Full chapter
Currents
All 4 topics and the revision summary on one page.
01
Charge flow and drift
Electric current measures the rate of charge flow through a chosen cross-section. A large current can coexist with a very small drift speed because the conductor contains many charge carriers.
If charge Q crosses a section in time t, the mean current is:
1 A = 1 C/s and 1 C = 1 A s
For steady current, this mean also equals the current throughout the interval. If the rate changes, Q/t describes that interval's average. Specify the direction when using signed current or charge transfer. Conventional current follows the direction in which positive charge would move.
The elementary charge e is a positive magnitude. Using the supplied value e = 1.60 × 10-19 C, an electron has charge -e. Electrons drifting one way in a metal give conventional current the other way. A resistor transfers energy while charge continues through the steady circuit; it does not use up the charge.
Count the carriers crossing a plane
Consider a uniform conductor of cross-sectional area A. Let n be its number density of mobile carriers, in m-3, and v their mean drift-speed magnitude. Assume one effective carrier type, uniform density and a uniform mean drift velocity over the chosen interval.
Count charge crossing one section
Electrons already occupy the whole metal. The shaded volume is a counting construction for their mean drift, not a packet moving through an otherwise empty wire. The cylinder and particle symbols are schematic.
For n = 8.00 × 1028 m-3, this volume contains 8.00 × 1019 carriers in the model. Their charge magnitude is 12.8 C in 5.00 s, giving I = 2.56 A.
Irregular motion and net drift are different
The sketch does not compare drift speed with current by arrow length. The circuit's electrical response and the much slower drift of an individual electron are also different processes.
In time Δt, the mean drift distance is vΔt. The associated volume is A vΔt, so the number of carriers crossing the section in the drift model is N = nA vΔt. Each carrier has charge magnitude |q|:
I = Q/Δt = nAv|q|
The commonly written form is I = nAvq, with q understood as the carrier-charge magnitude when I and v are magnitudes. To use a negative carrier charge in a signed equation, also define an axis and a signed drift velocity consistently. Do not insert electron charge -e while treating everything else as an unsigned magnitude.
Here n is a count per volume, not a total count N or an amount in moles. A means area, while the unit symbol A means ampere. The dimensions confirm the current unit:
Estimate the scale
For a rough metal-wire model, take n of order 1029/m3, area 10-6 m2, charge magnitude 10-19 C and drift speed 10-4 m/s. Their product suggests current of order 1 A. Over several seconds, charge can be of order 10 C while a carrier drifts only of order 1 mm. These are rough model scales; use the supplied values for a precise calculation.
Worked carrier model
Slow drift, appreciable current
Use n = 8.00 × 1028 m-3, A = 1.00 mm2 = 1.00 × 10-6 m2, v = 2.00 × 10-4 m/s and |q| = 1.60 × 10-19 C.
× (2.00 × 10-4)(1.60 × 10-19)
= 2.56 A
Over 5.00 s, check each step of the counting model:
Swept volume = Avt = 1.00 × 10-9 m3
N = nAvt = 8.00 × 1019 carriers
Q = N|q| = 12.8 C
Q/t = 12.8/5.00 = 2.56 A
At fixed current, n and |q|, a second wire with half the area needs twice the drift speed: 4.00 × 10-4 m/s. Holding voltage fixed instead would not automatically hold current fixed.
In a metal, rapid irregular electron motion has a much smaller net drift superimposed on it. Irregular motions in opposite directions do not alone produce a sustained net current. Charges already exist throughout the circuit, and its electrical response propagates separately from the slow drift of any one carrier. A particular electron need not travel from the source to the load before the load responds; there is no single universal signal speed for every circuit.
Optional check A wire carries a fixed current with drift-speed magnitude 2.00 x 10^-4 m/s. A second wire has half the cross-sectional area, with the same current, carrier density and carrier-charge magnitude. What is its drift speed?
02
Energy per charge
Potential difference describes energy transferred per charge between two points. A source's e.m.f. describes the energy it supplies per charge, including energy subsequently transferred inside the source.
If electrical work W is done as charge Q passes through a load, the potential difference across its named terminals is:
1 V = 1 J/C = kg m2 s-3 A-1
A load transferring 24.0 J as 4.00 C passes has p.d. 24.0/4.00 = 6.00 V. That is 6.00 J transferred for each coulomb, not 6.00 J regardless of how much charge passes. Current is through a section; voltage is between two points.
