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 _______.
5.87 | |
2.35 | |
11.5 | |
9.2 |
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.
1.121 | |
1.732 | |
2.254 | |
3.214 |
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.
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 _______.
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 _______.
340.5 | |
170.6 | |
588.9 | |
377.8 |
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?
9\Omega | |
99\Omega | |
990\Omega | |
9990\Omega |
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
2, 5.55 | |
2, 1 | |
2.18, 0.55 | |
11.1, 2 |
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}
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
0.532 | |
0.632 | |
0.707 | |
0.866 |
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}
\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 _______.
100 | |
1000 | |
10000 | |
100000 |
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
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
1.5kHz | |
2kHz | |
3kHz | |
4.5kHz |
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
\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} | |
\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} | |
\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} | |
\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} |
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:
W_1=1732 \; and \; W_2 =W_3=0 | |
W_1=0 , W_2 =1732 \; and \; W_3=0 | |
W_1=866, W_2 =0 , W_3=866 | |
W_1=W_2 =0 \; and \; W_3=1732 |
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}
\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.