Question 1 |
Two balanced three-phase loads, as shown in the figure, are connected to a 100\sqrt{3}V ,
three-phase, 50 Hz main supply. Given Z_1=(18+j24)\Omega and Z_2=(6+j8)\Omega . The
ammeter reading, in amperes, is _______. (round off to nearest integer)
15 | |
20 | |
18 | |
22 |
Question 1 Explanation:
First perform delta to star conversion
we know, for balanced load
Z_{star}=\frac{Z_{delta}}{3}=\frac{18+j24}{3}=6+j8\Omega
Draw the per phase diagram:
Z_{eq}=(6+j8)||(6+j8)=(3+j8)\Omega =5\angle 53.13^{\circ} \Omega
Therefore, Meter reading, I=\frac{100}{5}=20A
Z_{star}=\frac{Z_{delta}}{3}=\frac{18+j24}{3}=6+j8\Omega
Draw the per phase diagram:
Z_{eq}=(6+j8)||(6+j8)=(3+j8)\Omega =5\angle 53.13^{\circ} \Omega
Therefore, Meter reading, I=\frac{100}{5}=20A
Question 2 |
A balanced Wheatstone bridge ABCD has the following arm resistances:
R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is
R_{AB}=1k\Omega \pm 2.1%; R_{BC}=100\Omega \pm 0.5%, R_{CD} is an unknown resistance; R_{DA}=300\Omega \pm 0.4%; . The value of R_{CD} and its accuracy is
30\Omega \pm 3\Omega | |
30\Omega \pm 0.9\Omega | |
3000\Omega \pm 90\Omega | |
3000\Omega \pm 3\Omega |
Question 2 Explanation:
The condition for balanced bridge
\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}
\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}
Question 3 |
Inductance is measured by
Schering bridge | |
Maxwell bridge | |
Kelvin bridge | |
Wien bridge |
Question 3 Explanation:
Maxwell's bridge is used for measurement of inductance.
Wein's bridge is used for measurement of frequency
Kelvin's bridge is used for measurement of low value of resistance.
Schering bridge is used for measurement of capacitance, dilectric loss and permittivity etc.
Wein's bridge is used for measurement of frequency
Kelvin's bridge is used for measurement of low value of resistance.
Schering bridge is used for measurement of capacitance, dilectric loss and permittivity etc.
Question 4 |
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 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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 8 |
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 8 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 9 |
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 9 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 10 |
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 10 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
There are 10 questions to complete.