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

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) (104 × 82/25) V (B) (104 × 8/25) V (C) (104/25 × 82) V (D) (104/25 × 8) V

Tardigrade Admin · Accepted Answer

VP IP=VS IS=10000 W VS(25 A )=10000 W ⇒ VS=(104/25) (VP/VS)=(NP/NS)=(8/1) Here VP=(8 × 104/25) V

Q. 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

$\frac{1 0 ^{4} \times 8 ^{2}}{25} V$

$\frac{1 0 ^{4} \times 8}{25} V$

$\frac{1 0 ^{4}}{25 \times 8 ^{2}} V$

$\frac{1 0 ^{4}}{25 \times 8} V$

$V_{P} I_{P} = V_{S} I_{S} = 10000 W$
$V_{S} (25 A) = 10000 W \Rightarrow V_{S} = \frac{1 0 ^{4}}{25}$
$\frac{V _{P}}{V _{S}} = \frac{N _{P}}{N _{S}} = \frac{8}{1}$
Here $V_{P} = \frac{8 \times 1 0 ^{4}}{25} V$

Q. An ideally efficient transformer has a primary power input of 10kW. 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

25104×82​V

25104×8​V

25×82104​V

25×8104​V