A step down transformer increases the input current 4 A to 24 A at the secondary. If the number of turns in the primary coil is 330, the number of turns in the secondary coil is

Question

A step down transformer increases the input current 4 A to 24 A at the secondary. If the number of turns in the primary coil is 330, the number of turns in the secondary coil is (A) 60 (B) 50 (C) 65 (D) 45

Tardigrade Admin · Accepted Answer

By step-down transformer, It converts a high voltage of low current into a low voltage of high current. (NS/NP) =(IP/IS) (NS/330) =(4/24) NS =(4 × 330/24)=(330/6)=55

Q. A step down transformer increases the input current $4 A$ to $24 A$ at the secondary. If the number of turns in the primary coil is $330$ , the number of turns in the secondary coil is

60

50

65

45

55

By step-down transformer,
It converts a high voltage of low current into a low voltage of high current.
$\frac{N _{S}}{N _{P}} = \frac{I _{P}}{I _{S}}$
$\frac{N _{S}}{330} = \frac{4}{24}$
$N_{S} = \frac{4 \times 330}{24} = \frac{330}{6} = 55$

Q. A step down transformer increases the input current 4A to 24A at the secondary. If the number of turns in the primary coil is 330, the number of turns in the secondary coil is

60

50

65

45

55

By step-down transformer, It converts a high voltage of low current into a low voltage of high current. NP​NS​​=IS​IP​​ 330NS​​=244​ NS​=244×330​=6330​=55

Q. A step down transformer increases the input current $4 A$ to $24 A$ at the secondary. If the number of turns in the primary coil is $330$ , the number of turns in the secondary coil is

By step-down transformer,
It converts a high voltage of low current into a low voltage of high current.
$\frac{N _{S}}{N _{P}} = \frac{I _{P}}{I _{S}}$
$\frac{N _{S}}{330} = \frac{4}{24}$
$N_{S} = \frac{4 \times 330}{24} = \frac{330}{6} = 55$