In short: electrical measurements is about getting a number you can defend — knowing what the instrument does to the circuit, how big the error is, and what the reading actually means. This page works the standard calculations in full, with units at every step. FLN Tutors (Freelancer Network) also offers one-to-one help with the same topics. Matched to a verified specialist who knows your course, your university's marking criteria, and your deadline.
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Most of the paper comes down to four things: balance a bridge, design a meter range, quantify an error, or read something off a screen. All four are numerical, and all four are lost more often to method than to arithmetic.
The calculations below are worked in full. Circuit analysis itself — single-phase and three-phase power, transformer ratios — is on the electrical engineering page, and amplifier and filter calculations are on the analog electronics page, so they are not repeated here.
One habit worth forming before any of it: write down what you are measuring, what the instrument will do to the circuit, and how accurate the answer can possibly be. The third question is the one students skip, and it is frequently the one carrying the marks.
The circuit. Four resistances in a diamond, a supply across one diagonal and a galvanometer across the other. At balance the galvanometer reads zero, which means no current flows through it and the two halves divide the voltage identically.
The balance condition. With ratio arms P and Q, a known variable standard R and the unknown S:
P ÷ Q = R ÷ S, so S = R × Q ÷ P
The problem. P = 100 Ω, Q = 1000 Ω, and the bridge balances with R = 245 Ω. Find S.
S = 245 × 1000 ÷ 100 = 2,450 Ω
The trap. Swap the ratio arms by mistake and you get S = 245 × 100 ÷ 1000 = 24.5 Ω — a factor of a hundred out, from one transposition. The ratio arms set the multiplier, so label which arm is P and which is Q on your diagram before substituting, every time.
The part most answers leave out. The unknown is calculated from three measured quantities, so their errors accumulate. Since S = RQ/P is a product and a quotient, the relative errors add:
If P, Q and R are each accurate to ±0.1%, then S is accurate to about ±0.3%, which is roughly ±7 Ω on 2,450 Ω.
That is why a bridge is accurate: not because the galvanometer is good, but because a null reading removes the galvanometer's own calibration from the answer entirely. The galvanometer only has to detect zero, and detecting zero needs no calibration at all. State that and you have answered the standard "why is a null method preferred" question.
A moving-coil movement has a full-scale deflection current of 1 mA and a coil resistance of 100 Ω. That alone measures 1 mA and 0.1 V, which is useless. Both extensions are standard questions.
As an ammeter reading 1 A — add a shunt in parallel.
At full scale the movement takes 1 mA, so the shunt must carry the rest:
I_shunt = 1 − 0.001 = 0.999 A
The shunt and the movement are in parallel, so they share the same voltage:
R_sh = (I_m × R_m) ÷ I_shunt = (0.001 × 100) ÷ 0.999 = 0.1 ÷ 0.999 = 0.1001 Ω
Check it the other way. The multiplying power n = I ÷ I_m = 1 ÷ 0.001 = 1000, and R_sh = R_m ÷ (n − 1) = 100 ÷ 999 = 0.1001 Ω. Same answer, which confirms the working.
As a voltmeter reading 10 V — add a multiplier in series.
The total resistance must limit the current to full scale at 10 V:
R_total = V ÷ I_m = 10 ÷ 0.001 = 10,000 Ω
The movement already provides 100 Ω, so:
R_series = 10,000 − 100 = 9,900 Ω
Again by the other route: m = 10 ÷ (0.001 × 100) = 100, and R_series = R_m(m − 1) = 100 × 99 = 9,900 Ω.
The sentence that earns the mark: an ammeter goes in series and must have the lowest possible resistance, so the shunt is tiny. A voltmeter goes in parallel and must have the highest possible resistance, so the multiplier is large. Get the shunt and multiplier the wrong way round and the sizes give it away immediately — a 9,900 Ω shunt would read almost nothing, and a 0.1 Ω voltmeter would short the circuit out.
This is the most useful thing on the page and the hardest to find explained anywhere. Search for it and you get calculators, not reasons.
The circuit. A 10 V source with two 100 kΩ resistors in series. The voltage across the lower one is obviously 5 V.
The voltmeter. Sensitivity 20,000 Ω/V, used on the 10 V range, so its resistance is 20,000 × 10 = 200 kΩ.
What happens when you connect it. The voltmeter sits in parallel with the 100 kΩ it is measuring:
R_parallel = (100k × 200k) ÷ (100k + 200k) = 20 × 10⁹ ÷ 300 × 10³ = 66.667 kΩ
The divider is now 100 kΩ over 66.667 kΩ, so the reading is:
V = 10 × 66.667 ÷ (100 + 66.667) = 10 × 66.667 ÷ 166.667 = 4.00 V
The meter reads 4 V where the true value is 5 V — an error of −20%. Nothing is broken. The voltmeter is working exactly as designed, and it is still wrong, because connecting it changed the circuit it was measuring.
