Unit-2

                           Basic principles of Electric circuits

Question Bank

Q1) Explain the construction and working of moving iron type voltmeter?

Sol: Construction

The moving element is a plate or soft iron. The plate is placed such that is moves freely in the magnetic field of stationary coil. These stationary coils are electromagnets. These magnets are temporary magnets whose field strength varies according to the amount of current passing through it. They can be used for both AC as well as DC.

Operation

These instruments use moving iron to measure the flow of current or voltages. When the iron is placed near magnetic field it is attracted towards the field. The force of attraction depends upon the strength of field.

The stationary coil of copper or aluminium wire acts as electromagnet when current passes through it. The plate when passes through this coil increases the inductance of stationary coil. The electromagnets attract the plate. The plate passes through coil from the least reluctance path. This plate experiences a repulsion due to electromagnets. The repulsion force increases the strength of the coil inductance. This happens because the inductance and reluctances are inversely proportional to each other.

One iron vane is held fixed to the coil frame and other is free to rotate, carrying with it the pointer shaft. Two irons lie in the magnetic field produced by the coil that consists of only few turns if the instrument is an ammeter or of many turns if the instrument is a voltmeter. Current in the coil induces both vanes to become magnetized and repulsion between the similarly magnetized vanes produces a proportional rotation

The moving iron instrument can be classified as

i)                   Repulsive moving iron instrument: It has two vanes one is fixed and the other is moving. When current passes through stationary coil the vane gets magnetized and a repulsive force is generated between them. Due to this repulsion the moving coil moves away from the fixed vane.

The spring coil provides controlling torque. The damping torque induced opposes the movement of the coil. These instruments are not affected by the direction of current flow.

Ii)                Attractive moving iron instrument: In this the iron plate is attracted from the weaker field towards the stronger one. The stationary coil is flat and has narrow opening. The flat disc of iron core is moving element. The current through stationary coil produces the magnetic field and attracts the iron coil.

                The iron plate then gets attracted from low magnetic field to high magnetic field. The strength through which the iron plate is deflected is directly proportional to the current flowing through it.

Advantages:

They are low cost instruments. They are easy to assemble. They have less friction error. They are used in both AC and DC.

Q2) Explain induction type Ammeter?

Sol: In all induction meters we have two fluxes which are produced by two different alternating currents on a metallic disc. Due to alternating fluxes there is an induced emf, the emf produced at one point (as shown in the figure given below) interacts with the alternating current of the other side resulting in the production of torque.

Let F1 =flux at point 1

F2= flux at point 2

Then

F1 = Fm1 sinwt

F2= Fm2 sin (wt-B)

B=phase difference between two fluxes.

The induced emf for both points will be

E1= -

E2= -

The value of eddy current at point 1 is

I1 = E1/Z = K x f x F1

The phasor is shown below

The deflection torque at point 1 will be

Td1= K x F2 x I1x cos(90-B+A) = K x F1 x (f/Z) cos(90-B+A) x F2

The deflection torque at point 2 will be

Td2= K x F1 x (f/Z) x cos(90+B+A) x F2

Total torque is

Td1- Td2 = K x F1 x F2 x (f/Z) sinB cos A

They are of two types:

i)                   Split phase type: It has two magnets M1 and M2 in series. The shunt resistance R is connected to M2. The current in M2 lags line current. The deflection torque is Td = φ1m φ2m sinα

Ii)                Shaded Pole Type: In this case the exciting coil is placed on poles. The current proportion to current or voltage is measured.

In the air gap of electromagnet aluminium disc mounted on a spindle is inserted. The spring is attached to the pointer and both are mounted on spindle. The spring provides the controlling torque.

The deflection torque is Td = φ1φ2 sinα

Q3) Explain working of moving coil type voltmeter?

Sol: It is called as Permanent magnet moving coil (PMMC). They measure only DC. This instrument measures the current through coil by angular deflection in magnetic field. A coil is placed between two permanent magnets in these types of instrument to produce magnetic field. The current carrying conductor when placed in the field experiences a force which is proportional to the amount of current through it. When the torques are balanced the moving coil stops. The angular deflection can be measured.

The deflection torque equation is given as

Td=NBldI

N: Number of turns

B: Magnetic flux density

l: length of moving coil

d: width of moving coil

I: Electric current

The below circuit extends the range of voltmeter by connecting resistance in series with the meter.

The external resistance connected will be

R= [V-V1/V1] x RV

RV: Voltmeter resistance

R: external resistance

V1: voltage across voltmeter

An iron ring of mean diameter 15cm has a cross sectional area of 100cm2 is wounded with 400turns of wire. Calculate the exciting current required to establish a flux density of 1Wb/m2. If relative permeability is 1000.

Sol: B=

1= x 1000x400xI/0.15

I =0.94A

Q4) An iron ring of mean diameter 15cm has a cross sectional area of 100cm2 is wounded with 400turns of wire. Calculate the energy stored. The flux density is 1Wb/m2. If relative permeability is 1000.

Sol: E= L I2

B=

1= x 1000x400xI/0.15

I =0.94A

L= = x 1000x 100x10-4x4002/0.15=4.27H

E= L I2

   = x 4.27 x 0.94 = 2.01J

Q5) The amount of flux present in around magnetic bar was measured at 0.02 weber. If the material has a diameter of 10cm, calculate the flux density?

