JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 24)
A power transmission line feeds input power at 2300 V to a srep down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :
50 A
45 A
35 A
25 A
Explanation
Given,
Primary voltage (VP) = 2300 V
Primary current (IP) = 5A
Secondary voltage (VS) = 230 V
efficiency ($$\eta $$) = 90%
We know,
Efficiency ($$\eta $$) = $${{{\mathop{\rm Sec}\nolimits} ondary\,Power} \over {\Pr imary\,\,Power}}$$
$$ \therefore $$ $$\eta $$ = $${{{P_S}} \over {{P_P}}}$$ = $${{{V_S}\,{I_S}} \over {{V_P}\,{I_P}}}$$
$$ \Rightarrow $$ 0.9 = $${{230 \times {{\rm I}_S}} \over {2300 \times 5}}$$
$$ \Rightarrow $$ I = 45 A
Primary voltage (VP) = 2300 V
Primary current (IP) = 5A
Secondary voltage (VS) = 230 V
efficiency ($$\eta $$) = 90%
We know,
Efficiency ($$\eta $$) = $${{{\mathop{\rm Sec}\nolimits} ondary\,Power} \over {\Pr imary\,\,Power}}$$
$$ \therefore $$ $$\eta $$ = $${{{P_S}} \over {{P_P}}}$$ = $${{{V_S}\,{I_S}} \over {{V_P}\,{I_P}}}$$
$$ \Rightarrow $$ 0.9 = $${{230 \times {{\rm I}_S}} \over {2300 \times 5}}$$
$$ \Rightarrow $$ I = 45 A
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