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 |
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 |
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 3 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 4 |
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 4 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 5 |
A dc voltage with ripple is given by
v(t)=[100+10\sin(\omega t)-5\sin (3\omega t)] volts.
Measurements of this voltage v(t), made by moving-coil and moving-iron voltmeters, show readings of V_1 \; and \; V_2 respectively. The value of V_2 - V_1, in volts, is _________.
v(t)=[100+10\sin(\omega t)-5\sin (3\omega t)] volts.
Measurements of this voltage v(t), made by moving-coil and moving-iron voltmeters, show readings of V_1 \; and \; V_2 respectively. The value of V_2 - V_1, in volts, is _________.
0.1 | |
0.31 | |
0.66 | |
1 |
Question 5 Explanation:
V(t)=100+10 \sin (\omega t)-5 \sin (3\omega t) \;\;\text{volt}
moving coil,
V_1=V_{avg.}=100V
moving iron,
V_2=v_{rms}=\sqrt{100^2+\frac{1}{2}(10^2+5^2)}
\;\;=100.312
V_2-V_1=0.312
moving coil,
V_1=V_{avg.}=100V
moving iron,
V_2=v_{rms}=\sqrt{100^2+\frac{1}{2}(10^2+5^2)}
\;\;=100.312
V_2-V_1=0.312
There are 5 questions to complete.