Measurement Errors and Statistics
Absolute Error
\Delta A = A_{measured} - A_{true}
| Symbol | Description | Unit |
|---|---|---|
| \Delta A | Absolute error | same as quantity |
| A_{measured} | Measured value | |
| A_{true} | True (reference) value |
Worked example
A voltmeter reads 102.5 V. True value is 100 V. Find absolute and relative error.
Given: A_measured=102.5 V, A_true=100 V
- ΔA = 102.5 − 100 = 2.5 V
- Relative error = ΔA/A_true = 2.5/100 = 0.025 = 2.5%
Answer: Absolute error = 2.5 V, Relative error = 2.5%
Percentage Error
\% \text{ Error} = \frac{\Delta A}{A_{true}} \times 100
| Symbol | Description | Unit |
|---|---|---|
| \% \text{ Error} | Percentage error | % |
Worked example
True resistance = 500 Ω, measured = 495 Ω. Find % error.
Given: A_true=500, A_measured=495
- ΔA = 495 − 500 = −5 Ω
- % Error = (−5/500) × 100 = −1%
Answer: % Error = −1% (reading low)
Mean Value
\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i
| Symbol | Description | Unit |
|---|---|---|
| \bar{x} | Arithmetic mean | same as measurements |
| n | Number of readings |
Worked example
Five voltage readings: 10.2, 10.4, 10.1, 10.3, 10.5 V. Find mean.
Given: Readings: 10.2, 10.4, 10.1, 10.3, 10.5 V
- Sum = 10.2+10.4+10.1+10.3+10.5 = 51.5
- Mean = 51.5/5 = 10.3 V
Answer: Mean = 10.3 V
Standard Deviation
\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}}
| Symbol | Description | Unit |
|---|---|---|
| \sigma | Standard deviation (sample) | same as x |
| \bar{x} | Sample mean |
Worked example
Using readings 10.2, 10.4, 10.1, 10.3, 10.5 V (mean=10.3 V). Find σ.
Given: Deviations: −0.1, 0.1, −0.2, 0, 0.2
- Σ(xi−x̄)² = 0.01+0.01+0.04+0+0.04 = 0.10
- σ = √(0.10/(5−1)) = √(0.025) = 0.158 V
Answer: σ = 0.158 V
Bridges for Resistance, Capacitance, and Inductance
Wheatstone Bridge Balance Condition
\frac{P}{Q} = \frac{R}{S} \Rightarrow R_x = \frac{Q}{P} \cdot S
| Symbol | Description | Unit |
|---|---|---|
| P, Q | Ratio arms of bridge | Ω |
| R | Rheostat arm (known, variable) | Ω |
| S | Unknown resistance (Rx) | Ω |
Worked example
Wheatstone bridge: P=100 Ω, Q=200 Ω, R=360 Ω at balance. Find Rx.
Given: P=100, Q=200, R=360 Ω
- At balance: P/Q = R/Rx
- Rx = R × (Q/P) = 360 × (200/100) = 360 × 2 = 720 Ω
Answer: Rx = 720 Ω
Maxwell's Inductance Bridge
L_x = R_2 R_3 C_4, \quad R_x = \frac{R_2 R_3}{R_4}
| Symbol | Description | Unit |
|---|---|---|
| L_x | Unknown inductance | H |
| R_x | Resistance of coil | Ω |
| R_2, R_3, R_4 | Known bridge resistances | Ω |
| C_4 | Standard capacitor | F |
Worked example
Maxwell bridge at balance: R2=500 Ω, R3=600 Ω, R4=1000 Ω, C4=1 µF. Find Lx and Rx.