A potential level needs a reference
Electric potential is electric potential energy per unit positive test charge, relative to a chosen reference. Suppose point Q is assigned VQ = 0 and point P has VP = +6 V. Their p.d. is VP - VQ = 6 V. For a positive test charge in the given field, this represents a 6 J/C difference in electric potential energy.
Assigning Q a value of +10 V and P a value of +16 V leaves that difference unchanged. Adding the same constant to both potential levels changes the reference, not the p.d. A voltmeter compares its two connection points; it does not report a potential level without a reference.
What a source supplies
Electromotive force, or e.m.f., is the energy a source supplies to the circuit per unit charge through it, by converting another form of energy into electrical energy. It is measured in volts, despite the word force. It is not a mechanical force measured in newtons.
For example, a discharging cell converts chemical energy. Some energy can be transferred internally before the remainder reaches the external circuit. Its terminal p.d. can therefore be lower than its e.m.f. The symbol ε is used here for e.m.f.; E is also used for it in circuit equations. Define the quantity locally so it is not confused with electric field strength or total energy.
Worked source account
Keep the source and external boundaries distinct
A steady discharging source has e.m.f. 1.60 V, terminal p.d. 1.50 V and current 0.400 A. The source converts 1.60 J per coulomb; 1.50 J per coulomb reaches the external circuit and 0.10 J per coulomb is transferred internally.
Separate conversion inside the source from delivery to the load
For the same steady current, I = 0.400 A, the source e.m.f. is 1.60 V and its terminal p.d. is 1.50 V. All bars use the same power scale. They show energy rates, not current or voltage arrows.
For each coulomb passing through the source, 1.60 J is supplied, 0.10 J is transferred internally and 1.50 J reaches the external circuit. Current continues through the load.
Current 0.400 A means 0.400 C passes each second. Multiplying each energy-per-charge value by that rate gives:
| Transfer | J per C | Power / W |
|---|---|---|
| Total source conversion | 1.60 | 0.640 |
| External circuit | 1.50 | 0.600 |
| Inside the source | 0.10 | 0.040 |
The e.m.f. and terminal p.d. are supplied here; the account does not require an internal-resistance calculation. A voltmeter across the external load measures its p.d., not automatically the source's total energy conversion per charge.
The power equations express these energy-per-charge and charge-per-time ideas directly.
Optional check A discharging source has e.m.f. 1.60 V, terminal p.d. 1.50 V and steady current 0.400 A. Which power account is consistent with these readings?
03
Electrical power and energy
Electrical power is the rate of energy transfer. Multiply energy per charge by charge per second, using the voltage and current of the same component.
For steady p.d. V and current I, start with work W = VQ and power P = W/t:
For a resistor at an operating state described by V = IR, substitute that relation in two ways:
P = VI = V(V/R) = V2/R
Use the form that matches the given quantities. V is across the resistor and I is through it. The local resistor relation does not mean that a motor's useful mechanical output equals I2R: its terminal input is VI, and the useful output needs its own measurement.
State what stays fixed
- At fixed resistance, doubling voltage quadruples power: P is proportional to V2.
- At fixed current, increasing resistance increases power: P = I2R.
- At fixed voltage, increasing resistance decreases power: P = V2/R.
These comparisons hold different quantities constant. There is no contradiction between them.
Worked steady resistor
Check the result by all three forms
A few volts at a few amperes suggests power of order 10 W and a few hundred joules over tens of seconds. Now use the precise supplied model: 6.00 V across a constant 3.00 Ω resistor for 40.0 s.
P = VI = 6.00 × 2.00 = 12.0 W
P = I2R = 2.002 × 3.00 = 12.0 W
P = V2/R = 6.002/3.00 = 12.0 W
At this steady operating state:
The current and power are substantial for a small resistor. This is a supplied calculation model; a physical setup at these values would need suitable current and power ratings. The measurement example uses a separate, lower-power load.
Energy over time
For constant power, the area under a power-time graph is a rectangle, Pt. For changing power, the total area still gives transferred energy; the average power multiplied by the full interval gives the same energy. Preserve the actual time widths when estimating an area from samples.
