Hi:
A 6.0v alternating supply is connected to the 100-turn primary coil of a transformer. The 200-turn secondary coil runs a lamp which is dissipating 24watts. (Assume that the transformer is an ideal one, with no flux leakage, and a laminated iron core). Here are five values of currents:
A:4A B:2A C:1 D:0.5A E:0.25A

Please see uploaded image.

a) Which is the best estimate of the current in the secondary circuit?
Vp/Vs=Np/Ns
So,Vs=6x(100/200)=3v

P=IV, So Is=3v/24w=0.125A
Therefore C:1A is the best estimate.

b) Which is the best estimate of the current in the primary circuit?
Ip=Is x Ns/Np=(0.125 x 200) /100=0.25A
E:0.25A is the best estimate

c) What will happen to the lamp if the two halves of the transformer core are pulled apart slightly? Give a reason for your answer.

I don't know the answer to this. But if I were to hazard a guess, I would say that the induced emf would still travel through the core but at less value and the lamp would dim.

Would somebody be good enough to check these please

 

Your answers to a and b are wrong.

Vs=Vp*Ns/Np, try again.

Your guess for c is quite correct however, the emf does not flow through the core. It is the flux which flows through the core. If the core is pulled apart, the flux density will reduce much because air has a low permeability and thus the lamb will dim or stop working depending of the amount of flux.

 

Oh God! I'm so stupid and sloppy. Sorry mik3.

a) Vs=6*200/100=12v
P=IV, So Is=12v/24w=0.5A
Therefore D:0.5A is the best estimate.

b) Ip=Is x Ns/Np=(0.5 x 200) /100=1A
C:1A is the best estimate

 

I=P/V and not V/P! Try again!

 

ahh! I think I have short circuited - even basic algebra is suffering under the pressure.
Is=P/V=24/12=2A

 

ahh! I think I have short circuited - even basic algebra is suffering under the pressure.
Is=P/V=24/12=2A

Yes, now your are right.

 

phew: Almost there for Electronics for the day.
Thanks for your help in finishing this one with me mik3