Question

An ideally efficient transformer has a primary power input of $10\, kW$. The secondary current when the transformer is on load is $25$ ampere. If the primary: secondary turns ratio is $8: 1,$ then the potential difference applied to the primary coil is

• A. $\frac{10^{4} \times 8^{2}}{25} V$
• B. $\frac{10^{4} \times 8}{25} V$
• C. $\frac{10^{4}}{25 \times 8^{2}} V$
• D. $\frac{10^{4}}{25 \times 8} V$

Answer 1

Answer: (B)

$ V_{P} I_{P}=V_{S} I_{S}=10000 W$
$V_{S}(25 A )=10000 W \Rightarrow V_{S}=\frac{10^{4}}{25}$
$\frac{V_{P}}{V_{S}}=\frac{N_{P}}{N_{S}}=\frac{8}{1}$
Here $V_{P}=\frac{8 \times 10^{4}}{25} V$