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:

The correct option is B 2 A

Given: Number of turns in primary coil ($N_p$) = 140
Number of turns in secondary coil ($N_s$) = 280
Current in primary coil ($i_p$) = 4 A

Let, current in secondary coil = ($i_s$)
Voltage in primary coil = ($V_p$)
Voltage in secondary coil = ($V_s$)
We know that, in a transformer
$\frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{i_s}{i_p}$

$\therefore$ $\frac{N_p}{N_s}=\frac{i_s}{i_p}\Rightarrow i_s=4\times \frac{140}{280}=2$ A