Characteristics of Instruments and Measurement Systems

Question 1
A non-ideal Si-based pn junction diode is tested by sweeping the bias applied across its terminals from -5 V to +5 V. The effective thermal voltage, V_T, for the diode is measured to be (29\pm2) mV. The resolution of the voltage source in the measurement range is 1 mV. The percentage uncertainty (rounded off to 2 decimal plates) in the measured current at a bias voltage of 0.02 V is _______.
A
5.87
B
2.35
C
11.5
D
9.2
GATE EE 2020   Electrical and Electronic Measurements
Question 2
Two resistors with nominal resistance values R_1 \; and\; R_2 have additive uncertainties \Delta R_1 and \Delta R_2, 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}\Delta R_1 \right \}^2+\left \{ \frac{\partial R}{\partial R_2}\Delta R_2 \right \}^2}
B
\pm \sqrt{\left \{ \frac{\partial R}{\partial R_2}\Delta R_1 \right \}^2+\left \{ \frac{\partial R}{\partial R_1}\Delta R_2 \right \}^2}
C
\pm \sqrt{\left \{ \frac{\partial R}{\partial R_1}\right \}^2 \Delta R_2 +\left \{ \frac{\partial R}{\partial R_2} \right \}^2 \Delta R_1}
D
\pm \sqrt{\left \{ \frac{\partial R}{\partial R_1}\right \}^2 \Delta R_1 +\left \{ \frac{\partial R}{\partial R_2} \right \}^2 \Delta R_2}
GATE EE 2017-SET-2   Electrical and Electronic Measurements
Question 2 Explanation: 
\begin{aligned} \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} \\ &= \sqrt{\left ( \frac{\partial R}{\partial R_1} \right )^2 \Delta R_1^2+\left ( \frac{\partial R}{\partial R_2} \right )^2 \Delta R_2^2} \end{aligned}
Question 3
The following measurements are obtained on a single phase load:
V = 220V\pm1%, I = 5.0A\pm1% and W=555W\pm 2%.
If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ________.
A
20%
B
40%
C
4%
D
0.40%
GATE EE 2017-SET-1   Electrical and Electronic Measurements
Question 3 Explanation: 
\begin{aligned} V &= 220 \pm 1\% \\ I&= 5 \pm 1 \%\\ W &=555 \pm 2 \% \\ W&= VI \cos (\phi )\\ p.f. &=\cos (\phi )=\frac{W}{VI} \\ &= \frac{555 \pm 2 \%}{(220 \pm 1\%)(5 \pm 1 \%)}\\ &= \frac{555}{220 \times 5} \pm 4\%\\ p.f.&= 0.5 \pm 4\% \end{aligned}
Question 4
When the Wheatstone bridge shown in the figure is used to find the value of resistor R_X, the galvanometer G indicates zero current when R_1=50\Omega ,R_2=65\Omega and R_3=100\Omega. If R_3 is known with \pm 5% tolerance on its nominal value of 100 \Omega , what is the range of R_X in Ohms?
