#### introduction

We know that in this world nothing is completely efficient only in ideal in situations we suppose every thing completely efficient. In electrical engineering most commonly used electrical variables are current , voltage , frequency etc. To measure them we use different electronic meters and devices. Question is how can we know that the value provided by meter is completely accurate? while they are 100% efficient.

Sensor are used to convert any given physical quantity into measurable quantity while signal conditioning are used to convert signal from one form to another. The combination of sensor and signal conditioning is called **Transducer** while signal processing is used to provide the output in desired form while the combination of transducer and signal processing is called **Transmitter**.

**error**

error = desired value – actual value

#### gross error

the error caused by human are called gross error they are of different kinds e.g

Carelessness: if meter is providing value 3.8 but by mistake you note it 3.6 then it is human error, if you write the values in different column.

Reading mistake: if you want reading in mA but your meter prob is set on A then meter is providing accurate value.

#### systematic error

the error caused by system is called systematic error .If any component inside meter stop working then it will provide us wrong information we can correct it by replacing faulty component by new one.

#### absolute error

If the error is given in specific units then it is called absolute error e,g if resistance of 500Ω with a possible error of ±50Ω the ±50Ω is absolute error. This is because when we add these two values 500+50=550 and 500-50=450 we obtain a absolute quantity.

#### relative error

The error given in form of percentage is called relative error e.g tolerance of resistance R = 500Ω ±50% we can calculate it’s tolerance by

50 / 500 * 100 = 10%

another method of expressing error in form of parts per million (ppm). For example the temperature coefficient of 1MΩ resistor might be stated as 100ppm/ºC , which means 100 parts per million degree Celsius.

**accuracy,precision,resolution and significant figures**

#### accuracy

is how close a measured value is to actual/real value. Example an voltmeter with an error of ±1% shows 50 V. Then the voltage is somewhere between 49 V and 51 V.

#### precision

is how measured values are closed to each other. Example In case of football if you are striking continuously but succeed to do a goal then accuracy is low and precision is high.

#### resolution

the smallest change in measured quantity that can be observed is called resolution. Example instrument used to measure heavy things do not show any reading for 200g but the instrument used to weight gold can measure a very small change so instrument have more resolution.

#### significant figures

the number of significant figures used in measured quantity indicate the precision of measurement. Example determine resistance from digital measurement of current and voltage

R = V / I

R = 8.14 / 2.33 mA = 3.493562 kΩ

so, this answer is not effective because s.f are 7 but if

R = V / I = 8.14 / 2.33 mA = 3.49 kΩ only 3 s.f so the no. of significant figures define the precision of the quantity.

**measurement error combinations**

When we perform any arithmetic operation on any value if that value contain errors then we can also see change in that error. There different measurement errors combinations.

#### sum of quantities

Where a quantity is determined as a sum of two measurements the total sum is sum of absolute errors in each measurement.

E = (V1 ± ∇V1) + (V2 ± ∇V2)

where V1 and V2 are voltages but ∇V1 & ∇V2 are errors by simplifying we can also write them as

E=(V1+V2) ± (∇V1 + ∇V2)

**Example:** Calculate the maximum percentage error in the sum of two voltage measurements when V1 = 100V ± 1% and V2 = 80V ± 5%.

Solution: V1 = 100V ± 1% , V2 = 80V ± 5%

we can convert into absolute form for V1 it is (1 / 100) * 100 = 1 now it is V1=100V ± 1V

for V2 it is (5 / 100) * 80 = 4 now it is V2=80V ± 4V

E = V1 + V2

= (100V ± 1V) + (80V ± 4V)

= 180V ± 5V

#### difference of quantities

Same as above the equation take form

E = (V1 ± ∇V1) – (V2 ± ∇V2)

E = (V1 – V2) ± (∇V1 + ∇V2)

#### product of quantities

When a calculated quantity is the product of two or more quantities, the percentage error is the sum of percentage errors in each quantity. Example

P = E.I

= (E ± ∇E) . (I ± ∇I)

= E.I ± E.∇I ± I.∇E ± ∇E.∇I

∇E.∇I are very small so we can neglect them

P = E.I ± (E.∇I + I.∇E)

Percentage error = (E.∇I + I.∇E) / EI * 100%

= (∇I / I + ∇E / E) * 100%

% ERROR IN P = (% ERROR IN I) + (% ERROR IN E)

#### quotient of quantities

% error in E / I = (% error in E) + (% error in I)

#### quantity raised to a power

When a quantity A raised to a power B the percentage error A^{B }shown as

% error in A^{B} = B(% error in A)

Next page:

-Electromechanical instruments

https://www.electricalengineering4u.com/instrumentation-measurement/electromechanical-instruments/