Question

A transformer has 500 primary turns and 10
secondary turns. If the secondary has resistive
load of 15 ohm ,the currents in the primary and
secondary respectively are

Answer 1

Answer:

0.32 and 0.006 ampere.

Explanation:

If we take the voltage of the primary coil and the secondary coil to be Vs and Vp respectively and the current and the number of turns in the secondary and primary coil be Is and Ip with Ns and Np respectively. So, Vs = (Ns/Np)Vp when Vs = Is*R = 15*Is. So, 15*Is = 10/500 * Vp. Taking the Vp as 240V we will get that Is=0.32A.

While we also know that the Is = (Np/Ns)Ip or 0.32 = 500/10*Ip which on solving we will get the value of Ip as 0.006A.  

Answer 2

Answer:

According to the problem ,

the number of turns in the primary n1= 500 and number of turns in secondary , n2 = 10

Therefore we know , n2/n1 = i1/i2 = 10/500 = i1/i2 or, i2 = 50 i1

Now as the secondary has 15 ohm of resistive load,

let the Volt is primary is vp therefore the volt in secondary is

vs = n2/n1 x vp = 1/50 vp

Therefore current in secondary ,

i2 = vs/r = 1/50vp/15 = vp/750 mA

i1 = vs/vp x i2 = 1/50 vp/vp x  vp/750 = vp/50 x 750 mA