JEE MAIN - Physics (2002 - No. 14)
In a transformer, number of turns in the primary coil are $$140$$ and that in the secondary coil are $$280.$$ If current in primary coil is $$4A,$$ then that in the secondary coil is
$$4A$$
$$2A$$
$$6A$$
$$10A$$
ব্যাখ্যা
$${N_p} = 140,\,\,{N_s} = 280,\,\,{I_p} = 4A,\,\,{I_s} = ?$$
For a transformer $${{{I_s}} \over {{I_p}}} = {{{N_p}} \over {{N_s}}}$$
$$ \Rightarrow {{{I_s}} \over 4} = {{140} \over {280}} \Rightarrow {I_s} = 2A$$
For a transformer $${{{I_s}} \over {{I_p}}} = {{{N_p}} \over {{N_s}}}$$
$$ \Rightarrow {{{I_s}} \over 4} = {{140} \over {280}} \Rightarrow {I_s} = 2A$$