A
[123.50, 136.50]
B
[125.89, 134.12]
C
[117.00, 143.00]
D
[120.25, 139.75]
GATE EE 2015-SET-1   Electrical and Electronic Measurements
Question 4 Explanation: 
R_1=50\Omega
R_2=60\Omega
R_3=100\pm 5%\Omega
The value of R_3 \text{with} \pm 5% of tollerance,
\begin{aligned} R_3&= 100\pm 5\% \\ &= 100+100 \times \frac{5}{100}=105 \Omega \\ &= 100-100 \times \frac{5}{100}=95 \Omega \end{aligned}
In both condition, the bridge is balance, so under balance condition,
\begin{aligned} \frac{R_1}{R_3}&=\frac{R_2}{R_x} \\ R_x&=\frac{R_2R_3}{R_1} \\ (i) \; \text{when}, \; R_3&=105\Omega \\ \therefore \;\;R_x &= \frac{105 \times 65}{50}=136.50\Omega \\ (ii)\; \text{when}, \; R_3&=95 \Omega \\ \therefore \;\;R_x &= \frac{95 \times 65}{50}=123.50\Omega \end{aligned}
Question 5
The measurement system shown in the figure uses three sub-systems in cascade whose gains are specified as G_1,G_2 \; and \; \frac{1}{G_3}. The relative small errors associated with each respective subsystem G_1,G_2 \; and \; G_3 are \varepsilon _1,\varepsilon _2 \; and \; \varepsilon _3. The error associated with the output is :
A
\varepsilon _{1}+ \varepsilon _{2} +1/ \varepsilon _{3}
B
\frac{\varepsilon _{1}\varepsilon _{2}} {\varepsilon _{3}}
C
\varepsilon _{1}+ \varepsilon _{2} -\varepsilon _{3}
D
\varepsilon _{1}+ \varepsilon _{2} +\varepsilon _{3}
GATE EE 2009   Electrical and Electronic Measurements
Question 5 Explanation: 
\frac{dG_1}{G_1}=\varepsilon _1, \frac{dG_2}{G_2}=\varepsilon _2,and \frac{dG_3}{G_3}=\varepsilon _3
Output (y)_0=\frac{G_1G_3}{G_3}x
where, \;\; x=input
\ln y=\ln G_1+ \ln G_2- \ln G_3 +\ln x
Differentiating both side,
\frac{dy}{y}=\frac{dG_1}{G_1}+\frac{dG_2}{G_2}-\frac{dG_3}{G_3}+\frac{dx}{x}
No error is specified in input, so \frac{dx}{x}=0
\frac{dy}{y}==\varepsilon _1+\varepsilon _2-\varepsilon _3
Question 6
A variable w is related to three other variables x, y, z as w=xy/z. The variables are measured with meters of accuracy \pm0.5% reading, \pm1% of full scale value and \pm1.5% 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 uncertainty in the measurement of w will be
A
\pm 0.5% rdg
B
\pm 5.5% rdg
C
\pm 6.7% rdg
D
\pm 7.0% rdg
GATE EE 2006   Electrical and Electronic Measurements
Question 6 Explanation: 
Full scale reading of all three =100
Reading of x =80
Reading of y =20
Reading of z =50
\delta x=\pm 0.5% of reading =\pm \frac{0.5 \times 80}{100}=\pm 0.4
\delta y=\pm 1% of reading =\pm \frac{1 \times 100}{100}=\pm 1
\delta z=\pm 1.5% of reading =\pm \frac{1.5 \times 50}{100}=\pm 0.75
Given \omega =\frac{xy}{z}
taking log, we get,
\log \omega = \log x +\log y - \log z
differenting w.r.t. \omega we get
\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{0.1}{20}+\frac{0.75}{100} \right ) \times 100=\pm 7%
Question 7
Resistances R_{1} and R_{2} have, respectively, nominal values of 10\Omega and 5\Omega, and tolerances of \pm 5 % and \pm 10%. The range of values for the parallel combination of R_{1} and R_{2} is
A
3.077 \Omega to 3.636 \Omega
B
2.805 \Omega to 3.371 \Omega
C
3.237 \Omega to 3.678 \Omega
D
3.192 \Omega to 3.435 \Omega
GATE EE 2001   Electrical and Electronic Measurements
Question 7 Explanation: 
Range of R_1=10\pm 10 \times \frac{5}{100}
\;\;= 9.5\Omega \text{to} 10.5\Omega
Range of R_2=5\pm 5 \times \frac{10}{100}
\;\;= 4.5\Omega \text{to} 5.5\Omega
\because \;\; R_p=\frac{R_1R_2}{R_1+R_2}
\therefore Range of R_p=\frac{9.5 \times 4.5}{9.5+4.5} \text{to} \frac{10.5 \times 5.5}{10.5+5.5}
\;\;=3.05\Omega \text{to} 3.61 \Omega
There are 7 questions to complete.
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