EIM
Unit-2Measurement of Voltage Q1) Explain moving iron instrument? The most common ammeters and voltmeters for laboratory or switch board used at power frequencies are the moving iron instruments. When the coil is excited it becomes an electromagnet and the iron vane moves in such a way so as to increase the flux of the electromagnet. An expression for torque of the moving iron instrument may be derived by considering the energy relations when there is small current in current supplied to the instrument. When it happens, there is a small deflection d and the mechanical work will be done. Let Td = deflecting torque Mechanical work done = Td . d Suppose the initial current is I, the instrument inductance is L and the deflection . If the current increases by dI then the deflection changes by d and the inductance by dL. In order to affect an increment dI in current there must be an increase in the applied voltage given by e = d/dt (LI) = I dL/dt + L dI/dt The electrical energy supplied e Idt = I2 dL + IL dIThe stored energy changes from = ½ I 2 L to ½ (I +dI) 2 (L + dL) Hence the change in stored energy = ½ (I2 + 2IdI + dI2) (L + dL) -1/2 I2 L From the principle of conservation of energy Electrical energy supplied = increase in stored energy + mechanical work done I2 dL + I L dI =I L dI= ½ I2 dL + Td dTd d = ½ I2 dL Or Deflecting torque Td = ½ I 2 dL/d T is in newton-metre I in ampere L in henry in radian The moving system is provided with control springs and it turns the deflecting torque Td is balance by the controlling torque Tc Tc = K Where K = control spring constant deflection At equilibrium Tc = Td Or K = ½ I 2 dL /dOr =1/2 I2 /K dL/d Q2) Explain moving coil instrument?
Figure 1. Moving Coil InstrumentThe moving coil is wound with many turns of enamelled or silk covered copper wire. The coil is mounted on a rectangular aluminium former which is pivoted on jewellery bearings. The coils move freely in the field of permanent magnet. Most voltmeter coils are wound on metal frames to provide the required elctro-magnetic damping. Most ammeter coils however are wound on non-magnetic formers because coil turns are effectively shorted by the ammeter shunt. The coil itself therefore provides electromagnetic damping. Torque Equation The torque of moving coil instrument is given by Deflection torque is Td = NB l dI = GI Where G is constant The spring control provides a restoring (controlling torque Td = KFor final steady deflection Tc = Td Or GI = KFinal steady deflection = G/K I Current I = Kd Kd = K/G = constant Q3) A permanent magnet moving coil instrument has a coil of dimensions 15mmX12mm. The flux density in the air gap is 1.8 x 10 -3 wb/m2 and the spring constant is 0.14 x 10 -6 Nm/rad. Determine the number of turns required to produce an angular deflection of 90 degrees when a current of 5mA flows through the coil. Solution:
Q4) What is Thermal Induction?
Figure 2. Measurement of induced voltageIn the experiment the induction coils with different areas and numbers of turns are arranged in a cylindrical field coil through which alternating currents of various frequencies, amplitudes and signal forms flow. In the field coil, the currents generate the magnetic fieldB = μo. N2 / L2.I Where μo = 4π x 10 -r Vs/ Am (permeability)and I(t) is the time-dependent current level, N2 the number of turns and L2 the overall length of the coil. Q5) Explain Rectifier type? Rectifier instruments are used for ac measurements by using a rectifier to convert ac into unidirectional dc and then to use d.c meter to indicate the value of rectified a.c. This method is attractive because PMMC has higher sensitivity than either the electrodynamometer type or the moving iron instruments. Rectifier instruments are suited for measurements on communication circuits and other light current work where the voltages are low and resistance is high. Q6) What is dynamometer? It consists of a fixed coil FF (split into two parts) which carries the current of the circuit under test. So, the magnetic field produced by this coil is proportional to the main current.
