same material and temperature, Rnew/Rold = (Lnew/Lold) x (Aold/Anew).

Determine resistance and investigate wire length

Use a suitable low-voltage classroom supply, with an ammeter in the current path and a voltmeter across the component. Record the p.d. and current together after the readings settle, and calculate R = V/I. Check meter ranges and polarity before collecting readings.

To investigate length, use the same uniform wire and vary the distance between its electrical contacts. Measure this active length with a ruler. The unused wire beyond the contacts is not part of the measured section.

  1. Keep the wire's material and cross-sectional area fixed. Use several measured active lengths.
  2. At each length, record the current through and p.d. across that section. Calculate its resistance from the paired readings.
  3. Keep the wire temperature as steady as practical: use a suitably low current and switch off between readings where appropriate. Wait for it to return to the comparison conditions if it warms.
  4. Plot resistance vertically against active length horizontally. Proportionality predicts a straight line through the origin for the wire alone at constant cross-section and temperature.

If diameter is needed, use a micrometer or suitable calipers, check for zero error and take readings at several positions and orientations. Avoid squeezing or measuring a damaged part of the wire. Convert the diameter to consistent units before calculating area.

Contact and lead resistance may add to a measured resistance, depending on the voltage lead positions. Wire heating can change readings, and a varying diameter means the cross-section is not uniform. Repeating and averaging readings can reveal scatter, but it does not automatically remove these effects. Explain the specific limitation and how the method addresses it.

Twice the diameter is four times the area. Also keep the voltage and current paired: dividing a voltage across one branch by the current through a different branch does not generally give either branch's resistance.

Optional check A wire has resistance 8.0 ohm. A second wire of the same material and temperature has twice its length and twice its diameter. What is the second resistance?
A wire has resistance 8.0 ohm. A second wire of the same material and temperature has twice its length and twice its diameter. What is the second resistance?

04

Temperature and I-V characteristics

An I-V graph shows how a component's current responds to different potential differences. Its shape tells you whether the resistance stays constant.

At any nonzero-current operating point, R = V/I. Read both values from that same point. The graphs here put current I in amperes on the vertical axis and p.d. V in volts on the horizontal axis.

Why a hotter metal has greater resistance

When a metallic conductor's temperature rises, the ions in its structure vibrate more vigorously. Their increased vibration hinders the drift of mobile electrons, so the resistance increases in the normal operating range considered here.

At the same p.d., the hotter conductor therefore carries a smaller current. This explanation concerns metals; it is not a rule that every material or component has increasing resistance with temperature.

Read and sketch three characteristic shapes

Start with labelled axes and an origin. Positive and negative values represent opposite voltage polarities and current directions under a fixed reference convention.

Read current vertically and p.d. horizontally

The conductor and lamp use the same axes and scales. Dots mark the supplied positive readings; dashed negative branches show the corresponding idealised symmetric behaviour.

Metallic conductor at constant temperature

Metallic conductor at constant temperatureCurrent I is vertical in amperes and potential difference V is horizontal in volts. A straight line passes through the origin and the supplied positive points 2 volts, 0.10 ampere; 4 volts, 0.20 ampere; and 6 volts, 0.30 ampere. The negative branch is the corresponding idealised symmetric behaviour. V divided by I is 20 ohm at every nonzero point. The horizontal tick interval is 2 volts and the horizontal grid lines are 0.10 ampere apart. Current values are labelled at the left edge and voltage values below the grid, aligned with their grid lines. This keeps the labels clear of the negative branches. The two axes cross at zero volts and zero amperes.-6-4-20246-0.4-0.3-0.2-0.100.10.20.30.4Current I / APotential difference V / V

Marked readings (V, I):

  • (2 V, 0.10 A)
  • (4 V, 0.20 A)
  • (6 V, 0.30 A)

The straight line through the origin has constant I/V. At each nonzero point, V/I = 20 Ω.

Filament lamp: the curve becomes shallower

Filament lamp: the curve becomes shallowerCurrent I is vertical in amperes and potential difference V is horizontal in volts, using the same scales as the conductor graph. A smooth curve passes through the origin and the supplied positive points 2 volts, 0.20 ampere; 4 volts, 0.30 ampere; and 6 volts, 0.36 ampere. Its gradient decreases as voltage magnitude increases. The negative branch is the corresponding idealised symmetric behaviour. The three positive V divided by I values are 10, approximately 13.3 and approximately 16.7 ohm. The horizontal tick interval is 2 volts and the horizontal grid lines are 0.10 ampere apart. Current values are labelled at the left edge and voltage values below the grid, aligned with their grid lines. This keeps the labels clear of the negative branches. The two axes cross at zero volts and zero amperes.-6-4-20246-0.4-0.3-0.2-0.100.10.20.30.4Current I / APotential difference V / V

Marked readings (V, I):

  • (2 V, 0.20 A)
  • (4 V, 0.30 A)
  • (6 V, 0.36 A)

As the filament heats, the graph becomes shallower and V/I increases. Use the point's V/I for resistance; the local tangent gradient is a different quantity.

Semiconductor diode: forward and reverse behaviour differ

A qualitative diode characteristicCurrent I is vertical and potential difference V horizontal. The positive voltage direction is chosen as forward bias. Forward current rises increasingly steeply under the illustrated conditions. On the negative-voltage side, the reverse current is very small over the displayed range. It is drawn just below the axis so its sign can be seen. This graph has no numerical voltage or current scale and does not state a universal turn-on voltage or show reverse breakdown. It passes through the origin.IV0Negative VPositive VForwardcurrent risesVery smallreverse currentQualitative: no numerical scale

Positive V is the forward direction chosen for this graph. Reverse current is negligible over the range shown; the curve does not imply that it stays negligible at every possible reverse voltage.

