# Electrical and Electronic Measurements

 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      Characteristics of Instruments and Measurement Systems
 Question 2
Currents through ammeters A2 and A3 in the figure are $1\angle 10^{\circ} \; and \; 1\angle 70^{\circ}$ respectively. The reading of the ammeter A1 (rounded off to 3 decimal places) is ________ A.
 A 1.121 B 1.732 C 2.254 D 3.214
GATE EE 2020      Galvanometers, Voltmeters and Ammeters
Question 2 Explanation:
$I=1\angle 10^{\circ}+1\angle 70^{\circ}$
$I=1.732\angle 40^{\circ}$
The ready of ammeter is 1.732 A.
 Question 3
The voltage across and the current through a load are expressed as follows
$v(t)=-170sin(377t-\frac{\pi}{6})V$
$i(t)=8 cos(377t+\frac{\pi}{6})A$
The average power in watts (round off to one decimal place) consumed by the load is _______.
 A 340.5 B 170.6 C 588.9 D 377.8
GATE EE 2019      Measurement of Energy and Power
Question 3 Explanation:
\begin{aligned} v(t) &=-170 \sin \left ( 377t-\frac{\pi}{6} \right )V=V_{pc}\\ i(t) &= 8 \cos \left ( 377t+\frac{\pi}{6} \right )A=I_{cc}\\ i(t) &= 8 \sin \left ( 377t+\frac{2\pi}{3} \right )A=I_{cc} \\ P_{avg}&=\frac{1}{2} \times (-170)(8) \times \cos\left ( -\frac{\pi}{6}-\frac{2 \pi}{3} \right ) \\ &= \frac{1}{2} \times (-170)(8) \times \cos(150^{\circ}) \\ &=588.9W \end{aligned}
 Question 4
A moving coil instrument having a resistance of 10 $\Omega$, gives a full-scale deflection when the current is 10 mA. What should be the value of the series resistance, so that it can be used as a voltmeter for measuring potential difference up to 100 V?
 A 9$\Omega$ B 99$\Omega$ C 990$\Omega$ D 9990$\Omega$
GATE EE 2019      Galvanometers, Voltmeters and Ammeters
Question 4 Explanation:

\begin{aligned} V_m&=I_mR_m \\ &= 10 mA \times 10\Omega \\ &= 100mV\\ (0-100mV)\Rightarrow & (0-100V)\\ m &= \frac{V_{ext}}{V_m}=\frac{100V}{100mV}=1000\\ R_{se}&=R_m[m-1] \\ R_{se} &= 10(1000-1)\\ &= 9990\Omega \end{aligned}
 Question 5
A 0-1 Ampere moving iron ammeter has an internal resistance of 50 m$\Omega$ and inductance of 0.1 mH. A shunt coil is connected to extend its range to 0-10 Ampere for all operating frequencies. The time constant in milliseconds and resistance in m$\Omega$ of the shunt coil respectively are
 A 2, 5.55 B 2, 1 C 2.18, 0.55 D 11.1, 2
GATE EE 2018      Galvanometers, Voltmeters and Ammeters
Question 5 Explanation:
Given,
$I_m=1A,$
$R_m=50m\Omega$
$L_m=0.1 mH,$
$I=10A$
We know,
$R_{sh}=\frac{R_m}{(m-1)};$
$m=\frac{I}{I_m}=\frac{10}{1}=10$
$R_{sh}=\frac{50 \times 10^{-3}}{10-1}$
$\;\;=5.55m\Omega$
For all frequencies time constant of shunt and meter arm should be equal.
\begin{aligned} i.e. \;\; \frac{\omega L_m}{R_m} &=\frac{\omega L_{sh}}{R_{sh}} \\ \frac{L_m}{R_m} &= \frac{L_{sh}}{R_{sh}}\\ \frac{L_m}{R_m} &= \frac{0.1 \times 10^{-3}}{50 \times 10^{-3}}\\ &=0.002=2ms \end{aligned}
 Question 6
Two wattmeter method is used for measurement of power in a balanced three-phase load supplied from a balanced three-phase system. If one of the wattmeters reads half of the other (both positive), then the power factor of the load is
 A 0.532 B 0.632 C 0.707 D 0.866
GATE EE 2018      Measurement of Energy and Power
Question 6 Explanation:
In two wattmeter method,
$\tan \phi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}$
\begin{aligned} \text{Given}, \; W_2 &=\frac{W_1}{2} \\ \tan \phi &=\frac{\sqrt{3}\left (W_1-\frac{W_1}{2} \right ) }{\left (W_1+\frac{W_1}{2} \right )} \\ \phi &= 30^{\circ}\\ \cos \phi &=\cos 30^{\circ}=0.866 \end{aligned}
 Question 7
A $10\frac{1}{2}$ digit timer counter possesses a base clock of frequency 100 MHz. When measuring a particular input, the reading obtained is the same in: (i) Frequency mode of operation with a gating time of one second and (ii) Period mode of operation (in the x 10 ns scale). The frequency of the unknown input (reading obtained) in Hz is _______.
 A 100 B 1000 C 10000 D 100000
GATE EE 2017-SET-2      CRO and Electronic Measurements
Question 7 Explanation:
$1.\;\;10\frac{1}{2}$ digital time counter:
Frequency mode of operation: $f=\frac{n}{t}$

