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°
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?
ReplyDeleteWe have not created a page specific for this yet.
DeleteBut 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.
Since the magnetic flux is inversely proportional to current. How the graph has same shape and phase as Ip
ReplyDeleteNote 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.
DeleteWe 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.
Thank you!
DeleteDo 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)?
It's logical.
DeleteIf 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.
WHY ISN'T THE emf graph starting from the max displacement in positive side of the graph::??? kindly explain the shape???
ReplyDeleteas explained above, differentiating sine gives cos.
DeleteBut since there is a negative sign, the graph is inverted.
SO, the graph is actually -cos