Galvanometers Voltmeters And Ammeters Practice Questions

$I=1\angle 10^{\circ}+1\angle 70^{\circ}$ $I=1.732\angle 40^{\circ}$ The ready of ammeter is 1.732 A.

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.

$V(t)=100+10 \sin (\omega t)-5 \sin (3\omega t) \;\;\text{volt}$ moving coil, $V_1=V_{avg.}=100V$ moving iron, $V_2=v_{rms}=\sqrt{100^{2+}\frac{1}{2}(10^{2+}5^2)}$ $\;\;=100.312$ $V_2-V_1=0.312$

$I_{fs}=50A,$ $V_m=0.1V,$ $R_m=\frac{0.1}{50}=2 \times 10^{-3}\Omega$ $\because \; m=10$ $R_{sh}=\frac{R_m}{(m-1)}$ $\;\;\;=\frac{2 \times 10^{-3}}{9}=0.22\Omega$

Currently no explanation available

$V_{BUS} \text{is} 100 kV_{rms}$ $C_1=1\mu F \pm 10\%$ $C_2=9\mu F \pm 10\%$ 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, $V_{out\;rms}=\left ( \frac{C_1}{C_1+C_2} \right )V_{BUS_{rms}}$ $\;\;\;=11.95kV$

Currently no explanation available

For a PMMC ammeter, at equilibrim, we have : controlling torque ( $T_c$ )=Deflecting torque( $T_d$ )

For an electrodynometer ammeter, at equilibrium, Controlling torque=Deflecting torque

For the deflection in the two meter to be same,

Putting the values in I, we get

$\therefore$ Required value of current, $I=3.2Amp.$

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,

$\therefore$ Current in ammeter X=10.28A

Moving iron voltmeter reads rms value of voltage and current. From the given waveform of voltage,

$\therefore$ Voltage reading by moving iron voltmeter =57.73 Volts

In the half cycle, D is ON $\Rightarrow \;\;V_0=0V$ In negative half cycle, D id OFF, PMMC voltmeter measures average value of $V_0$ . In case of half wave rectification,

As PMMC meter reads only DC voltage or average value and average value is equal to $V_{avg}=\frac{\text{Area under the curve}}{\text{Total time}}$ $V_{avg}=\frac{\left ( \frac{1}{2}\times 10 \times 10 \right )+(-5 \times 2)+(5 \times 8)}{20}$ $V_{avg}=\frac{80}{20}=4V$

To extend the range of ammeter, a resistance $R_{sh}$ is connected across the meter. $I_m=$ full scale deflection current=5A $I=25A$ Multiplying power $=m=\frac{I}{I_m}=\frac{25}{5}=5$ $I_m=\frac{IR_{sh}}{R_{sh}+R_m}$ $\frac{I}{I_m}=\frac{R_{sh}+R_m}{R_{sh}}$ $m=1+\frac{R_m}{R_{sh}}$ $R_{sh}=\frac{R_m}{m-1}=\frac{0.2}{5-1}=0.05$

$I=-8+6\sqrt{2} \sin (\omega t+30^{\circ})$ $I_1=-8A$ $I_2=6\sqrt{2} \sin (\omega t+30^{\circ})$ Average value of $I_1=-8A$ Average value of $I_2=0$ So Average value of $I=-8A$ PMMC reads only average value of current Therefore PMMC reads =-8A RMS value of $I=\sqrt{(-8)^{2+}\left ( \frac{6\sqrt{2}}{\sqrt{2}} \right )^2}=10A$ RMS meter and moving iron meter both reads rms value of the current. So, both $m_2$ and $m_3$ reads 10A.

Total voltage across PMMC $=V_T=V_1+V_2$ $=2+3 \sin (4t)V$ PMMC reads average value Average value of $V_1=2V$ Average value of $V_2=0$ Average value of $V_T=2V$ So PMMC reads $=2 V$

Resistance of voltmeter $(R_V)$ appear parallel to $4 M\Omega$ Effective resistance between A and B,

Voltmeter reads, $IR_{AB} =\frac{1000}{6.05} \times 10^{-6} \times 0.05 \times 10^6 \approx 8V$

$I_m=$ fill scale deflection current =100 A $I=$ current to be measure =500A Multiplying power, $m=\frac{I}{I_m}=\frac{500}{100}=5$ $R_m=0.1\Omega$ $I_m=\frac{IR_{sh}}{R_{sh}+R_m}$ $\Rightarrow \; m=1+\frac{R_m}{R_{sh}}$ $\Rightarrow \; R_{sh}=\frac{R_m}{m-1}=\frac{0.1}{5-1}=0.025\Omega$