00:01
Hello students in this question we are given here in the a part we are given that the primary of a transformer has twice as many turns as a secondary.
00:11
So we are given n1 is equal to 2 times n2 where n1 is the number of turns in primary coil and n2 is the number of turns in secondary coil and the primary voltage that is v1 is equal to 4 ,150 volts over a rate.
00:30
And a load of 170 .5 oms is connected across the secondary.
00:42
So we have to calculate the power delivered by the transformer as well as the primary and secondary currents.
00:48
So here we have v2 over v1 is equal to n2 over n1.
00:58
So from here we have v2 is v1 multiplied by n2 over n1.
01:03
N1 so from here v2 is v1 that is 4 ,150 multiplied by n2 and n2 over in place of n1 we can write 2n2 v2 will be 4 ,150 over 2 the water across the secondary coil will come out to be 20275 volts now we have to find the current also so current i2 will be equal to v2 over r2.
01:39
This is basically we are given r2 the load in the secondary coil.
01:43
So from here we have i2 is equal to v2 that is 2075 over r2 which is 170 .5.
01:53
So from here the value of i2 that is the secondary current is equal to 12 .17 ampers.
02:00
Now we can write that v1 i1 is equal to v2 i2 so from here we have i1 as v2 over v1 multiplied by i 2 so here i 1 is equal to v2 here that is 2705 over v1 that is v1 that is v1 was 4 ,000 here the v1 150 multiplied by i2 which is 12 .17 here i1 will come out to be 6 .085 amperors now we have found out the power we have found out the current in primary secondary and now the power delivered to the circuit by the transformer that is p is equal to v2 i2 so it will be here 2075 multiplied by 12 .170 so this will mean kilowatt.
03:08
So power will be 25 .252 kilowatt over there...