Figure 3. Dynamometer The identical pressure coils A and B pivoted on a spindle constitutes the moving system. Pressure coil A has a non-inductive resistance R which is connected in series with it, and coil B has a highly inductive choke coil L which is connected in series with it. The two coils are connected across the voltage of circuit. The value of R and L are so adjusted that the two coils carry the same value of current at normal frequency i.e R= WL.The current through the coil A is in phase with the circuit voltage while that through the coil B legs the voltage by an angle ẟ which is nearly equal to 90°. The angle between the planes of coils is equal to δ. Theory:Let the current through coil B lags the voltage by exactly 90˚.The field of the two fixed coils is uniform and in the direction of arrow. The torque on each coil for a given coil current will be maximum when the coil is parallel to the filed that is along XX. When the system power factor angle is φ, the coils take up a position of equilibrium displaced θ from the vertical. Then the torque due to the two coils must be equal and opposite.Now since the current in coil A is in phase with the system voltage, the field in which it moves is proportional to the system current, then coil A is essentially a wattmeter movement displaced 90˚- θ from the maximum torque position. Then the torque of A is given by
Q7) Explain single phase power meter? The general circuit diagram of single- phase electrodynamometer power factor meter is given below.
Figure 4. Single Phase Power factor meterThe pressure coil is split into two parts one is purely inductive another is purely resistive as shown in the diagram. At present the reference plane is making an angle A with coil 1 and the angle between both the coils 1 and 2 is 90o. Thus, the coil 2 is making an angle (90o + A) with the reference plane. Consider the electrical resistance connected to coil 1 be R and inductor connected to coil 2 be L. During measurement of power factor the values of R and L are adjusted so that R = wL so that both coils carry equal magnitude of current. Therefore, the current passing through the coil 2 is lags by 90o with reference to current in coil 1 as coil 2 path is highly inductive in nature.
Derivation for power factor meter. Now there are two deflecting torques one is acting on the coil 1 and another is acting on the coil 2. The coil winding are arranged such that the two torques produced, are opposite to each other and therefore pointer will take a position where the two torques are equal. The mathematical expression for the deflecting torque for coil 1-T1 = KVI M cos A sin B
Where M is the maximum value of mutual inductance between the two coils,
B is the angular deflection of the plane of reference.
Now the mathematical expression for the deflecting torque for coil 2 is-T2 = KVIM cos (90-A) sin(90+B) = KVIM sin A cos B
At equilibrium we have both the torque as equal thus on equating T1=T2 we have A = B. The deflection angle is the measure of phase angle of the given circuit. The phasor diagram is also shown for the circuit such that the current in the coil 1 is approximately at an angle of 90o to current in the coil 2.
Figure 5. Phasor Diagram Q8) Explain three phase power factor meters? A dynamometer type three-phase power factor meter gives correct readings only when the load is balanced. It consists of two fixed coils FF connected in series in one of the phases and carries the line current as shown in the figure.
Figure 6. Three Phase Power Meter The two moving coils A and B are fixed with their planes 120° apart and connected across the two remaining phases respectively through resistances which are high as shown in the figure. When the three-phase power factor meter is connected in the circuit, under balanced load conditions, the angle through which the pointer is deflected from the unity power factor position is equal to the phase angle of the circuit, because the two moving coils are fixed 120° apart.
The deflections in three phase power factor meters are independent of frequency and waveform since the currents in the two moving coils are affected in the same way by any change of frequency. Q9) What is resonance?
Figure 7. Electrical resonance typeIt consists of fixed coil which is connected across the supply whose frequency is to be measured. This coil is called magnetizing coil. The magnetizing coil is mounted on laminated iron core. The iron core has cross-section which varies gradually over length being maximum near the end where the magnetizing coil is mounted and minimum at the other end. A moving coil is pivoted over this iron core. A pointer is attached to the moving coil. The terminals of the moving coil are connected to suitable capacitor C. There is no controlling force. The operation of the instrument can be understood from three phasor diagrams (a), (b), (c). The magnetizing coil carries current I and this current produce flux ɸ. Flux ɸ being alternating in nature induces emf e in the moving coil. This emf lags behind the flux by 900. The emf induced circulates current I in the moving coil. The phase of this current I depend on L and capacitance.
