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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Wednesday, December 5, 2018

Fig. 6.2 shows the variation with time t of the current Ip in the primary coil. There is no current in the secondary coil.


Question 2
An ideal iron-cored transformer is illustrated in Fig. 6.1.

Fig. 6.1

(a) Explain why
(i) the supply to the primary coil must be alternating current, not direct current, [2]

(ii) for constant input power, the output current must decrease if the output voltage
increases. [2]


(b) Fig. 6.2 shows the variation with time t of the current Ip in the primary coil. There is no current in the secondary coil.


Fig. 6.2


Fig. 6.3



Fig. 6.4

(i) Complete Fig. 6.3 to show the variation with time t of the magnetic flux Φ in the
core. [1]

(ii) Complete Fig. 6.4 to show the variation with time t of the e.m.f. E induced in the
secondary coil. [2]

(iii) Hence state the phase difference between the current Ip in the primary coil and the
e.m.f. E induced in the secondary coil. [1]





Reference: Past Exam Paper – June 2005 Paper 4 Q6





Solution:
(a)
(i) The flux/field in the core must be changing so that an e.m.f./current is induced in the secondary.

(ii)
Power = VI
Since the output power is constant, if the (secondary) output voltage VS increases, the (secondary) current IS must decrease.


(b)
(i)
Graph has same shape and phase as IP




(ii)
Graph:
same frequency, correct phase w.r.t. Fig. 6.3



{emf induced E = - dϕ / dt
The graph of ϕ is a sine curve. Differentiating sine gives cos. But since we have a negative sign, the cosine curve is inverted.}


(iii)
½π rad or 90°

8 comments:

  1. Is there something like for each topic you have worked solutions, the same way as in the Doubts section but topical and a lot of questions?

    ReplyDelete
    Replies
    1. We have not created a page specific for this yet.

      But you can find this under the title 'A Level 9702 Topic by Topic' in the sidebar on the right.

      Note that these do not yet contain the over 1000 questions already solved. As more questions are being solved, the number of worked solutions under each topic will increase.

      Delete
  2. Since the magnetic flux is inversely proportional to current. How the graph has same shape and phase as Ip

    ReplyDelete
    Replies
    1. Note that the equation F = BIL applies for the force in a current-carrying conductor when placed in a magnetic field (which is not due to the wire itself). This causes the wire to move. But it is NOT the case here as the wire in NOT moving.

      We know that a current flows through a wire, a magnetic field is formed due to the wire. If a greater current flows, the magnetic field is greater.

      Delete
    2. Thank you!
      Do we have any other relation than [[[(Induced emf)∝ I! As (Induced emf ∝ Φ), Φ ∝ I?]]]
      Which can prove the (greater the current , greater the magnetic field)?

      Delete
    3. It's logical.

      If there is a current in the wire, there is a magnetic field formed around it.
      So, a greater current results into a greater field strength.

      Delete
  3. WHY ISN'T THE emf graph starting from the max displacement in positive side of the graph::??? kindly explain the shape???

    ReplyDelete
    Replies
    1. as explained above, differentiating sine gives cos.

      But since there is a negative sign, the graph is inverted.
      SO, the graph is actually -cos

      Delete

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