Alternating Current Question 3
Question 3 - 2024 (01 Feb Shift 2)
A transformer has an efficiency of $80 %$ and works at $10 \mathrm{~V}$ and $4 \mathrm{~kW}$. If the secondary voltage is $240 \mathrm{~V}$, then the current in the secondary coil is :
(1) $1.59 \mathrm{~A}$
(2) $13.33 \mathrm{~A}$
(3) $1.33 \mathrm{~A}$
(4) $15.1 \mathrm{~A}$
Show Answer
Answer: (2)
Solution:
Efficiency $=\frac{\mathrm{E}{\mathrm{s}} \mathrm{I}{\mathrm{s}}}{\mathrm{E}{\mathrm{p}} \mathrm{I}{\mathrm{p}}}$
$0.8=\frac{240 \mathrm{I}_{\mathrm{s}}}{4000}$
$\mathrm{I}_{\mathrm{S}}=\frac{3200}{240}=13.33 \mathrm{~A}$
