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## Transformers, Physics tutorial

:TransformersE.m.f. related by a changing magnetic flux is employed to convert a varying voltage in one circuit to the large or smaller voltage in the other circuit. This is accomplished by employing a transformer comprising of two coils electrically insulated from one other and wound on the similar ferromagnetic core.

The power is supplied to one coil termed as the primary coil. A varying current in this coil sets up a varying magnetic flux, largely confined to the ferromagnetic core. The other coil, termed as the secondary coil, therefore encloses a varying flux. As this flux is changing at the rate dΦ/dt, each of the Ns turns in the secondary coil experiences, according to the Faraday's flux rule, an induced e.m.f. equivalent to dΦ/dt, and the e.m.f for the whole secondary coil is:

E

_{S}= - Ns dΦ/dtWe ignore the small quantity of leakage flux and suppose that the magnetic flux at any instant is similar via the primary and secondary coils. Then there is an e.m.f. dΦ/dt induced in each of the NP turns of the primary coil and the net e.m.f. for this coil is:

E

_{P}= -N_{P}dΦ/dtDividing the above two equations, we get

E

_{S}/E_{P}= N_{S}/N_{P}According to the above result, if there are varying currents, then the ratio of the e.m.f. induced in the secondary and primary coils at any instant is equivalent to the 'turn ratio', N

_{S}/N_{P}.If N

_{S}is more than N_{P}, the secondary e.m.f surpasses the primary e.m.f and the transformer is termed as a step-up transformer. In a step-down transformer, N_{S}is less than N_{P}.This is desirable to transmit electrical power at high voltages and small currents in order to minimize the I

^{2}R poor dissipation in the transmission line. However this power should be produced and finally delivered at relatively low voltages to avoid the problems of insulation and safety. An extremely helpful characteristic of alternating current is the fact that the voltage and the current can be transformed with simplicity and effectiveness by the use of transformers.:Energy Losses in a TransformerHowever transformers are very proficient devices, small energy losses do take place in them due to four main causes.

1) Resistance of windings: The copper wire employed for the windings consist of resistance and so ordinary (I

^{2}R) heat losses take place. In high-current, low potential difference windings these are minimized by employing thick wire.2) Eddy currents: The alternating magnetic flux induces the eddy currents in the iron core and causes heating. The effect is decreased by containing a laminated core.

3) Hysteresis: The magnetization of the core is repeatedly reversed through the alternating magnetic field. The resultant expenses of energy in the core appear as heat and are kept to a minimum by employing a magnetic material (like mametal) which consists of a low hysteresis loss.

4) Flux leakage: The flux due to the primary might not all links the secondary when the core is poorly designed or has air gaps in it.

:Mutual InductanceTo foliate the analysis of couple circuits like the primary and secondary circuits of a transformer, we must know the relationship between the changing current in one circuit and the e.m.f that this induces in the other circuit.

Assume that a current Ip in the primary circuit generates flux Φ

_{s}via each turn of the secondary coil. The mutual inductance 'M' of the primary and secondary circuits is stated by:Ns Φ

_{s}= M IpWhen there are no ferromagnetic materials present, Φ

_{s}is proportional to Ip and m is then a constant having a value which based only on the geometry of the circuits. For a transformer having a ferromagnetic core, 'M' depends as well on the permeability of the core that is a function of the magnetic field 'B' established in the core.The SI unit of mutual inductance is the Weber per ampere (Wb. A

^{-1}), that is termed as the Henry (H).IH = I Wb A

^{-1}A current I

_{s }in the secondary will make a flux Φ_{p}enclosed by each and every turn of the primary. By using an equation analogous to above equation, we define a mutual inductance M by:N

_{p}Φ_{p}= M I_{s}In general M = M, so that there is a single mutual inductance of any two circuits.

If M is constant, N

_{p}d Φ_{p}/dt = M dI_{s}/dt, and Faraday's flux rule gives:E

_{p}= - M dI_{s}/dtLikewise, Es = - m dI

_{p}/dtThese two equations precisely state the coupling between any two different circuits in the absence of the ferromagnetic materials.

:Self-InductanceIf the current via a coil is changing, the flux generated through this current and enclosed by the turns of the coil is as well changing, as a result there is an e.m.f induced in the coil. Such an e.m.f. is termed as self-induced e.m.f.

