JEE Main 10 Apr 2019 Paper 1 Solved Paper

Number of turns in the primary side Np=300
Number of turns in the secondary side Ns=150
The output power or the secondary side power is Po=2.2kW=2200W
The current in the secondary coil is Is=10A
We know that Ohm's law is given by the formula
PO=VOIO−−(i)
Hence by using Ohm's law, we can calculate the output voltage of the transformer
VO=‌

PO

IO

⟹VO=‌

2200

10

∴VO=220V
Hence the secondary voltage or the output voltage of the transformer is =220V
Now we know that the ratio of the number of turns in the secondary coil to the number of turns in the primary coil is equal to the ratio of the output voltage to the input voltage, hence we can write n=‌

Ns

NP

=‌

Vo

Vi

Now we substitute the values
‌

Ns

Np

=‌

V0

Vi

‌

150

300

=‌

220

Vi

Vi=‌

300

150

×220
=220×2
=440V
Hence the input voltage or the primary voltage of the transformer is =440V
We also know that the ratio n is equal to the ratio of current in the secondary side to the current on the primary side, so we can write
n=‌

Ns

Np

=‌

Io

Ii

Hence by substituting the values
‌

Ns

NP

=‌

Io

Ii

‌

150

300

=‌

10

Ii

⟹Ii=‌

300

150

×10
=20A
So, the current in the primary coil =20A