Looking for a Tutor Near You?

Post Learning Requirement »
x

Choose Country Code

x

Direction

x

Ask a Question

x

Hire a Tutor
Doha

Electronic Circuit

Loading...

Published in: Electrical | Electronics
20 Views

Exam Revision

Hamida S / Doha

13 years of teaching experience

Qualification: Bachelor in Electrical Engineering

Teaches: Electrical, Electronics, Instrumentation, Maths, Computer / IT, Advanced Excel, Artificial Intelligence, C / C++, C# (C Sharp), Embedded Systems, Machine Learning, Matlab, MS Office, Python Programming, Robotics, Wordpress Training, Computer Basics, Control System, Data Analysis, Telecommunications, Science, Mathematics, Computing, Physics, Statistics, Engineering & Technology, Arabic, English, Economics, Sciences, Advanced Maths, Basic Computer, Science Projects, Coding & Programming, Engineering Subjects, Computer Science/IT

Contact this Tutor
  1. Problem 3: (35 pts) For the core shown below. The winding on the left leg of the core (NI) has 500 turns and the winding on the right (N2) has 300 turns. Assume 1500 and constant. ölfthe dimensions are as shown, then what would be the total reluctance in the core? If il = 0.6 A, What is the flux in the core due to il? Find the value and direction of the applied current i2 to produce zero net flux in the core (mark the direction of i2 on the figure below)? an Re-LQance•. : -g Lc (7. S*7.sæSöX+ = A c. 225 cot 61304.126 X 10-7 ) * , F=Nlil
  2. N/A
  3. # 130..eES : 1 y-ce : Relucå-aoce. ngshace.- RelNOe pervaL;lib : FOP Mead Ceil •turn crus sec+ioo.l (7.5C7.5+5)X+
  4. Floe•s id al (AC Si'. bek at- ? ah EHR 52: indica*u.
  5. The figure below shows a ferromagnetic core. There is a small gap of 0.05 cm in the structure of the core. The depth ofthe core is 5 cm, the relative permeability of the core is 4000, and the coil of wire on the core has 1000 turns. Assume that fringing in the arr gap increases the effective cross-sectional area of the air gap by 5 percent. Given this in ormation, nd a) b) c) d) The total reluctance of the flux path (iron plus air gap). (12 pts) The current required to produce a flux density of 0.5 T in the air gap. Show the direction of the flux in the core. (12pts) The flux intensity in the air gap. (5 pts) The inductance of the coil.(6 pts) = 45757.04 4æo R. I-cz pct- -11791.2 1-020.0504 -r A3 I.os A
  6. oas Lio Air reluckance: 22912+.ß7 b) The current required to produce a flux density of 0.5 T in the air gap. Show the direction of the flux in the core. ( 12pts) fie(L c) The flux intensity in the air gap. (5 pts)
  7. d) The inductance of the coil.(6 pts)
  8. Problem 2: Answer the next, (15 pts) a) Give all types of losses in the transformer losses U-- ; s lo%es b) Draw the Exact equivalent circuit (electric circuit) of below real transformer c) Plot the phasor diagram of the single-phase transformer circuit referred to the secondary side for lagging PF.
  9. + Req. Is co •oe-l C.o•aeß leading power factor apacitive load -2 Vs
  10. Problem 5: (25pts) A linear machine has a magnetic flux density of 0.7 T directed into the page, a resistance of 0.5 Q, a bar length I = I .0 m, and a battery voltage of 120 V. VB — 120 V _ a) b) c) d) e) What is the initial force on the bar at starting? What is the initial current flow? What is the initial torque? (5pts) What is the no-load steady-state speed of the bar and what is the direction of motion? (5pts) If the bar is loaded with a force of 10 N in the direction of motion, what is the new steady state speed and what is its direction? Is the machine working as motor or as generator and wh ? (5pts) If the bar is loaded with a force of 20 N opposite to the direction o motion, t IS t e new steady state speed and what is its dilection? Is the machine working as motor or as generator and why? (5pts) for the case in part (d) it is required to control the speed to have the same value as in part (c), what can you change to achieve this? (5pts) 0.5 Q i B -0.7 T a TLß lai+ialb no Wt appo-s OCTSS ßar•. All vs 12.0 •z.A-o-t••, appeays acsss 0 -9 side b) 120 speed
  11. Far = -!2— - A K 127. S V app) led dineQ4-bn 127, Is d) lithe bar is loaded with a force of 20 N opposite to steady state speed and what is its dirction? Is the machine working as motor or as generator and why? (5pts) * 20 - (28.6) (0.9—105-7
  12. (d) With a load or 20 N oppthe direction of moon. steady-state current flow in the bar Will be given by 20 N m - (0.7 m) SOLUTION (a) TEX' current in tip bat at starting is v 120 v -240A R 0.50 TtEreI«e. on the bar at starting is 168 N. tothenght (b) no-load steady-state speed or this bar can be found frcgn the equation 120 v -171.4 m = (0.7 m) (c) With a load Of 10 N in direction or n»tion. the steady-state current now in bar will 28.6 A be given by ION -14.3A TIE iruluced voltage in bar will be % —v, +iR- 120 v + Q) and tip velocity of the bar will be 127.15 v —181 nvs 127.15 v TIP induced vdtage in tip bar Will be iR- 120V-( Q) - and trp velocity or ttp bar will be 105.7 (0.7 m) - 105.7 (e) Either change VB to 141 V, or change the flux density to 0.565T, or 0.0977 T
  13. Problem 5: (35 ptsy A linear machine has a magnetic flux density of0.4 T directed into the page, a resistance of 0.75 Q, a bar length I = 0.8 m, and a battery voltage of 150 V. a) What is the initial force on the bar at starting? What is the initial current flow? What is the initial torque? (1 5 pts) b) What is the no-load steady-state speed of the bar and what is the direction of motion? (8pts) c) If the bar is loaded with a force of 10 N in the direction of motion, what is the new steady state speed and what is its direction? Is the machine working as motor or as generator and why? (12 pts) VB-150; R-0.75 1=0.8 t—Os 0.75 Q i i i a) The current O=VB/R; = 200 starting therefore, the strt*i X B •0.4T 0.8 m % Battery voltage (V) % Resistance (ohms) % Bar length (m) Flux density (T) into page in the bar at starting force on the bar at starting is FC starting — 64 N} tothe might Initial torque 02, % b) the no—load steady—state speed Of this bar can be found from the equation VB—e ind=vB1 sS-VB/ (B*ff, 60 steady—state speed = 468.75 m/ s, to the right. % c) With a load Of 10 N in the direction Of mo ion, the steady— state current Fapp= 10; i-Fapp/ 1) ; - 31.25 A the induced flow in voltage the bar will be in the bar will given by be
  14. e ind = 173.4375 V and the velocity of the bar will be ind/ (B* 1); = 541.9922 m/s, the machine is working as a generator, and providing power to the battery 10 moving at a higher speed MOTDe ?

Need a Tutor or Coaching Class?

Post an enquiry and get instant responses from qualified and experienced tutors.

Post Requirement

Related Study Notes

  • Ecen 370
    Electrical, Electronics
    23 Views
  • ECEN 370
    Electrical, Electronics
    24 Views

Upload Study Notes

If you have your own Study Notes which you think can benefit others, please upload on MyPrivateTutor. For each approved Study Note you will get 100 Credit Points and 100 Activity Score which will increase your profile visibility.

Upload Now
Home > Study Notes > Electronic Circuit