JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 8)
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be ;
50 A
45 A
25 A
20 A
Explanation
Efficiency n = 0.9 = $${{{P_s}} \over {{P_p}}}$$
as $$\,\,\,\,\,\,$$ P = VI
$$\therefore\,\,\,$$ Ps = 0.9 $$ \times $$ Pp
$$ \Rightarrow $$ $$\,\,\,\,$$ Vs Is = 0.9 $$ \times $$ Vp Ip
$$ \Rightarrow $$ $$\,\,\,\,$$ Is = $${{0.9 \times 2300 \times 5} \over {230}}$$
= $$\,\,\,\,$$ 45 A
as $$\,\,\,\,\,\,$$ P = VI
$$\therefore\,\,\,$$ Ps = 0.9 $$ \times $$ Pp
$$ \Rightarrow $$ $$\,\,\,\,$$ Vs Is = 0.9 $$ \times $$ Vp Ip
$$ \Rightarrow $$ $$\,\,\,\,$$ Is = $${{0.9 \times 2300 \times 5} \over {230}}$$
= $$\,\,\,\,$$ 45 A
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