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Vidaara.orgClass 12 · Physics
CodeVID-P12-03-TLC-01
Transformers & LC Oscillations — Assignment
Chapter: Electromagnetic Induction and Alternating Current
Topic: Transformers & LC Oscillations
Maximum Marks: 30
Time: 60 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • All questions are compulsory.
  • Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
  • Show all steps in numericals and draw neat labelled diagrams wherever asked. For full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
A step-up transformer has:
  • A.$N_s < N_p$
  • B.$N_s > N_p$
  • C.$N_s = N_p$
  • D.no secondary
2.
A transformer works on:
  • A.DC only
  • B.AC only
  • C.both equally
  • D.neither
3.
Energy in an LC circuit oscillates between:
  • A.two capacitors
  • B.electric and magnetic fields
  • C.two resistors
  • D.heat and light
4.
Hysteresis loss is reduced by using a core of:
  • A.steel
  • B.soft iron
  • C.copper
  • D.wood
5.
Peak EMF of an AC generator is:
  • A.$NBA$
  • B.$NBA\omega$
  • C.$NBA/\omega$
  • D.$BA\omega$
Section B — Short Answer (2 marks) 3 × 2 = 6 marks
6.
State the principle of a transformer.
7.
A transformer steps 220 V up to 4400 V. Find the turns ratio $N_s/N_p$.
8.
Write the frequency formula for an LC circuit.
Section C — Short Answer (3 marks) 2 × 3 = 6 marks
9.
Why is electric power transmitted at high voltage? Explain.
10.
List three causes of energy loss in a transformer.
Section D — Long Answer (5 marks) 1 × 5 = 5 marks
11.
Describe the construction and working of a transformer with a labelled diagram, derive the turns-ratio relation, and explain how power is conserved in an ideal transformer.

Answer Key

Section A — Multiple Choice Questions
  1. (B) $N_s > N_p$
  2. (B) AC only
  3. (B) electric and magnetic fields
  4. (B) soft iron
  5. (B) $NBA\omega$
Section B — Short Answer (2 marks)
  1. Mutual induction: changing flux in the primary induces an EMF in the secondary.
  2. $N_s/N_p = V_s/V_p = 4400/220 = 20$.
  3. $f = 1/(2\pi\sqrt{LC})$.
Section C — Short Answer (3 marks)
  1. High voltage means low current, so $I^2R$ losses in the cables are minimised.
  2. Copper loss, eddy-current loss and hysteresis loss (also flux leakage).
Section D — Long Answer (5 marks)
  1. Two coils on a laminated soft-iron core; mutual induction gives $V_s/V_p = N_s/N_p$; ideal: $V_pI_p = V_sI_s$, so current ratio is inverse of turns ratio.
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