Skip to main content

EMF EQUATION OF A TRANSFORMER

Figure 1.22 shows the representation of alternating flux, varying sinusoidally, which increases from its zero value to maximum value (Φm) in one-quarter of the cycle, that is in one-fourth of a second where f is the frequency of AC input in hertz.
The average rate of change of flux is given by images, that is 4m Wb/s or V.
images
Figure 1.22 Representation of Alternating Flux
This rate of change of flux per turn is the induced emf in V.
Therefore, average emf/turn = 4mm V.
Let N1 and N2 be the number of turns in primary and secondary.
images
The rms value of induced emf in primary winding is given by
E1 = (4.44m m) × N1 = 4.44m mN = 4.44f BmArN1 (1.1)
where images is the maximum value of flux density having unit Tesla (T) and Ar is the area of cross-section.
Similarly, RMS value of induced emf in secondary winding is
E 2 = (4.44fΦm )x N2 = 4.44mN2 = 4.44f BmArN2 (1.2)
From Equations (1.1) and (1.2), we have
images
i.e., images
where ‘a’ is the turns ratio of the transformer,
i.e., images
Equation (1.3) shows that emf induced per turn in primary and secondary windings are equal.
In an ideal transformer at no load, V1 = E1 and V2 = E2, where V2 is the terminal voltage of the transformer. Equation (1.3) becomes
images
Example 1.1 The voltage ratio of a single-phase, 50 Hz transformer is 5,000/500 V at no load. Calculate the number of turns in each winding if the maximum value of the flux in the core is 7.82 mWb.
Solution
Here
E1 = V1 = 5,000 V
E2 = V2 = 500 V
φmax = 7.82 m Wb = 7.82 × 10−3 Wb, f = 50Hz
Let N1 and N2 be the number of turns of the primary and secondary windings, respectively.
Since
E1 = 4.44 f φmN1
images
i.e., images
Again, images
images

Comments

Popular posts from this blog

Transformer multiple choice questions part 1

Hello Engineer's Q.[1] A transformer transforms (a) frequency (b) voltage (c) current (d) voltage and current Ans : D Q.[2] Which of the following is not a basic element of a transformer ? (a) core (b) primary winding (c) secondary winding (d) mutual flux. Ans : D Q.[3] In an ideal transformer, (a) windings have no resistance (b) core has no losses (c) core has infinite permeability (d) all of the above. Ans : D Q.[4] The main purpose of using core in a transformer is to (a) decrease iron losses (b) prevent eddy current loss (c) eliminate magnetic hysteresis (d) decrease reluctance of the common magnetic circuit. Ans :D Q.[5] Transformer cores are laminated in order to (a) simplify its construction (b) minimize eddy current loss (c) reduce cost (d) reduce hysteresis loss. Ans : B Q.[6] A transformer having 1000 primary turns is connected to a 250-V a.c. supply. For a secondary voltage of 400 V, the number of secondary turns should be (a) 1600 (b) 250 (c) 400 (d) 1250 A

Condition for Maximum Power Developed In Synchronous Motor

The value of δ for which the mechanical power developed is maximum can be obtained as, Note : Thus when R a is negligible, θ = 90 o for maximum power developed. The corresponding torque is called pull out torque. 1.1 The Value of Maximum Power Developed        The value of maximum power developed can be obtained by substituting θ = δ in the equation of P m .        When R a is negligible,     θ = 90 o  and cos (θ) = 0 hence, . . .               R a = Z s cosθ   and X s = Z s sinθ        Substituting   cosθ = R a /Z s in equation (6b) we get,         Solving the above quadratic in E b we get,        As E b is completely dependent on excitation, the equation (8) gives the excitation limits for any load for a synchronous motor. If the excitation exceeds this limit, the motor falls out of step. 1.2 Condition for Excitation When Motor Develops ( P m ) R max        Let us find excitation condition for maximum power developed. The excitation

Electrical Engineering interview questions and answers Part 17

Why star delta starter is preferred with induction motor? Star delta starter is preferred with induction motor due to following reasons: • Starting current is reduced 3-4 times of the direct current due to which voltage drops and hence it causes less losses. • Star delta starter circuit comes in circuit first during starting of motor, which reduces voltage 3 times, that is why current also reduces up to 3 times and hence less motor burning is caused. • In addition, starting torque is increased and it prevents the damage of motor winding. State the difference between generator and alternator Generator and alternator are two devices, which converts mechanical energy into electrical energy. Both have the same principle of electromagnetic induction, the only difference is that their construction. Generator persists stationary magnetic field and rotating conductor