Figure 8. Phasor diagram for electrical resonanceIn fig(a) the circuit of the moving coil is assumed to be inductive and therefore current I lag behind emf e by angle α. The torque acting on the moving coil is Td α Ii cos(90+α)In fig(b) the moving coil is assumed to be capacitive and therefore current I lead emf by angle β and the deflecting torque is Td α Ii cos(90-β)In fig (c) the inductive reactance is supposed to be equal to capacitive reactance therefore the circuit is in resonance. Thus, the moving coil circuit is purely resistive and so the current I is in phase with emf e. This is because inductive reactance XL = 2 πfL =Xc= 1/2πfC. Therefore, the net reactance of the circuit is zero. The deflecting torque is therefore Td α Ii cos 90 =0 Hence the deflecting torque on the moving coil is zero when the inductive reactance equals capacitive reactance. Q10) Explain moving iron frequency meter? Moving Iron Frequency Meter
Figure 9. Moving Iron Frequency meter This meter depends on the variations in an electric current drawn by inductive and non-inductive circuits connected in parallel. The current flows from these circuits when the frequency changes its value.This meter consists of two fixed coils A and B that their magnetic axes are perpendicular to each other. A long and soft iron needle in pivoted at their centres. This circuit remains balanced at the supply frequency to be measured. Coil A consists of a series resistance Ra and a reactance La in parallel and the coil B consists of a series resistance Rb and a reactance Lb in parallel. The series inductance helps to suppress higher harmonics in the current waveform which helps to minimize the waveform errors in the indication of the instruments. When the supply is connected to the meter, the current pass through the coils A and B and these two coils produce opposing torques. When the supply frequency increases then the current of the coil A increases and decreases in the coil B. The iron needle lies more nearly to the magnetic axis of the coil A. For low frequencies, the current of coil B increases and the current of the coil A decreases.
Deflection =90 = π/2 At equilibrium Td = Tc NBI dl = K Number of turns N = K/ Bl. dI = 0.14 x 10 -6 x π/2 / 1.8 x 10 -3 x 15 x 10 -3 x 12 x 10-3 x 5 x 10 -3 = 136 |
TA=KVI cos φ cos (90˚- θ) where k is a constant. Similarly, since the current in coil B lags 90° on the system voltage, coil B is in a sine meter movement displaced θ from the maximum torque position. Then torque on B is. TB = KVI sin φ cos θ Hence at equilibrium TA = TB i.e KVI cos φ cos (90˚- θ) KVI sin φ cos θ or sin φ cos θ = cos φ sin θ or tan θ = tan φ or θ = φ therefore, the deflection of the instrument is a phase angle of the circuit. |
Derivation for power factor meter. Now there are two deflecting torques one is acting on the coil 1 and another is acting on the coil 2. The coil winding are arranged such that the two torques produced, are opposite to each other and therefore pointer will take a position where the two torques are equal. The mathematical expression for the deflecting torque for coil 1-T1 = KVI M cos A sin B
Where M is the maximum value of mutual inductance between the two coils,
B is the angular deflection of the plane of reference.
Now the mathematical expression for the deflecting torque for coil 2 is-T2 = KVIM cos (90-A) sin(90+B) = KVIM sin A cos B
At equilibrium we have both the torque as equal thus on equating T1=T2 we have A = B. The deflection angle is the measure of phase angle of the given circuit. The phasor diagram is also shown for the circuit such that the current in the coil 1 is approximately at an angle of 90o to current in the coil 2.
The deflections in three phase power factor meters are independent of frequency and waveform since the currents in the two moving coils are affected in the same way by any change of frequency. Q9) What is resonance?
0 matching results found