# 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,
$V_2=v_{rms}=\sqrt{100^2+\frac{1}{2}(10^2+5^2)}$
$\;\;=100.312$
$V_2-V_1=0.312$

There are 5 questions to complete.