A kilowatt-hour is another energy unit:
480 J = 480/(3.6 × 106) kWh
= 1.33 × 10-4 kWh
kW is power; kWh is energy. In W = Pt, the quantity W denotes work or transferred energy in joules, whereas the upright unit W after a number denotes watts.
For a passive resistor, I2R is non-negative even when a chosen current reference makes I negative. With current defined into the terminal used as positive for the voltage, VI also describes the power absorbed by the load.
Optional check A constant 3.00 ohm resistor has 6.00 V across it for 40.0 s. What electrical energy is transferred to it?
04
Measure input and useful output
Measure voltage at the chosen component and current through it. To calculate efficiency, measure its useful output separately over the same start and end events.
Electrical energy delivered to a resistor
A supplied bench model uses 30.0 Ω at about 6.00 V. A quick estimate gives I about 6/30 = 0.2 A and power of order 1 W. A 0-0.1 A range is insufficient. A suitable 0-1 A range may work if its resolution is adequate; check the actual instrument rather than assuming a range also guarantees sufficient resolution.
Measure the chosen resistor, between P and Q
The ammeter is in the load's current path; the voltmeter is connected across P and Q. Current I enters P, and V = VP - VQ. The model assumes negligible voltmeter current and negligible ammeter voltage drop. Displayed values are supplied readings, not instrument ratings.
The selected load receives 1.20 W, or 48.0 J over 40.0 s if both readings stay steady. This is the 30.0 Ω bench model; it is separate from the 3.00 Ω power-calculation example.
Use an appropriately rated low-voltage supply and a resistor rated above the expected dissipated power. Connect the ammeter in series and the voltmeter across the load. Check the meter zero, polarity, range and resolution, then record actual V, I and elapsed time with units.
If the supplied readings remain steady for 40.0 s:
P = VI = 6.00 × 0.200 = 1.20 W
W = VIt = 1.20 × 40.0 = 48.0 J
This is a different load from the 3.00 Ω, 12.0 W calculation. Check for changed readings as the resistor warms. Switching off between measurements reduces warming; it does not prove resistance remained constant. If leads or other components take appreciable power, the measured load input differs from total source output.
V, I and t can determine the energy delivered electrically to the resistor under the steady-reading approximation. They do not alone measure useful mechanical output or establish a heating efficiency.
Compare electrical input with a measured lift
Choose a boundary around a motor and gearing. Measure p.d. at the motor terminals and current in its path. Assume the voltmeter draws negligible current and the ammeter has negligible resistance. The motor's electrical input power is measured VI; a source rating or total source power is not automatically the input at those terminals.
During a steady vertical lift, useful work increases the gravitational potential energy of the load-Earth system by mg(hfinish - hstart). Use the same start and end events for the height difference, time and electrical readings. Equal endpoint load speeds and a steady motor/gear state avoid unaccounted kinetic-energy changes. The power and efficiency notes explain this useful-work comparison.
Measure input at the motor terminals
The ammeter is in the load's current path; the voltmeter is connected across P and Q. Current I enters P, and V = VP - VQ. The model assumes negligible voltmeter current and negligible ammeter voltage drop. Displayed values are supplied readings, not instrument ratings.
The dashed boundary selects the motor and gearing. The measured input is 6.00 V × 0.100 A = 0.600 W; VI alone does not identify useful mechanical output.
Use the same two events for height, time and input
With g = 9.81 N/kg, useful work is 0.981 J. Motor-terminal input over the same 5.00 s is 3.00 J. Equal endpoint speeds avoid an extra change in the load's kinetic energy; the supplied model gives 32.7% efficiency.
- Set up a suitable low-voltage lift. Use equipment rated for the motor, including its starting current. Secure a small load and arrange two vertical height markers within the steady part of the motion.
- Record the quantities and references. Measure the total raised mass, both vertical positions and the elapsed time between the marker events. Record motor-terminal voltage and motor current during that same interval, with units, instrument ranges/resolutions and the supplied local g.
- Check the operating state. Confirm that voltage, current and speed remain sufficiently steady and that the endpoint speeds agree. Keep the start/end definitions unchanged between the electrical and mechanical measurements.
- Compare input with useful output. Calculate VIt and mgΔh for the same run. Repeat comparable runs while retaining their actual variation and any unusual observations, with reasons for any exclusion.
