Question

A transformer has $100$ turns in the primary coil and carries $8 \, A$ current. If the input power is $1 \, kW,$ the number of turns in secondary coil to have $500 \, V$ of output will be

• A. $100$
• B. $200$
• C. $400$
• D. $300$

Answer 1

Answer: (C)

Given that,
$N_{p}=100, \, I_{p}=8 \, A$
$V_{s}=500 \, V, \, P_{i}=1 \, kW=100 \, W$
Input power, $P_{i}=V_{P}I_{P}$
$1000=V_{p}\times 8$
$V_{P}=125 \, V$
Again, we know that
$\frac{N_{s}}{N_{P}}=\frac{V_{s}}{V_{p}}$
$\Rightarrow \, \, \frac{N_{s}}{100}=\frac{500}{125}$
$N_{s}=\frac{500 \times 100}{125}$
$=400$