Let f = frequency of input signal
n = number of cycles of repetitive signal
$\Rightarrow \;\;100 \times 10^6$
Let $t \Rightarrow$ Gate time $\Rightarrow \;t=1 sec.$
$f=\frac{100 \times 10^6}{1 sec}=10^8 Hz$
$\;\;=10^8 cycles/sec$
On $10\frac{1}{2}$ digit display
$100000000.00Hz$

$2.\;\; \text{Period mode of operation:}$
$P=\frac{1}{f}=\frac{t}{n}=\frac{1sec}{100 \times 10^6}$
Let P=Period of input signal
$P=0.01 \times 10^{-6}$
$\;\;=10 \times 10^{-9}$
$\;\;=1 \times 10n-sec$
$\;\;=10.000000000 n-sec$
 Question 8
A stationary closed Lissajous pattern on an oscilloscope has 3 horizontal tangencies and 2 vertical tangencies for a horizontal input with frequency 3 kHZ. The frequency of the vertical input is
 A 1.5kHz B 2kHz C 3kHz D 4.5kHz
GATE EE 2017-SET-2      CRO and Electronic Measurements
Question 8 Explanation:
\begin{aligned} \frac{f_y}{f_x} &= \frac{\text{Horizontal Tangencies}}{\text{VerticalTangencies}}\\ \Rightarrow \; \frac{f_y}{3} &=\frac{3 }{2}\\ \Rightarrow \; f_y&= 4.5kHz \end{aligned}
 Question 9
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      Characteristics of Instruments and Measurement Systems
Question 9 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 10
The load shown in the figure is supplied by a 400 V (line to line) 3-phase source (RYB sequence). The load is balanced and inductive, drawing 3464 VA. When the switch S is in position N, the three watt-meters $W_1,W_2 \; and \; W_3$ read 577.35W each. If the switch is moved to position Y, the readings of the watt-meters in watts will be:
 A $W_1=1732 \; and \; W_2 =W_3=0$ B $W_1=0 , W_2 =1732 \; and \; W_3=0$ C $W_1=866, W_2 =0 , W_3=866$ D $W_1=W_2 =0 \; and \; W_3=1732$
GATE EE 2017-SET-1      Measurement of Energy and Power
Question 10 Explanation:
When switch is on Neutral side
\begin{aligned} W &= W_1+W_2+W_3\\ &=577.35+577.35+577.35 \\ &=1732 \; Watts \\ \text{VA load} &= 3464 =3 V_{ph}I_{ph}\\ \Rightarrow \; V_{ph}I_{ph} &=1154.66 \end{aligned}
Each $\Omega$ meter reading
\begin{aligned} &= V_{RN}I_{R} \cos (\angle V_{RN} \; \text{and}\; I_R)\\ 577.35&= V_{ph}I_{ph} \cos (\phi )\\ \cos (\phi ) &=\frac{577.35}{1154.66}=05 \\ \phi &= \cos ^{-1} (0.5)=60^{\circ} \end{aligned}
When switch is on Y-phase side,
\begin{aligned} W_1&= V_{RY}I_R \cos (\angle V_{RY} \; \text{and}\; I_R)\\ &= \sqrt{3} V_{ph}I_{ph} \cos (30^{\circ}+\phi ) \\ &= \sqrt{3} \times 1154.66 \times \cos (30^{\circ}+ 60^{\circ})\\ &= 0 \text{Watt}\\ W_3 &= V_{BY}I_R \cos (\angle V_{BY} \; \text{and}\; I_R)\\ &=\sqrt{3} V_{ph}I_{ph} \cos (30^{\circ}-\phi )\\ &=\sqrt{3} \times 1154.66 \times \cos (30^{\circ}- 60^{\circ}) \\ &= 1732 \text{Watt}\\ W_1&=0\\ W_2&=0\\ W_3&=1732 \text{Watt} \end{aligned}
There are 10 questions to complete.