6.20

·         6.20 recall and use the relationship (for 100% efficiency):

        input power = output power

                     V_p I_p = V_s I_s

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6.19 and 6.20 Plenary answers

01 December 2011

18:08

a.       V_s/V_p=n_s/n_p

V_s/2=80/20

V_s=8V

 

b.      V_s/V_p=n_s/n_p

V_s/10=20/100

V_s=2V

 

c.       V_s/V_p=n_s/n_p

V_s/240=20/400

V_s=12V

 

d.      V_s/V_p=N_s/N_p

V_s/4=2000/50

V_s=160V

 

2.      V_s/V_p=n_s/n_p

3/24=n_s/480

n_s=60turns

 

a.       Iron  (Why?  Iron is a magnetically soft material - it can be magnetised and lose its magnetism easily.  This is necessary in a transformer as the magnetic field needs to change repeatedly)

b.      electrical energy in the primary coil => magnetic energy in the core =>electrical energy in the secondary coil

c.       dc current produces a steady magnetic field in the core.  To induce a current in the secondary coils there must be a changing magnetic field in the core.

 

a.       Step down (because the voltage decreases)

 

b.      V_s/V_p=n_s/n_p

12/240=n_s/1000

n_s=50turns

 

c.       P = V x I

24 = 12 x I

I = 2A

 

d.      Power in secondary = 24W

 

e.       Power in primary = 24W

 

f. P = V x I

24 = 240 x I

I = 0.1A

 

g.      The current would be greater/double

 

Why?

Efficiency = Useful P_out/Total P_inx100

50 = 24/Total P_in x 100

Total P_in = 48W

 

P = V x I

48 = 240 x I

I = 0.2A

 

 

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