Sol: Area=r2

Diameter=2r

r=10/2=5cm=0.05m

Area=3.14 x 0.052=0.008m2

Flux Density B==A=0.02/0.008=2.5 Tesla

Q6) Calculate the radius of the material having flux density of 0.7 T and flux present around the magnetic bar is 0.05T?

Sol: Flux Density B=A

        A= B=0.05/0.7=0.07 m2

Area=r2

r=0.151m

Q7) What are magnetic lines of forces. List their properties?

Sol: All magnets have two regions called magnetic poles with the magnetism both in and around a magnetic circuit producing a definite chain of organised and balanced pattern of invisible lines of flux around it. These lines of flux are called as the magnetic field of the magnet.  At each end of a magnet is a pole.  Magnetic poles are always present in pairs, there is always a region of the magnet called the North-pole and there is always an opposite region called the South-pole. The lines which go to make up a magnetic field showing the direction and intensity are called Lines of Force or Magnetic Flux and represented as Phi ( Φ ).

• Lines of force never cross.
• Lines of force are continuous.
• Lines of force always form individual closed loops around the magnet.
• Lines of force have a definite direction from North to South.
• Lines of force that are close together indicate a strong magnetic field.
• Lines of force that are farther apart indicate a weak magnetic field.

Like poles of the magnet repel each other and unlike poles attract each other.

Magnetic flux density =Magnet flux/Area

             B=A                Tesla

Q8) Define as well as derive magnetic flux? 

Sol: The number of magnetic lines of forces set up in a magnetic circuit is called Magnetic Flux. It is analogous to electric current, I in an electric circuit. Its SI unit is Weber (Wb) and its CGS unit is Maxwell. It is denoted by φB.

ΦB=B.S

ΦB=B.S Cos

B – the magnitude of the magnetic field
S – area of surface
θ – angle between the magnetic field lines and perpendicular distance normal to the surface area
Magnetic flux for a closed surface

ΦB==0

Magnetic flux for open surface is

E=

E=

E – electromotive force
v – velocity of the boundary
E – electric field
B – magnetic field
øB - magnetic flux through the open surface

The magnetic flux through a closed surface is always zero, but in the open surface, it is not zero.

Q9) An iron wire ring of 15 cm mean diameter having a cross-section of 90cm2 is wound with 400 turns of wire. Calculate the exciting current required to establish a flux density of 1Wb/m2 if relative permeability of iron is 1000. What is the value of energy stored?

Sol: B=

1=4 x x 1000 x 400 x I/0.15

I=0.94A

L=AN2/l=4 x x 1000 x (90x 10-4) x (400)2/0.15=3.84H

E=LI2

   =x 3.84 x (0.94)2=1.696J

Q10) State Faraday’s Law of electromagnetic induction?

Sol: Faraday’s Laws of Electromagnetic Induction consists of two laws. The first law describes the induction of emf in a conductor and the second law quantifies the emf produced in the conductor

i)                    Faradays First law of EMI: Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced. If the conductor circuit is closed, a current is induced which is called induced current.

Ii)                 Faradays Second law of EMI: The induced emf in a coil is equal to the rate of change of flux linkage.

= electromotive force

N = Number of turns

= magnetic flux

• Increase in the number of turns in the coil increases the induced emf
• Increasing the magnetic field strength increases the induced emf
• Increasing the speed of the relative motion between the coil and the magnet, results in the increased emf

Q11) Explain working of galvanometer coil?

Sol: A moving coil galvanometer is an instrument which is used to measure electric current. A current-carrying coil when placed in an external magnetic field experiences magnetic torque. The angle through which the coil is deflected due to the effect of the magnetic torque is proportional to the magnitude of current in the coil. The moving coil galvanometer is made up of a rectangular coil that has many turns and it is usually made of thinly insulated or fine copper wire that is wounded on a metallic frame.

The coil is free to rotate about a fixed axis. A phosphor-bronze strip that is connected to a movable torsion head is used to suspend the coil in a uniform radial magnetic field. A cylindrical soft iron core is symmetrically positioned inside the coil to improve the strength of the magnetic field and to make the field radial. The lower part of the coil is attached to a phosphor-bronze spring having a small number of turns. The other end of the spring is connected to binding screws.

We know that torque τ = force x perpendicular distance between the forces

τ = F × b

Substituting the value of F we already know,

Torque τ acting on single-loop of the coil = BIl × b

Where lx b is the area A of the coil,

Hence the torque acting on n turns of the coil is given by

τ = nIAB

The magnetic torque thus produced causes the coil to rotate, and the phosphor bronze strip twists. In turn, the spring S attached to the coil produces a counter torque or restoring torque kθ which results in a steady angular deflection.

Under equilibrium condition:

kθ = nIAB

Here k is called the torsional constant of the spring (restoring couple per unit twist). The deflection or twist θ is measured as the value indicated on a scale by a pointer which is connected to the suspension wire.

θ= ( nAB / k)I

Therefore θ ∝ I

The quantity nAB / k is a constant for a given galvanometer. Hence it is understood that the deflection that occurs the galvanometer is directly proportional to the current that flows through it.