8867 / 2027
Currents overview

Topic 4 of 4

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.

A series ammeter and parallel voltmeter measure the thirty-ohm loadA complete low-voltage d.c. circuit runs from the positive source terminal through ammeter A to terminal P, through the chosen load to terminal Q, and back to the negative source terminal. The voltmeter is on a separate branch connected exactly at P and Q, with positive reference at P. The resistor is thirty ohms in this supplied model. The ammeter reads 0.200 ampere and the voltmeter 6.00 volts, giving 1.20 watts in the resistor. Brown arrows indicate conventional load current. The voltmeter branch does not short the load: the V symbol is a high-resistance meter. Ideal-meter approximations are stated in the figure text.+-A0.200 A30.0 ΩPQV+-6.00 VILow-voltaged.c. sourceCompletereturn path

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.

The ammeter is in the load path and the voltmeter compares resistor terminals P and Q. The displayed 6.00 V and 0.200 A are supplied model readings, with negligible meter loading; they are not specifications of an instrument.

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:

I = 6.00/30.0 = 0.200 A
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.

Series current and motor-terminal voltage measure motor and gearing inputA complete low-voltage d.c. circuit runs from the positive source terminal through ammeter A to terminal P, through the chosen load to terminal Q, and back to the negative source terminal. The voltmeter is on a separate branch connected exactly at P and Q, with positive reference at P. The ammeter reads 0.100 ampere and the voltmeter 6.00 volts. A dashed boundary contains the motor and gearing, whose mechanical output is shown in the next panel. Input is measured across these motor terminals, not inferred from the source's rating. The gearing symbol and its shaft are mechanical parts, not additional electrical branches. Brown arrows indicate conventional load current. The voltmeter branch does not short the load: the V symbol is a high-resistance meter. Ideal-meter approximations are stated in the figure text.+-A0.100 AMMotor + gearingPQV+-6.00 VILow-voltaged.c. sourceCompletereturn path

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

One 0.200-kilogram load rises half a metre in the matched five-second intervalThe same motor and gearing winds a cord and lifts one load. A solid upper load and a dashed lower load are alternative positions, not two loads. The same marked point on the load has start height 0.100 metre and finish height 0.600 metre. The vertical scale is five hundred drawing units per metre, with zero at y four hundred, start at y 350 and finish at y one hundred. Thus the 0.500-metre rise occupies 250 units. A blue upward arrow shows the change of position, not a force. The five-second interval starts and ends at these two crossings; endpoint load speeds are equal and the motor and gearing have a steady operating state. The total raised mass is 0.200 kilogram. Electrical input and useful output refer to this same interval.h / m00.1000.600MotorFinishStartRise0.500 mSame interval: 5.00 sTotal raised mass: 0.200 kg

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.

The electrical view measures input at motor terminals P and Q. The mechanical view shows alternative positions of the same load at the two height markers, with one matched 5.00 s lifting interval. The height and motion markers are not force arrows.
  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.

Supplied model, not observed data

One matched interval

Supplied lifting model with steady readings and equal endpoint speeds
QuantityValue
Total raised mass0.200 kg
g9.81 N/kg
Start / finish height0.100 / 0.600 m
Matched interval5.00 s
Motor-terminal p.d.6.00 V
Motor current0.100 A
Δh = 0.600 - 0.100 = 0.500 m
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?
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.