Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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
Question 5 |
A capacitive voltage divider is used to measure the bus voltage V_{bus} in a high-voltage 50 Hz AC system as shown in the figure. The measurement capacitors C_1 \; and \; C_2 have tolerances of \pm10% on their nominal capacitance values. If the bus voltage V_{bus} is 100 kV rms, the maximum rms output voltage V_{out} (in kV), considering the capacitor tolerances, is __________.
8.52 | |
11.95 | |
16.35 | |
22.25 |
Question 5 Explanation:
V_{BUS} \text{is} 100 kV_{rms}
C_1=1\mu F \pm 10\%
C_2=9\mu F \pm 10\%
To get maximum output voltage we need minimum C_2 and maximum C_1,
So, C_2=8.1\mu F and C_1=1.1\mu F
So, V_{out\;rms}=\left ( \frac{C_1}{C_1+C_2} \right )V_{BUS_{rms}}
\;\;\;=11.95kV
Question 6 |
Match the following.
P-1 Q-2 R-1 S-3 | |
P-1 Q-2 R-1 S-3 | |
P-1 Q-2 R-3 S-3 | |
P-3 Q-1 R-2 S-1 |
Question 7 |
A (0-50 A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0-500 A) is _______.
0.11 | |
0.22 | |
0.45 | |
0.68 |
Question 7 Explanation:
I_{fs}=50A,
V_m=0.1V,
R_m=\frac{0.1}{50}=2 \times 10^{-3}\Omega
\because \; m=10
R_{sh}=\frac{R_m}{(m-1)}
\;\;\;=\frac{2 \times 10^{-3}}{9}=0.22\Omega
Question 8 |
A periodic waveform observed across a load is represented by
V(t)=\left\{\begin{matrix} 1+sin \omega t & 0\leq \omega t \lt 6 \pi\\ -1+sin \omega t & 6 \pi \leq \omega t \leq 12 \pi \end{matrix}\right.
The measured value, using moving iron voltmeter connected across the load, is
V(t)=\left\{\begin{matrix} 1+sin \omega t & 0\leq \omega t \lt 6 \pi\\ -1+sin \omega t & 6 \pi \leq \omega t \leq 12 \pi \end{matrix}\right.
The measured value, using moving iron voltmeter connected across the load, is
\sqrt{\frac{3}{2}} | |
\sqrt{\frac{2}{3}} | |
\frac{3}{2} | |
\frac{2}{3} |
Question 8 Explanation:
Question 9 |
Two ammeters X and Y have resistances of 1.2 \Omega and 1.5 \Omega respectively and they give full-scale deflection with 150 mA and 250 mA respectively. The ranges have been extended by connecting shunts so as to give full scale deflection with 15 A. The ammeters along with shunts are connected in parallel and then placed
in a circuit in which the total current flowing is 15 A. The current in amperes indicated in ammeter X is_____.
10.28 | |
5.45 | |
15.85 | |
20.45 |
Question 9 Explanation:
Given, R_{m_X}=1.2\Omega ,
R_{m_Y}=1.5\Omega
I_{m_X}=0.15A,
I_{m_Y}=0.25A
and I= full scale deflection current=15A
\therefore Shunt multiplying factor for ammeter X is
m_X=\frac{I}{I_{m_X}}=\frac{15}{0.15}=100
and shunt multiplier resistance,
R_{{sh}_X}=\frac{R_{m_X}}{(m_X-1)}=\frac{12}{99}=0.012\Omega
Also, shunt multiplying factor for ammeteer Y is
m_Y=\frac{I}{I_{m_Y}}=\frac{15}{0.25}=60
and shunt multiplier resistance,
R_{{sh}_Y}=\frac{R_{m_Y}}{(m_Y-1)}=\frac{1.5}{59}=0.025\Omega
when these ammeters are connected in parrelel as shown in the figure below,
Let current in ammeter X be I_X then,
\begin{aligned} R_X&=\left ( \frac{ R_{{sh}_X} \cdot R_{m_X}} { R_{{sh}_X}+R_{m_X}} \right )\\ &= \left ( \frac{0.012 \times 1.2}{1.212} \right )\\ &= 0.011 \Omega \\ R_Y &=\left ( \frac{R_{{sh}_Y}\cdot R_{m_Y}}{R_{{sh}_Y}+R_{m_Y}} \right )\\ &= \left ( \frac{0.025 \times 1.5}{1.525} \right )\\ &=0.024\Omega \\ \therefore \; I_X &=\left (\frac{R_Y}{R_X+R_Y} \right )\times 15 \\ &=\left ( \frac{0.024}{0.035} \right ) \times 15 \\ &= 10.28A \end{aligned}
\therefore Current in ammeter X=10.28A
Question 10 |
The saw-tooth voltage waveform shown in the figure is fed to a moving iron
voltmeter. Its reading would be close to __________
26.26 | |
57.73 | |
82.96 | |
96.48 |
Question 10 Explanation:
Moving iron voltmeter reads rms value of voltage and current. From the given waveform of voltage,
\begin{aligned} V(t) &= \left\{\begin{matrix} 50000t, & 0 \lt t \lt 0.02s\\ 50000(t-0.02). & 0.02 \lt t \lt 0.04s \end{matrix}\right.\\ \therefore \; V_{rms} &= \sqrt{\frac{1}{T}\int_{0}^{T}(V(T)^2 \; dt}\\ &= \sqrt{\frac{1}{0.04} \int_{0}^{0.02} (50000t)^2 \; dt +\int_{0.02}^{0.04}(50000(t-0.02))^2 \; dt}\\ &= 25000\sqrt{\left [ \frac{t^3}{3} \right ]_0^{0.02}+\left [ \frac{(t-0.02)^3}{3} \right ]_{0.02}^{0.04}}\\ &= 25000\sqrt{\frac{1}{3}\left [ (0.02)^3+(0.02)^3 \right ]}\\ &= 57.73 \text{Volt} \end{aligned}
\therefore Voltage reading by moving iron voltmeter =57.73 Volts
\begin{aligned} V(t) &= \left\{\begin{matrix} 50000t, & 0 \lt t \lt 0.02s\\ 50000(t-0.02). & 0.02 \lt t \lt 0.04s \end{matrix}\right.\\ \therefore \; V_{rms} &= \sqrt{\frac{1}{T}\int_{0}^{T}(V(T)^2 \; dt}\\ &= \sqrt{\frac{1}{0.04} \int_{0}^{0.02} (50000t)^2 \; dt +\int_{0.02}^{0.04}(50000(t-0.02))^2 \; dt}\\ &= 25000\sqrt{\left [ \frac{t^3}{3} \right ]_0^{0.02}+\left [ \frac{(t-0.02)^3}{3} \right ]_{0.02}^{0.04}}\\ &= 25000\sqrt{\frac{1}{3}\left [ (0.02)^3+(0.02)^3 \right ]}\\ &= 57.73 \text{Volt} \end{aligned}
\therefore Voltage reading by moving iron voltmeter =57.73 Volts
There are 10 questions to complete.