What follows from that: - The error is worst when the meter resistance is comparable to the circuit resistance. Across a low-resistance circuit the same meter would be fine. - A higher sensitivity meter, say 200,000 Ω/V, gives 2 MΩ on this range and the reading rises to about 4.88 V. Better, still not exact. - Choosing a higher voltage range raises the meter resistance, which reduces loading but costs you scale accuracy. The two pull in opposite directions, and saying so is the complete answer.
The general principle is worth memorising in one line: every instrument takes energy from the circuit it measures, so every measurement disturbs what it measures. The skill is knowing when the disturbance matters.
The three categories.
| Type | Cause | What to do about it |
|---|---|---|
| Gross | Human — misreading, wrong recording, wrong range | Take several readings, by more than one person |
| Systematic | Instrumental, environmental or observational, including parallax | Calibrate, control conditions, read square on |
| Random | Residual, unpredictable causes | Take many readings and treat statistically |
The key distinction: a systematic error pushes every reading the same way and averaging will never remove it. A random error scatters both ways and averaging does reduce it. An answer that says "take more readings" for a systematic error has missed the point.
The basic definitions, with a measured 205 V against a true 200 V:
Now the trap that is asked almost every year. Instrument accuracy is usually quoted as a percentage of full-scale deflection, not of the reading.
Take a 0–150 V voltmeter specified as ±1% of full scale. Its limiting error is fixed at:
±1% of 150 = ±1.5 V, anywhere on the scale.
The same meter is two and a half times worse at 30 V than at 75 V. This is the reason for the practical rule that you choose a range on which the reading falls in the upper part of the scale, and being able to show that with numbers is what the question is testing.
Bridges extend naturally to alternating current, where balance requires both magnitude and phase to match — so two conditions must be satisfied, not one.
| Bridge | Measures | Best suited to |
|---|---|---|
| Maxwell's | Inductance, against a standard capacitor | Coils of medium quality factor |
| Hay's | Inductance | High quality factor coils |
| Anderson's | Inductance | Low quality factor coils |
| Schering | Capacitance and dissipation factor | Insulation and dielectric testing |
| De Sauty | Capacitance | Comparing two capacitors |
| Wien | Frequency | Frequency measurement, and oscillators |
Why a standard capacitor rather than a standard inductor? Because a capacitor of known value is far easier to make accurate and stable: it has no significant stray magnetic field, is little affected by nearby metal, and does not suffer the losses a coil does. That is the reasoning behind Maxwell's bridge and it is a favourite short-answer question.
Deriving Maxwell's bridge balance. Put the unknown coil (inductance L₁, resistance R₁) in arm 1, plain resistors R₂ and R₃ in arms 2 and 3, and the standard capacitor C₄ in parallel with R₄ in arm 4. AC balance requires Z₁Z₄ = Z₂Z₃. Arm 4 is a parallel combination, so it is easier to use its admittance, Y₄ = 1/R₄ + jωC₄:
Z₁ = Z₂Z₃Y₄, so R₁ + jωL₁ = R₂R₃(1/R₄ + jωC₄)
Equate the real parts and then the imaginary parts:
R₁ = R₂R₃ ÷ R₄ and L₁ = R₂R₃C₄
Neither result contains ω, so the balance does not depend on frequency. That is the point an examiner is looking for after the derivation.
Worked numbers. R₂ = 500 Ω, R₃ = 1000 Ω, R₄ = 2000 Ω and C₄ = 0.5 µF at balance:
A Q of about 6 is inside the range Maxwell's bridge handles well. For a high-Q coil, R₄ would need to be impractically large, and that is why Hay's bridge exists.
Frequency from the timebase. One complete cycle spans 4 divisions with the timebase at 0.5 ms/division:
T = 4 × 0.5 = 2 ms, so f = 1 ÷ 0.002 = 500 Hz
Voltage from the vertical scale. The trace spans 3.2 divisions peak-to-peak at 2 V/division:
V_pp = 3.2 × 2 = 6.4 V, so V_peak = 3.2 V
and for a sine wave V_rms = 3.2 ÷ √2 = 2.263 V
The two errors here are constant. First, reading peak-to-peak and then quoting it as if it were RMS. Second, dividing by √2 for a waveform that is not sinusoidal — that factor is a property of the sine wave and of nothing else. A square wave's RMS equals its peak.
