EE 303 Homework on Transformers, Dr. McCalley.

Transcription

1 EE 303 Homework on Transformers, Dr. ccaey.. The physica construction of four pairs of magneticay couped cois is shown beow. Assume that the magnetic fux is confined to the core materia in each structure (no eakage). Show two possibe ocations for the dot markings on each pair of cois.. Write the equations for v (t) and v (t) for the circuit beow. i (t) i (t) _ v (t) v (t) - 3. Write the equations for (a) v a(t) and v b(t) and (b) v c(t) and v d(t) for the circuit beow.

2 i (t) i (t) v a (t) - v c (t) v d (t) v b (t) - 4. Write a set of mesh current equations that describe the circuit beow in terms of i, i, and i 3. ote the ast equation (right oop) shoud be equa to zero.

3 5. A pair of couped inductors is connected in two different ways as shown beow. In each case, find the differentia equation reating v(t) and i(t), and then find the equivaent inductance seen at the terminas ooking into the circuit. a) v(t)= di/dtdi/dt di/dtdi/dt v(t)=[ ]dt/dt eq= b) v(t)= di/dt-di/dt di/dt-di/dt v(t)=[ -]dt/dt eq= - 6. Use phasors to find the votage v (t) in the circuit beow. Since we need not do any power cacuations in this probem, we may express a phasors using the maximum vaue as the magnitude. One important concept you must use here is that the differentia equations, which characterize the reationships for any time under any type of excitation, may be converted to phasor reations for performing anaysis under steady-state conditions when the excitation is sinusoida. The transformation requires that you repace differentiation by j in the phasor domain. One can see why if you just differentiate i(t)= I sint. Then you get di/dt= I costt, which is just the origina function, scaed by and rotated forward by 90 degrees, i.e., di/dt= I sin(t90). So you see that the phasor for i(t) is just I= I 0, and the phasor for di/dt is just I 90= I 090=jI. et s use the time domain expression first. We can use a mesh current for each side of the transformer, as shown beow.

4 I I ote the above cacuation has two errors in the ast two ines. These errors are: a. next to ast ine: shoud be I =.38 /_ b. ast ine: shoud be V =-I =.656/_ The ohmic vaues of circuit parameters of a transformer, having turns ration of / =5, are =0.5, =0.0, X =3., X =0., c=350, and X m=98. (, X, c, and X m are given referred to the primary side; and X are given referred to the secondary). Draw the approximate equivaent circuit of the transformer, with a quantities referred to (a) the primary and (b) the secondary. Show the numerica vaues of the circuit parameters.

5 8. The exact equivaent circuit parameters of a 50-kVA, 400vot/40vot transformer are =0., =m, X =0.45, X =4.5m, c=0k, and X m=.55k. (, X, c, and X m are given referred to the primary side; and X are given referred to the secondary). Using the circuit referred to the primary, determine the (a) percent votage reguation and (b) efficiency of the transformer operating at rated oad (50 kva) with 0.8 agging power factor. Assume that V =40 vots and note that percent votage reguation is given by (V no-oad-v oad)/v oad (where these are votage magnitudes).

6 9. Using the approximate equivaent circuit #, shown beow, repeat the cacuations of probem 8 and compare the resuts (note that a is the turns ration / ).

7 0. The coefficient of couping for couped cois is defined as k (a) Show that the idea case of no-eakage fux resuts in k=. (b) Determine for the actua case, where eakage fux exists, whether k> or k<. (c) The couped circuit beow has a coefficient of couping of, Determine the energy stored in the mutuay couped inductors at time t=5 msec, where =.653 mh and =0.6 mh.

8 ωωωωωωωω ωωωωωωωω Ω i(t) i(t) 4cos377t - 4Ω Soution: (a) We derived in cass that when there is no eakage fux, case their ratio must be.0. (b) From our notes, we derived that and i Under the no-eakage fux condition, we then used A i and i. i A. Therefore in this ow we consider the case with eakage fux. Here, we assume that the percentage eakage fux from one coi is the same as the percentage eakage fux from the other coi. In this case, these reations shoud be written as ra ra r and r i i. where r is the percentage eakage fux from each coi, 0<r<. Then we can proceed as in the cass notes. Substitution yieds: ra i A r i ikewise ra i A r i Thus, we have that r r

9 r From the notes, we sti have that Substitution yieds: r r And so r=k, which impies, k<. (c) Soution is beow: r