Given: R2=500, R3=600, R4=1000, C4=1×10^−6 F
- Lx = R2 × R3 × C4 = 500 × 600 × 1×10^−6 = 0.3 H
- Rx = R2×R3/R4 = 500×600/1000 = 300 Ω
Answer: Lx = 0.3 H, Rx = 300 Ω
Schering Bridge (Capacitance)
C_x = \frac{C_2 R_4}{R_3}, \quad R_x = \frac{C_4 R_3}{C_2}
| Symbol | Description | Unit |
|---|---|---|
| C_x | Unknown capacitance | F |
| C_2 | Standard capacitor | F |
| R_3, R_4 | Bridge resistances | Ω |
Worked example
Schering bridge: C2=100 pF, R3=1000 Ω, R4=2500 Ω. Find Cx.
Given: C2=100×10^−12 F, R3=1000, R4=2500
- Cx = C2×R4/R3 = 100×10^−12 × 2500/1000
- Cx = 100×10^−12 × 2.5 = 250 pF
Answer: Cx = 250 pF
Instrument Sensitivity and Loading
Galvanometer Sensitivity
S_v = \frac{\theta}{V} \quad (\text{div/V}), \quad S_i = \frac{\theta}{I} \quad (\text{div/A})
| Symbol | Description | Unit |
|---|---|---|
| S_v | Voltage sensitivity | div/V |
| S_i | Current sensitivity | div/µA |
| \theta | Deflection in divisions | div |
Worked example
Galvanometer deflects 30 divisions for 6 µA current. Find Si.
Given: θ=30 div, I=6 µA=6×10^−6 A
- Si = θ/I = 30/(6×10^−6) = 5×10^6 div/A = 5 div/µA
Answer: Si = 5 div/µA
Voltmeter Loading Error
V_{measured} = V_{true} \cdot \frac{R_V}{R_V + R_{source}}
| Symbol | Description | Unit |
|---|---|---|
| V_{measured} | Voltmeter reading | V |
| R_V | Voltmeter input resistance | Ω |
| R_{source} | Source/Thevenin resistance | Ω |
Worked example
True voltage across a 50 kΩ source resistance = 10 V. Voltmeter Rv = 100 kΩ. Find reading.
Given: V_true=10 V, Rv=100 kΩ, Rs=50 kΩ
- V_meas = 10 × 100/(100+50) = 10 × 100/150 = 6.67 V
Answer: V_measured = 6.67 V (−33.3% error)
Ammeter Loading Effect
I_{measured} = I_{true} \cdot \frac{R_{circuit}}{R_{circuit} + R_A}
| Symbol | Description | Unit |
|---|---|---|
| I_{measured} | Ammeter reading | A |
| R_A | Ammeter resistance | Ω |
| R_{circuit} | Circuit resistance without ammeter | Ω |
Worked example
10 V across 1000 Ω. Ammeter Ra=10 Ω inserted. Find I_meas vs I_true.
Given: V=10, R_ckt=1000 Ω, Ra=10 Ω
- I_true = 10/1000 = 10 mA
- I_meas = 10/(1000+10) = 10/1010 = 9.9 mA
- Error = (10−9.9)/10 × 100 = 1%
Answer: I_meas = 9.9 mA (1% low)
Transducers — LVDT, Strain Gauge, Thermocouple
Strain Gauge Gauge Factor
GF = \frac{\Delta R / R}{\varepsilon} = \frac{\Delta R / R}{\Delta L / L}
| Symbol | Description | Unit |
|---|---|---|
| GF | Gauge factor (typically 2 for metallic) | |
| \Delta R / R | Fractional change in resistance | |
| \varepsilon | Mechanical strain | m/m |
Worked example
Strain gauge: R=120 Ω, GF=2.0, applied strain ε=500 µε. Find ΔR.
Given: R=120 Ω, GF=2.0, ε=500×10^−6
- ΔR/R = GF × ε = 2.0 × 500×10^−6 = 0.001
- ΔR = 0.001 × 120 = 0.12 Ω
Answer: ΔR = 0.12 Ω
LVDT Sensitivity
S_{LVDT} = \frac{V_{out}}{x} \quad (\text{mV/mm})
| Symbol | Description | Unit |
|---|---|---|
| S_{LVDT} | LVDT sensitivity | mV/mm |
| V_{out} | Differential output voltage | mV |
| x | Core displacement | mm |
Worked example
LVDT gives 250 mV output for 5 mm core displacement. Find sensitivity and output for 3 mm.
