# 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$\pm$2) 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$\pm$1%, I = 5.0A$\pm$1% and W=555W$\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 $\pm$0.5% reading, $\pm$1% of full scale value and $\pm$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 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.