FIG 2a ACME POWER TRANSFORMERS Model No DS3412T 240 V 50 Hz

FIG. 2(a) ACME POWER TRANSFORMERS Model No. DS-34-12T 240 V 50 Hz 28 VA T130 FIG. 2(b)

Solution

i) Ans It is shell type transformer where core surrounds the winding,as we can see from image both primary and secondary windings wound on a central limb,the completes flux path through outer limbs.

As the transformer is stepdown from 240 to 12 V with VA rating of 28 VA gives primary current Ip=28/240=0.117 A

and secondary current Is=28/12=2.33 A.As the current rating od secondary winding is 2.33 A it needs largers dia,meter than primary

ii) Ans T130 means if temperature rise is above 130 deg then the insulation of transformer cannot withstand and it leads to failure of the transformer

iii) Rp=125 ohms and Rs=0.7 ohms

copper loss =(Ip^2)*Rp +(Is^2)*Rs=5.51 W by substituting Ip and Is calculated above

b)

i) Ans N1:N2=1:5 Vp=400 V and Is=100 A given

Vs=(N2/N1)*Vp=5*400=2000 V; Vs=2000 V

RL=Vs/Is=2000/100=20 ohms; RL=20 ohms

RL seen from primary RL\'=RL*(N1/N2)^2=20/5^2=20/25

RL\'=0.8 ohms

To get Maximum power transfer Rs=RL\'=0.8 ohms

so source resistance Rs=0.8 ohms

ii) losses =4 kW=4000 W

Output= Vs*Is=2000*100=20 kW

Input=Output+losses=20 kW+4 kW=24 kW

Efficiency =output/Input=20/24=0.8333=83.3%