Supplied model, not observed data
One matched interval
| Quantity | Value |
|---|---|
| Total raised mass | 0.200 kg |
| g | 9.81 N/kg |
| Start / finish height | 0.100 / 0.600 m |
| Matched interval | 5.00 s |
| Motor-terminal p.d. | 6.00 V |
| Motor current | 0.100 A |
Pinput = VI = 0.600 W
Winput = VIt = 3.00 J
Wuseful = mgΔh = 0.200 × 9.81 × 0.500
= 0.981 J
Puseful = 0.981/5.00 = 0.1962 W
Efficiency = 0.981/3.00 = 0.327 = 32.7%
Useful output power is about 0.196 W, not the 0.600 W terminal input. Friction, warming and other energy transfers can account for the difference.
Optional check During the same 5.00 s steady lift, a motor has 6.00 V and 0.100 A at its terminals and raises 0.200 kg by 0.500 m. Use g = 9.81 N/kg and equal endpoint speeds. What are the input energy, useful output and efficiency?
Evaluate the measurement, not just the percentage
If V or I varies materially, form matched samples Pi = ViIi and estimate the power-time area over the same lifting interval. Multiplying separately averaged voltage and current does not generally give their mean product.
Under otherwise correct readings, understating the vertical rise lowers the inferred useful work and efficiency; overstating input V or I raises the inferred input and lowers the efficiency. Identify the affected quantity and cause. Repetition does not repair mismatched events, a wrong measurement boundary or an incorrect height reference.
An apparent efficiency above one calls for a check of the data, uncertainties and model: for example, unmatched times or a loss of stored kinetic energy could invalidate the simple comparison. Preserve the readings while investigating the reason; do not edit them to make the ratio look plausible.
Revision
Currents: revision summary
Identify the charge-flow section, voltage endpoints and energy boundary before choosing an equation.
Charge, drift and energy transfer
I = nAvq (magnitude form)
V = W/Q
P = VI = I2R = V2/R
W = Pt (constant power)
- Current: Q/t is a mean over the interval, and is also the steady current when the flow rate is constant.
- Drift: n is carrier number density, A is cross-sectional area and v is mean drift speed. In the magnitude equation, q is positive carrier-charge magnitude; electron charge itself is -e. The count follows from volume AvΔt.
- Direction: electron drift is opposite conventional current. Charge is not used up in a resistor, and slow drift does not require a slow response of the whole circuit.
- Voltage: name both endpoints and a reference for potential levels. Adding the same constant to both levels leaves their difference unchanged.
- E.m.f.: source energy supplied per charge can exceed the external terminal transfer per charge during discharge. Account for the internal part separately.
- Power: use voltage and current of the same component. The resistor forms use V = IR; state whether current, voltage or resistance is fixed in a comparison.
For changing power, energy is the power-time area. Use 1 kWh = 3.6 × 106 J; kWh is energy and kW is power.
Measurement conditions
Put the ammeter in the component's path and the voltmeter across its terminals, with appropriate ranges, resolution and loading assumptions. Measured VI is electrical input; useful mechanical output requires a separate measurement.
Useful lifting work = mg(hfinish - hstart)
Efficiency = useful output / input
Use matched events and a steady motor/gear state with equal endpoint load speeds. For varying V and I, sum the time area of matched products ViIi, rather than multiplying separate means. Keep actual readings, model values and derived results distinguishable; investigate an impossible efficiency without altering the observations.
Return to the meter and measured-lift methods for the full procedure and error reasoning.
| Quantity | Symbol | Unit / meaning |
|---|---|---|
| Current | I | A = C/s |
| Transferred charge | Q | C = A s |
| Carrier charge | q | C; magnitude in I = nAvq |
| Elementary-charge magnitude | e | C; electron charge -e |
| Carrier number density | n | m-3; not moles |
| Cross-sectional area | A | m2 |
| Drift speed | v | m/s |
| Time interval | t or Δt | s |
| Potential level / difference | V | V = J/C; state reference / endpoints |
| Source e.m.f. | ε or E | V; source energy per charge |
| Resistance | R | Ω = V/A |
| Work / transferred energy | W or E | J; also kWh |
| Power | P | W = J/s |
| Raised mass / vertical rise | m, Δh | kg; m |
| Gravitational field strength | g | N/kg |
The symbol A can mean area while the unit A means ampere. W can mean work while the unit W means watt. E needs its local definition: it may denote e.m.f. in volts or energy in joules.
Return to charge flow and drift