Given: V_out=250 mV, x=5 mm
- S = 250/5 = 50 mV/mm
- For x=3 mm: V_out = 50×3 = 150 mV
Answer: S=50 mV/mm; V_out(3mm)=150 mV
Thermocouple Seebeck EMF (Linearised)
E = S_{AB} \cdot (T_{hot} - T_{cold})
| Symbol | Description | Unit |
|---|---|---|
| E | Thermocouple EMF | mV |
| S_{AB} | Seebeck coefficient of thermocouple pair | µV/°C |
| T_{hot} | Hot junction temperature | °C |
| T_{cold} | Cold junction temperature | °C |
Worked example
Type K thermocouple: SAB=41 µV/°C, Thot=500°C, Tcold=25°C. Find EMF.
Given: SAB=41 µV/°C, ΔT=500−25=475°C
- E = 41×10^−6 × 475 = 19.475×10^−3 V = 19.475 mV
Answer: E = 19.475 mV
Signal Conditioning — Op-Amp and Filters
Instrumentation Amplifier Gain
A_v = \left(1 + \frac{2R_1}{R_G}\right) \frac{R_3}{R_2}
| Symbol | Description | Unit |
|---|---|---|
| A_v | Differential voltage gain | |
| R_G | Gain-setting resistor | Ω |
| R_1 | Input stage feedback resistors (equal pair) | Ω |
Worked example
In-amp: R1=10 kΩ, RG=1 kΩ, R2=R3=10 kΩ. Find gain.
Given: R1=10000, RG=1000, R2=R3=10000
- First stage gain = 1 + 2×10000/1000 = 1+20 = 21
- Second stage gain = R3/R2 = 10000/10000 = 1
- Av = 21 × 1 = 21
Answer: Av = 21 V/V
First-Order Low-Pass Filter Cutoff
f_c = \frac{1}{2\pi RC}
| Symbol | Description | Unit |
|---|---|---|
| f_c | −3 dB cutoff frequency | Hz |
| R | Filter resistance | Ω |
| C | Filter capacitance | F |
Worked example
RC filter: R=10 kΩ, C=15.9 nF. Find cutoff frequency.
Given: R=10000 Ω, C=15.9×10^−9 F
- fc = 1/(2π × 10000 × 15.9×10^−9)
- fc = 1/(2π × 1.59×10^−4)
- fc = 1/9.99×10^−4 = 1001 Hz ≈ 1 kHz
Answer: fc ≈ 1 kHz
Oscilloscopes and Display Instruments
Time Period from Oscilloscope
T = \text{Divisions (horizontal)} \times \text{Time/div}
| Symbol | Description | Unit |
|---|---|---|
| T | Time period of waveform | s |
| \text{Divisions} | Number of horizontal graticule divisions per cycle | div |
| \text{Time/div} | Sweep speed setting | s/div |
Worked example
Oscilloscope sweep = 5 ms/div. One cycle spans 4 divisions. Find frequency.
Given: Sweep=5 ms/div, Divisions=4
- T = 4 × 5×10^−3 = 20 ms = 0.02 s
- f = 1/T = 1/0.02 = 50 Hz
Answer: f = 50 Hz
Voltage from Oscilloscope
V_{pp} = \text{Divisions (vertical)} \times \text{Volts/div}
| Symbol | Description | Unit |
|---|---|---|
| V_{pp} | Peak-to-peak voltage | V |
| \text{Volts/div} | Vertical sensitivity setting | V/div |
Worked example
Oscilloscope: 2 V/div sensitivity. Waveform spans 6 vertical divisions peak-to-peak. Find Vrms (sine).