Assume that a coil current 'I' generates a flux Φ

_{ }via each of 'N' turns of a coil. The self-inductance 'L' of the coil is stated by the equation:NΦ = LI

When there are no ferromagnetic materials present, Φ is proportional to I. Then L is a constant having a value determined by the coil geometry. The self-inductance of a given coil can be raised greatly by giving a ferromagnetic core; though, due to the core's magnetic permeability 'μ' based on B the self-inductance is then a complex function of the coil current.

When 'L' is constant, NdΦ/dt = L dI/dt, and Faraday's flux rule represents:

E = L dI/dt

The equation provides a helpful expression for the self-induced e.m.f in terms of the changing current.

Lenz's law means that the direction of the self-induced e.m.f is such as to oppose the change in current which gives rise to the change in flux. Thus, the induced e.m.f is in the direction of the current when the current is decreasing, however in the opposite direction of the current whenever increasing. After a direction in an electric circuit is chosen as the positive direction for both currents and e.m.f, the minus sign in the above equation makes sure that E and I have the proper relative directions at all times.

Circuit elements which are specifically designed to encompass appreciable self-inductance are termed as inductors.

The unit of inductance is Henry (H), stated as the inductance of a coil (or circuit) in which an e.m.f. of 1 volt is induced if the current changes at the rate of 1 ampere per second. That is, IH = 1 Vs A

^{-1}:Inductance of SolenoidLet us suppose a long, air-cored solenoid of length l, cross-sectional area 'A' having 'N' turns and carrying current 'I'. The flux density 'B' is nearly constant over A and neglecting the ends, is represented by:

B = μ

_{o}(NI/l)The flux Φ via each turn of the solenoid is BA and for the flux-linkage, we have:

NΦ = BAN

NΦ = [(μ

_{o}N I)/l] (AN)NΦ = (μ

_{o}AN^{2}. I)/lWhen the current changes by dI in time dt causing a flux-linkage change d(NΦ) then by means of Faraday's law the induced e.m.f is:

E = d/dt (NΦ)

E = - [(μ

_{o}AN^{2})/l] .dI/dtWhen 'L' is the inductance of the solenoid, then from the defining equation we have:

E = -L(dI/dt)

Comparing such two expressions at follows that:

L = μ

_{o}AN^{2}/lExhibiting that 'L' based only on the geometry of the solenoid.

When N = 400 turns, l = 25cm = 25 x 10

^{-2}m, A 50cm^{2}= 50 x 10^{-4}m^{2}and μ_{o}= 4 π x 10^{-7}Hm^{-1}, then L = 4.0 x 10^{-3}H = 4.0mH.A solenoid encompassing a core of magnetic material would encompass a much greater inductance however the value would vary based on the current in the solenoid.

:Energy Stored by an InductorIf an e.m.f. E = - L dI/dt, is induced in the inductor through a changing current I, the rate at which energy is supplied by the inductor is EI. If E is in the direction of I, the inductor supplies energy to the external circuit. And if E and I encompass opposite directions, energy is supplied to the inductor. If the current I in an inductor is in the positive direction and is rising, energy is being transferred to the inductor at a rate:

dv/dt = (LdI/dt)I

The net energy supplied as the current rise from 0 to I is thus,

U =

_{o}∫^{1}L I^{1}dI^{1}(1/2) LI^{2}We sum up that the energy stored in the magnetic field of an inductor carrying a current I is:

U = (1/2) L I

^{2}As each and every current generates a field, every circuit should have some self-inductance. On switching on any circuit some time is essential to give the energy in the magnetic field and therefore no current can be brought instantly to a non-zero value. Likewise, on switching off any circuit, the energy of the magnetic field should be dissipated somehow, thus the spark.

:Transients in R-L CircuitsAn inductor in which there is a rising current is a seat of e.m.f whose direction is opposite to that of the current. As an effect of this 'back' e.m.f, the current in the inductive circuit will not mount to its final value at the instant if the circuit is closed, however will grow at a rate which is based on the inductance and resistance of the circuit.

In this part we shall consider transients in a circuit which includes a resistor and inductor (R-L circuit). Consider the circuit shown above which includes an ideal source (ν

_{s}), a resistor (R), an inductor (L) and a switch (S). It is supposed that the switch is open and is closed at an instant of time t = 0. This means that the current 'i' is zero before closing of the switch.Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

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