A power transformer is used to step up an alternating e.m.f. of 220 V to 11 kV to transmit 4.4 kW of power. If the primary coil has 1000 turns, what is the current rating of the secondary? (Assume 100 % efficiency for the transformer)

Question

A power transformer is used to step up an alternating e.m.f. of 220 V to 11 kV to transmit 4.4 kW of power. If the primary coil has 1000 turns, what is the current rating of the secondary? (Assume 100 % efficiency for the transformer) (A) 4 A (B) 0.4 A (C) 0.04 A (D) 0.2 A

Tardigrade Admin · Accepted Answer

is=(Ps/Vs)=(4.4 × 103/11 × 103)=0.4 A

Q. A power transformer is used to step up an alternating e.m.f. of $220 V$ to $11 kV$ to transmit $4.4 kW$ of power. If the primary coil has $1000$ turns, what is the current rating of the secondary? (Assume $100%$ efficiency for the transformer)

4 A

0.4 A

0.04 A

0.2 A

$i_{s} = \frac{P _{s}}{V _{s}} = \frac{4.4 \times 1 0 ^{3}}{11 \times 1 0 ^{3}} = 0.4 A$

Q. A power transformer is used to step up an alternating e.m.f. of 220V to 11kV to transmit 4.4kW of power. If the primary coil has 1000 turns, what is the current rating of the secondary? (Assume 100% efficiency for the transformer)

4 A

0.4 A

0.04 A

0.2 A

is​=Vs​Ps​​=11×1034.4×103​=0.4A

Q. A power transformer is used to step up an alternating e.m.f. of $220 V$ to $11 kV$ to transmit $4.4 kW$ of power. If the primary coil has $1000$ turns, what is the current rating of the secondary? (Assume $100%$ efficiency for the transformer)

$i_{s} = \frac{P _{s}}{V _{s}} = \frac{4.4 \times 1 0 ^{3}}{11 \times 1 0 ^{3}} = 0.4 A$