Given: V/div=2, divisions=6
- Vpp = 6 × 2 = 12 V
- Vpeak = 12/2 = 6 V
- Vrms = 6/√2 = 4.24 V
Answer: Vrms = 4.24 V
CRO Sensitivity
S = \frac{\text{Deflection (m)}}{\text{Deflecting voltage (V)}} \quad (\text{m/V})
| Symbol | Description | Unit |
|---|---|---|
| S | CRO sensitivity | m/V or mm/V |
| \text{Deflection} | Spot deflection on screen | m |
Worked example
CRO screen: 25 V deflects spot by 50 mm. Find sensitivity.
Given: V=25 V, deflection=50 mm
- S = 50 mm / 25 V = 2 mm/V
Answer: S = 2 mm/V
Power and Energy Measurement
Two-Wattmeter Method (3-Phase Power)
P_{total} = W_1 + W_2, \quad \tan\phi = \frac{\sqrt{3}(W_1 - W_2)}{W_1 + W_2}
| Symbol | Description | Unit |
|---|---|---|
| P_{total} | Total three-phase active power | W |
| W_1, W_2 | Readings of two wattmeters | W |
| \phi | Phase angle | ° |
Worked example
Two wattmeters read W1=8 kW and W2=4 kW. Find total power and power factor.
Given: W1=8000 W, W2=4000 W
- P_total = 8000 + 4000 = 12000 W = 12 kW
- tan(φ) = √3 × (8000−4000)/(8000+4000) = √3 × 4000/12000 = 1.732 × 0.333 = 0.577
- φ = arctan(0.577) = 30°
- pf = cos(30°) = 0.866
Answer: P = 12 kW, pf = 0.866 lagging
Quick reference
| Formula | Expression |
|---|---|
| Absolute Error | \Delta A = A_{meas} - A_{true} |
| Percentage Error | \%E = (\Delta A/A_{true})\times 100 |
| Standard Deviation | \sigma = \sqrt{\sum(x_i-\bar{x})^2/(n-1)} |
| Wheatstone Balance | Rx = R \cdot (Q/P) |
| Maxwell Bridge Lx | L_x = R_2 R_3 C_4 |
| Maxwell Bridge Rx | R_x = R_2 R_3/R_4 |
| Gauge Factor | GF = (\Delta R/R)/\varepsilon |
| LVDT Sensitivity | S = V_{out}/x |
| Thermocouple EMF | E = S_{AB}(T_H - T_C) |
| Instrum. Amp Gain | A_v = (1+2R_1/R_G)(R_3/R_2) |
| LP Filter Cutoff | f_c = 1/(2\pi RC) |
| Voltmeter Loading | V_{meas} = V_{true} \cdot R_V/(R_V+R_s) |
| Oscilloscope Period | T = \text{divs} \times \text{time/div} |
| Two-Wattmeter Power | P = W_1 + W_2 |
| Two-Wattmeter PF | \tan\phi = \sqrt{3}(W_1-W_2)/(W_1+W_2) |
Exam tips
- GATE and ESE bridge circuit problems always give four arm values — check whether the condition P/Q = R/S is satisfied before solving; if it is, the galvanometer reads zero.
- Two-wattmeter method power factor questions require remembering that when W2 goes negative, the total power is still W1+W2, not W1−W2 — sign must be preserved.
- Gauge factor numerical questions are straightforward if you remember GF ≈ 2 for metallic gauges and ΔR is typically very small (mΩ range); a large ΔR answer is a calculation error flag.
- Loading effect questions test whether you know that voltmeters should have high resistance and ammeters low resistance — the error formula direction follows from this principle.
- In LVDT questions the output is zero at null position and varies linearly with displacement only within the linear range — examiners often add a question about phase reversal across null.
- Thermocouple cold-junction compensation questions appear in PSU interviews; always subtract the cold junction reference from the hot junction reading using the standard table values.