Currents

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:

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.

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|:

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.

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.

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:

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 V_Q = 0 and point P has V_P = +6 V. Their p.d. is V_P - V_Q = 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.

The power equations express these energy-per-charge and charge-per-time ideas directly.

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:

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.

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:

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.

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.

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:

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(h_finish - h_start). 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.

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

Evaluate the measurement, not just the percentage

If V or I varies materially, form matched samples P_i = V_iI_i 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.

05

Describe an alternating supply

An alternating current reverses direction. An alternating voltage reverses polarity relative to named terminals. A negative graph value identifies the opposite direction or polarity to that reference.

A sinusoid is one possible alternating waveform. Its period T is the time for one complete repeat, its frequency f is the number of complete cycles per second, and its peak value x₀ is the maximum magnitude of the quantity.

T is in seconds, f in hertz (Hz = s^-1) and angular frequency ω in rad/s. One cycle is 2π radians. Angular frequency is not numerically the same as frequency in Hz.

For a sine wave whose time origin is an upward zero crossing:

Here x stands for either current or voltage, not a displacement. The argument ωt is in radians. A different starting phase needs a corresponding phase shift in the equation; do not impose an upward zero crossing on a graph that starts elsewhere.

Read a measured or supplied trace

Identify the zero reference, voltage or current scale and time-base scale. Measure the time between equivalent points moving in the same sense, such as two upward zero crossings. Measuring several complete cycles can reduce the fractional error in timing, provided the frequency remains stable. Measure peak magnitude from zero, not from the negative trough.

A suitable low-voltage source and waveform instrument, or a supplied trace, can show the pattern. Preserve the instrument settings and stated calibration with measured readings. A meter's a.c. display must be interpreted using its specified waveform and rms response; not every a.c. meter gives a true rms reading for an arbitrary waveform.

The rms value describes equivalent heating over a cycle. It is different from a peak, a signed mean or a chosen instantaneous reading.

06

Mean power and rms values

Opposite current directions can produce the same heating in a resistor. Mean signed current therefore does not tell you its mean heating power.

Use a constant resistive load R and a sinusoidal current i = I₀ sin(ωt). With the passive reference v = iR, instantaneous power is:

Voltage and current reverse together, so their product stays non-negative. Over a complete cycle the positive and negative current contributions cancel, giving zero mean signed current. Squaring them removes that sign cancellation.

Deduce the mean power

The identity sin²(θ) + cos²(θ) = 1 holds at every angle. The two squared functions have the same full-cycle mean because one is a quarter-cycle shift of the other. Their means must therefore each be 1/2.

This deduction uses the full sinusoidal variation. It is not an instruction to average only the maximum and minimum readings for any waveform.

The equivalent steady current

The root-mean-square current is the steady d.c. current producing the same mean heating power in the same resistor. It is the square root of the mean of the squared instantaneous current:

For the specified sinusoid, substitute P_mean = I₀²R/2 and take the non-negative square root. The voltage relation follows in the same way from p = v²/R:

In the example, I_rms = 2√2 A ≈ 2.83 A and V_rms = 12√2 V ≈ 17.0 V. For this in-phase, purely resistive load:

The rms value belongs to a full-cycle squared average. The fact that the sinusoidal current happens to equal +2.83 A at 2.50 ms does not define rms by that instant.

Keep instantaneous value, signed mean, peak, peak-to-peak and rms distinct. Dividing a peak by √2 applies to a sinusoid; the general rms definition does not give every periodic waveform that ratio. The product V_rmsI_rms above also uses the stated in-phase resistive condition.

07

Half-wave rectification

A single diode can allow load current during one half of an alternating input and block it during the other. The result is current in one direction, with zero-current intervals.

Use an ideal diode: zero forward voltage drop while conducting and no reverse current. Its cathode is identified by the bar in the circuit symbol. This model gives a simple rule for the two circuit states; a real diode's forward drop can reduce the measured output peak.

Fix the circuit and voltage references

The source is between A and return C, with input v_in = V_A - V_C. The diode connects A, its anode, to B, its cathode. A 6.00 Ω resistor connects B to C. Output is measured across that resistor as v_out = V_B - V_C; positive load current goes B to C.

Take v_in = 24.0 sin(100πt) V, with t in seconds and the initial upward zero crossing at t = 0. Choosing C as zero potential makes it easier to compare the other two nodes.

• Positive input: A is above C. The diode conducts, so B has A's potential in the zero-drop model. Current passes from A through the diode to B, then through the load to C and back through the source. The resistor output follows the positive input.
• Negative input: A is below C. The diode is reverse biased and blocks current. With zero current through the resistor, its p.d. is zero and B is at C's potential. The source input can be negative while the load output is zero.

The ideal positive output peak is still 24.0 V, giving peak load current 24.0/6.00 = 4.00 A. Positive pulses repeat every 20.0 ms, so their repetition frequency is 50.0 Hz. This is one pulse per input cycle. It differs from the unrectified resistor's two power peaks per current cycle.

The rectified current is unidirectional but not steady. A single diode does not hold the last peak across the resistor during a blocked half-cycle. No energy-storage or smoothing component is included in this circuit.

Reversing the physical diode produces negative output pulses under the same V_B - V_C reference. Reversing only a voltmeter's leads changes the displayed sign, not which half-cycle the diode conducts. Keep the circuit orientation and measurement reference separate.

For measured waveforms, preserve the time-base, voltage scale and diode/load details. A real forward drop and instrument response can explain differences from the ideal peak; an a.c. meter calibrated for a sinusoid may not correctly report this rectified waveform's rms value.

Revision

Currents: revision summary

Identify the charge-flow section, voltage endpoints and energy boundary before choosing an equation.

Charge, drift and energy transfer

• 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.

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 V_iI_i, 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.

Sinusoidal quantities and equivalent heating

The half-peak mean-power and divide-by-√2 results require a sinusoid and the stated constant resistive load. The general rms meaning is equivalent d.c. heating in the same resistor. Zero mean signed current can still give positive mean power.

For i = 4 sin(100πt) A through 6 Ω: T = 20 ms, f = 50 Hz, voltage peak 24 V, power peak 96 W and mean power 48 W. Current/voltage reverse together; power repeats every 10 ms. Energy over a full current cycle is 0.960 J.

Single-diode output

With diode anode A, cathode B and resistor B-to-C, positive V_A - V_C conducts and the negative half-cycle blocks. Output V_B - V_C is zero when blocked. The ideal positive pulses retain the input peak but repeat once per full input period; they are not steady current. Use the full waveform when finding rms, rather than reusing a sinusoidal peak rule.

Review the equal-area rms deduction or trace the two diode states.

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. Lowercase i and p denote instantaneous current and power. Lowercase v on the alternating-voltage graphs denotes instantaneous voltage, rather than the drift speed v used in the carrier model.