Galvanometers, Voltmeters and Ammeters

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.
A
1.121
B
1.732
C
2.254
D
3.214
GATE EE 2020   Electrical and Electronic Measurements
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.
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?
A
9\Omega
B
99\Omega
C
990\Omega
D
9990\Omega
GATE EE 2019   Electrical and Electronic Measurements
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
A
2, 5.55
B
2, 1
C
2.18, 0.55
D
11.1, 2
GATE EE 2018   Electrical and Electronic Measurements
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}
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 _________.
A
0.1
B
0.31
C
0.66
D
1
GATE EE 2016-SET-1   Electrical and Electronic Measurements
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
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 __________.
A
8.52
B
11.95
C
16.35
D
22.25
GATE EE 2015-SET-2   Electrical and Electronic Measurements
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.
A
P-1 Q-2 R-1 S-3
B
P-1 Q-2 R-1 S-3
C
P-1 Q-2 R-3 S-3
D
P-3 Q-1 R-2 S-1
GATE EE 2015-SET-2   Electrical and Electronic Measurements
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 _______.
A
0.11
B
0.22
C
0.45
D
0.68
GATE EE 2015-SET-1   Electrical and Electronic Measurements
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
A
\sqrt{\frac{3}{2}}
B
\sqrt{\frac{2}{3}}
C
\frac{3}{2}
D
\frac{2}{3}
GATE EE 2014-SET-3   Electrical and Electronic Measurements
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_____.
A
10.28
B
5.45
C
15.85
D
20.45
GATE EE 2014-SET-2   Electrical and Electronic Measurements
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 __________
A
26.26
B
57.73
C
82.96
D
96.48
GATE EE 2014-SET-2   Electrical and Electronic Measurements
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
There are 10 questions to complete.
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