# Galvanometers, Voltmeters and Ammeters

 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)

 A 15 B 20 C 18 D 22
GATE EE 2022   Electrical and Electronic Measurements
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, $I=\frac{100}{5}=20A$
 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.
 A 1.121 B 1.732 C 2.254 D 3.214
GATE EE 2020   Electrical and Electronic Measurements
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.
 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?
 A 9$\Omega$ B 99$\Omega$ C 990$\Omega$ D 9990$\Omega$
GATE EE 2019   Electrical and Electronic Measurements
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
 A 2, 5.55 B 2, 1 C 2.18, 0.55 D 11.1, 2
GATE EE 2018   Electrical and Electronic Measurements
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}
 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 _________.
 A 0.1 B 0.31 C 0.66 D 1
GATE EE 2016-SET-1   Electrical and Electronic Measurements
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,
$V_2=v_{rms}=\sqrt{100^2+\frac{1}{2}(10^2+5^2)}$
$\;\;=100.312$
$V_2-V_1=0.312$
 Question 6
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 $\pm$10% 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 6 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 7
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 8
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 8 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 9
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 9 Explanation:

 Question 10
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 10 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
There are 10 questions to complete.