Lissajous figures. Feeding an unknown frequency to one pair of plates and a known one to the other gives a closed figure whose shape fixes the ratio. Counting the points where the figure touches a horizontal line and a vertical line:
f_vertical ÷ f_horizontal = (tangencies on the horizontal) ÷ (tangencies on the vertical)
A figure with 2 horizontal tangencies and 3 vertical gives a ratio of 2 to 3. The method also shows phase difference between two signals of the same frequency, which is what it is most often used for in a lab.
Four words used loosely in conversation and strictly in the exam.
| Term | Meaning | A way to remember it |
|---|---|---|
| Accuracy | How close a reading is to the true value | Hitting the bullseye |
| Precision | How close repeated readings are to each other | Hitting the same spot each time |
| Resolution | The smallest change the instrument can show | The gap between adjacent digits |
| Sensitivity | Output change per unit input change | Ohms per volt, for a voltmeter |
An instrument can be precise but inaccurate — tightly grouped readings, all 5 V too high, which is a systematic error. It cannot usefully be accurate but imprecise, because a reading you cannot repeat is not evidence of anything.
A four and a half digit display showing 12.3456 V has a resolution of 0.0001 V. That says nothing whatever about whether the figure is right. Confusing a long display with accuracy is one of the commonest errors in a lab report.
See electrical engineering help for circuit analysis and power, analog electronics help for amplifiers and filters, digital electronics help, process instrumentation help and electromagnetic field theory help.
If a bridge or error calculation keeps coming out wrong, send it over with your own working and you will get an answer from someone who teaches the subject.
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Electrical measurements covers the instruments and methods used to measure quantities such as current, voltage, resistance, inductance, capacitance and frequency. It also covers the analysis that goes with them: how accurate a reading can be, and how the instrument itself disturbs the circuit it is connected to.
Four resistances form a diamond with a supply across one diagonal and a detector across the other. The variable arm is adjusted until the detector reads zero, meaning both halves divide the voltage identically. The unknown then follows from the ratio of the other three.
The ratio of the two arms on one side equals the ratio on the other, so P divided by Q equals R divided by S. Rearranged, the unknown equals the known standard multiplied by the ratio arms, which is why transposing those arms is so costly.
Because at balance the detector reads zero, so its own calibration never enters the result. It only has to indicate the absence of current, which requires no accuracy at all. The answer depends instead on the standard resistances, which can be made very accurate.
The shunt carries the current the movement cannot. Multiply the full-scale current by the coil resistance to get the voltage across the movement, then divide by the shunt current. Equivalently, the shunt equals the coil resistance divided by the multiplying power minus one.
Add a resistor in series with the movement, called a multiplier. Divide the desired full-scale voltage by the movement's full-scale deflection current to find the total resistance required, then subtract the coil resistance already present. The remainder is the multiplier value needed for that particular range.
Because it is connected in series with the circuit, so its resistance adds to the circuit resistance and reduces the very current it is trying to measure. The lower its resistance, the smaller that disturbance. A voltmeter is parallel, so it needs the opposite.
It is the change a measuring instrument causes in the circuit simply by being connected. A voltmeter draws some current, so it lowers the voltage it reads. The effect is worst when the instrument's resistance is comparable with the resistance across which it is connected.
Use an instrument whose resistance is far higher than the circuit resistance, which for a voltmeter means high sensitivity in ohms per volt. Selecting a higher range also raises its resistance, though at some cost in scale accuracy, so the two considerations are balanced.
Accuracy is closeness to the true value; precision is closeness of repeated readings to one another. An instrument can be precise but inaccurate, giving tightly grouped readings that are all wrong by the same amount, which indicates a systematic error rather than random scatter.
Gross errors come from human mistakes such as misreading or recording wrongly. Systematic errors come from the instrument, the environment or the observer, including parallax. Random errors come from unpredictable residual causes and are the only type that averaging many readings genuinely reduces.
Limiting error is the maximum deviation an instrument may show, usually specified as a percentage of full-scale deflection rather than of the reading. Because it is fixed across the scale, it becomes a larger percentage of smaller readings, which is why low readings are less reliable.
Because accuracy is specified on full-scale deflection, so the absolute error stays constant while the reading changes. A meter accurate to one per cent of a 150 volt scale is within two per cent at 75 volts but only five per cent at 30 volts.
Because a capacitor of known value is easier to make accurate and stable than a standard inductor. It produces no significant external magnetic field, is largely unaffected by nearby metal objects, and does not suffer the resistive losses that a real coil inevitably has.
Yes. A one-to-one tutor can work through bridge balance, shunt and multiplier design, loading and error calculations with you, and check how you read and interpret instrument specifications. You complete your own assessed work in line with your institution's academic rules.
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