Basics & Errors - Grad Plus

### Electrical Engineering (EE) | Topic-wise Previous Solved GATE Papers | Electrical and Electronic Measurements

GATE-EE Solutions for Electrical and Electronic Measurements | Basics & Errors
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GATE-EE Solutions for Electrical and Electronic Measurements | CRO
GATE-EE Solutions for Electrical and Electronic Measurements | Transformers

# Basics & Errors

[1994 1M] A precise measurement guarantees accuracy of the measured quantity.[True/False]

Ans: False.

Explanation:

A precise measurement is attempted as close as accurate but it does not guarantee the accuracy.

[1995 : 1 M] Four ammeters M1, M2, M3 and M4 with the following specifications are available.

A current of 1 A is to be measured. To obtain minimum error in the reading, one should select meter.

(a) M1

(b) M2

(c) M3

(d) M4

Ans: (d)

Explanation:

Error in the measuring instrument is estimated as the product of Full Scale Deflection and Accuracy.

Error in Reading = FSD × Accuracy

Hence, error in reading of first meter = 20 \times \frac{\pm 0.1}{100}= \pm 0.02

Error in reading of second meter = 10 \times \frac{\pm 0.2}{100}= \pm 0.02

Error in reading of third meter = 5 \times \frac{\pm 0.5}{100}= \pm 0.025

Error in reading of fourth meter = 1 \times \frac{\pm 1.00}{100}= \pm 0.01

Fourth meter has least error, hence it would be the choice.

[1999 : 2 M] A current of \left [ 2+\sqrt{2}\; sin (sin\; 314t+30^{o})+2+\sqrt{2} cos (952t+45^{o})\right ] A is measured with a thermocouple type, 5A full scale, class-1 meter. The meter reading would lie in the range,

(a) 5 A ± 1%

(b) (2+3\sqrt2)A\pm1\%

(c) 3 A ± 1.67%

(d) 2 A ± 0.5%

Ans. (c)

Explanation:

We know that, thermocouple type instruments read RMS values. Hence the meter reading is nothing but the RMS value of the current.

I_{rms}=\sqrt{2^{2}+\left ( \frac{\sqrt{2}}{\sqrt{2}} \right )^{2}+\left ( \frac{2\sqrt{2}}{\sqrt{2}} \right )^{2}}

= \sqrt{4+1+4}

Hence, IRMS = 3A

Given meter is class 1 meter and for class 1 meter, accuracy is 1% for full scale range i.e. for 5 A.

Hence, for 3A, accuracy would be around 5/3 % = 1.67 %

[2001 : 2 M] Resistance R1 and R2, have, respectively, nominal values of 10Ω and 5Ω and tolerances of ±5% and ±10%. The range of values for the parallel combination of R1 and R2 is_______.

(a) 3.077 Ω to 3.636 Ω

(b) 2.805 Ω to 3.371 Ω

(c) 3.237 Ω to 3.678Ω

(d) 3.192 Ω to 3.435 Ω

Ans. (a)

Explanation:

As we are given the tolerances for each resistance, let us find out their ranges. And considering lower and upper limits separately we can find the range for their parallel combination.

Range of R1 = 10\pm 10\times \frac{5}{100} = 9.5Ω to 10.5Ω

Range of R2= 5\pm 5\times \frac{10}{100}=4.5Ω to 5.5 Ω

R_{p}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}

Therefore,  Range for Rp

R_p=\frac{9.5\times4.5}{9.5\;+\;4.5}\;to\;\frac{10.5\times5.5}{10.5\;+\;5.5}=3.05\Omega\;to\;3.61\Omega

[2006: 2 M] A variable w is related to three other variables x, y, z as w = xy/z. The variables are measured with meters of accuracy ± 0.5% reading, ± 1% of full scale value and ±1.5% reading. The actual readings of the three meters are 80, 20 and 50 with 100 being the full scale value for all three. The maximum limiting error in the measurement of w will be______.

a) ±0.5 % rdg

b) ±5.5% rdg

c) ±6.7% rdg

d) 7.0% rdg

Ans. (d)

Explanation:

From the given data we can find the errors encountered while taking the readings.

δx=±0.5 % of reading= \pm \frac{0.5\times 80}{100} =±0.4.

δy=±1 % of full reading= \pm \frac{1\times 100}{100} =±1.

δz=±1.5 % of reading= \pm \frac{1.5\times 50}{100} =±0.75.

