In a transformer, the number of turns in the primary coil are 140 and that in the secondary coil are 280. If the current in primary coil is 4 A, then the current in the secondary coil is:

Q. In a noiseless transformer an alternating current of

2 A

is flowing in the primary coil. The number of turns on the primary and secondary coils are

100

and

20

respectively. The value of the current in the secondary coil is

Join BYJU'S Learning Program

In a transformer the number of turns of primary coil and secondary coil are 2.5 and 2 respectively and 140 V is applied on primary coil, then find the ratio of current in primary and secondary coils :

The correct option is A Ip:Is=4:5NpNs=2.52=VpVsBut, VpVs=IsIp∴IsIp=NpNs=2.52(or) Ip:Is=4:5

The correct option is A $I_{p} : I_{s} = 4 : 5$
$\frac{N_{p}}{N_{s}} = \frac{2.5}{2} = \frac{V_{p}}{V_{s}}$
But, $\frac{V_{p}}{V_{s}} = \frac{I_{s}}{I_{p}}$
$∴ \frac{I_{s}}{I_{p}} = \frac{N_{p}}{N_{s}} = \frac{2.5}{2}$
(or) $I_{p} : I_{s} = 4 : 5$