The constant-temperature conductor has a straight line through the origin. The lamp's current increases less rapidly as it heats. Their negative branches show the corresponding idealised symmetric behaviour. The diode panel is qualitative: appreciable forward current and negligible reverse current in the illustrated range.
Metallic conductor at constant temperature
Sketch a straight line through the origin, extending into both positive and negative quadrants. Current is proportional to p.d., so V/I is constant. This is an ohmic conductor under those conditions. A wire that heats significantly is no longer being compared at constant temperature.
Filament lamp
Sketch a curve through the origin that becomes less steep as the voltage magnitude increases, with a corresponding branch for reversed polarity. Greater current heats the metal filament, increasing its resistance. Doubling the p.d. therefore need not double the current.
Semiconductor diode
Identify the forward and reverse polarities. In the forward direction, current rises strongly over the illustrated higher-voltage part of the curve. In reverse, it is negligible over the range shown. Sketch this asymmetric behaviour; there is no single exact turn-on voltage for every diode.

The diode's forward and reverse directions describe its orientation in the circuit. They are not a claim that all positive voltages in every drawing are forward: check the stated reference and component orientation. The displayed small reverse-current range does not establish zero current at every possible reverse voltage.

Calculate resistance at a point

Compare V/I

A constant-temperature wire and a lamp

The wire's supplied readings are 2.0 V with 0.10 A, 4.0 V with 0.20 A, and 6.0 V with 0.30 A. Each gives R = 20 Ω.

For the illustrative lamp readings:

  • At 2.0 V and 0.20 A: R = 2.0/0.20 = 10 Ω.
  • At 4.0 V and 0.30 A: R = 4.0/0.30 = 13.3 Ω.
  • At 6.0 V and 0.36 A: R = 6.0/0.36 = 16.7 Ω.

The lamp's resistance rises as the filament heats. At the corresponding reversed point, -6.0 V and -0.36 A still give a positive resistance: (-6.0)/(-0.36) = 16.7 Ω.

For the straight I-V line through the origin, the gradient is I/V = 1/R. Here 0.30/6.0 = 0.050 A/V, whose reciprocal is 20 Ω. A steeper straight line on the same scales means smaller resistance.

For a curved graph, use the point's V/I. A tangent gradient describes how current changes close to that point; its reciprocal is not the resistance V/I there. Likewise, a ratio of changes between two separated points does not generally give either point's resistance. If the axes are exchanged to V vertically and I horizontally, the constant-resistance line instead has gradient R.

Collect readings that can distinguish the behaviours

Measure current through and p.d. across the component while varying a suitable low-voltage supply. Record paired readings, then plot I against V. For the constant-temperature wire comparison, limit heating and allow the wire to return to the same conditions between readings where needed.

For a lamp characteristic, let each reading settle at its operating condition; the changing filament temperature is part of the behaviour being observed. Use suitable component ratings and current limits. To investigate the opposite polarity, switch off before reversing the source connections and record the resulting signs consistently.

State the graph's axes and conditions. A straight line through the origin supports constant resistance over the range shown. A changing gradient on a lamp curve must be interpreted with the filament's changing temperature.

Optional check A filament lamp carries 0.20 A at 2.0 V and 0.36 A at 6.0 V. What is its resistance at 6.0 V, and what do these readings show?
A filament lamp carries 0.20 A at 2.0 V and 0.36 A at 6.0 V. What is its resistance at 6.0 V, and what do these readings show?

Revision summary

Charge and current

Charge, Q or q
Unit: coulomb, C; 1 C = 1 A s (one ampere-second). For steady current, Q = It. Convert time to seconds and milliamperes to amperes: 1 mA = 0.001 A.
Current, I
Rate of flow of charge; unit: ampere, A = C/s. I = Q/t gives the average over the stated interval. An ammeter connects in series with the measured path.
Direction in an external metal circuit
Conventional current goes from the discharging cell's positive terminal towards its negative terminal. Electron drift is opposite. A component transfers energy without consuming the charge.

Energy per charge

E.m.f., E
Work done per unit charge by the source in driving charge around the complete circuit; unit: volt, V = J/C.
Potential difference, V
Work done per unit charge in driving charge through a component; also measured in volts. 1 mV = 0.001 V. Work done = p.d. x charge. A voltmeter connects across the component's two terminals.
Ideal sources in series
Choose a reference direction and add signed e.m.f.s. Cross negative to positive: add. Cross positive to negative: subtract. Three 1.5 V cells all aiding give 4.5 V; reversing one gives 1.5 V.

Resistance and wire comparisons

Resistance, R
R = V/I; V = IR; I = V/R. Unit: ohm, Ω = V/A. Use the voltage across and current through the same component at the same operating point.
Wire dimensions
Same material and temperature: R ∝ L and R ∝ 1/A. For a circular cross-section, A ∝ d2. Twice the length and twice the diameter give half the resistance.
Measurement
Use suitable meter ranges and d.c. polarities. Keep wire temperature steady in length comparisons and measure the active length between contacts. Contact resistance, heating and diameter variation need specific attention.

Temperature and I-V shapes

  • Metallic conductor: increasing temperature increases resistance in the stated range. At constant temperature, its ohmic I-V graph is a straight line through the origin.
  • Filament lamp: resistance rises as the filament heats. With I vertical and V horizontal, the curve becomes less steep as voltage magnitude increases.
  • Diode: appreciable current flows forward under suitable conditions; reverse current is negligible within the illustrated range.
  • Read a point: R = V/I. For a straight I-V line through the origin, gradient = 1/R; a curved graph's tangent gradient is not the same ratio.
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