Given that, w=\frac{xy}{z}

taking log, we get

log w= log x + log y – log z

differentiating w.r.t ω we get

\frac1\omega=\frac1x\frac{\delta x}{\delta\omega}+\frac1y\frac{\delta y}{\delta\omega}-\frac1z\frac{\delta z}{\delta\omega} \frac{\delta \omega }{\omega }=\frac{\delta x }{x }+\frac{\delta y }{y }-\frac{\delta z }{z }

for maximum limiting error,

\frac{\delta \omega }{\omega }=\pm \left ( \frac{0.4}{80} + \frac{1}{20}+ \frac{0.75}{100}\right )\times 100

Hence, Ans = ±7 %

[2015 : 1 M, Set-1] When the Wheatstone bridge shown in the figure  is used to find the value of resistor Rx, the galvanometer G indicates zero current when R1 = 50 Ω. R2 = 65 Ω and R3 = 100 Ω. If R3 is known with ±5% tolerance on its nominal value of 100 Ω, what is the range of Rx in Ohms?

(a) [123.50, 136.50]

(b) [125.89, 134.12]

(c) [117.00, 143.00]

(d) [120.25, 139.75]

Ans. (a)

Explanation:

R1=50Ω and R2=65Ω

The value of R3 is given with ±5% tolerance over the nominal value

∴ R3=100 ± [5% of 100]

R_{3}=100 + 100\times \frac{5}{100}=105\Omega

or, R_{3}=100 - 100\times \frac{5}{100}=95\Omega

In both conditions, the bridge is balanced,

\frac{R_{1}}{R_{3}}=\frac{R_{2}}{R_{x}} R_{x}=\frac{R_{2}R_{3}}{R_{1}}

So, let us find range for R3 by considering range for Rx

i) when R3=105Ω

R_{x}=\frac{105\times 65}{50} =136.50Ω

ii) when R3=95Ω

R_{x}=\frac{95\times 65}{50} =123.50Ω

[2017: 1 M, Set-1] The following measurement are obtained on a single phase load: V = 220 V ±1%. I = 5.0 A ± 1% and W=555 W ± 2%. If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ____ (Give answer up to one decimal place.)

Ans. 0.5 ± 4 %

Explanation:

We know that, W=V×I×cos (Φ)

p.f= cos (\phi )=\frac{W}{V.I} \therefore P.F.=\frac{555\pm2\%}{(220\pm1\%)(5\pm1\%)}=\frac{555}{220\times5}\pm4\%

Hence, p.f=cos(Φ)=0.5 ±4%

[2017: 1 M, Set-2] Two resistors with nominal resistance values R1, and R2, have additive uncertainties ∆R1 and ∆R2 respectively. When these resistances are connected in parallel, the standard deviation of the error in the equivalent resistance R is _____

a) \pm \sqrt{\left ( \frac{\partial R}{\partial R_{1}} \bigtriangleup R_{1}\right )^{2} +{\left ( \frac{\partial R}{\partial R_{2}} \bigtriangleup R_{2}\right )}^{2}}

b) \pm \sqrt{\left ( \frac{\partial R}{\partial R_{2}} \bigtriangleup R_{1}\right )^{2} +{\left ( \frac{\partial R}{\partial R_{1}} \bigtriangleup R_{2}\right )}^{2}}

c) \pm \sqrt{\left ( \frac{\partial R}{\partial R_{1}}\right )^{2} \bigtriangleup R_{2} +{\left ( \frac{\partial R}{\partial R_{2}} \right )}^{2}\bigtriangleup R_{1}}

d) \pm \sqrt{\left ( \frac{\partial R}{\partial R_{1}}\right )^{2} \bigtriangleup R_{1} +{\left ( \frac{\partial R}{\partial R_{2}} \right )}^{2}\bigtriangleup R_{2}}

Ans. (a)

Explanation:

\sigma _{res}=\sqrt{\left ( \frac{\partial R}{\partial R_{1}} \right )^{2} \sigma _{1}^{2}+{\left ( \frac{\partial R}{\partial R_{2}} \right )^{2}}}\sigma _{2}^{2} \sigma_{res}=\sqrt{\left(\frac{\partial R}{\partial R_1}\right)^2\bigtriangleup R_1^2+\left(\frac{\partial R}{\partial R_2}\right)^2\bigtriangleup R_2^2}
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