A transformer is to be used to provide power for a computer drive. The number of

Question

A transformer is to be used to provide power for a computer drive. The number of turns in the primary is 1000, and it delivers a secondary current of 600 milliamperes (mA) at a secondary voltage of 8.0 V for the drive. A voltage of 400 volts is applied across the primary.

Part A
What kind of transformer is this?
A . AC
B . Load
C. Regulated
D. Step-up
E. Step-down

Part B
Determine the number of turns in the secondary of the transformer

Part C
Determine the primary current in millamps of the transformer

Part D
A transformer is said to be ideal if the power developed across the primary equals the power developed across the secondary. Assuming the transformer used in Question 4 is ideal, find the power (P=IV) in watts developed across either the primary or secondary.

Explanation / Answer

A)Vp/Vs = 400/8 = 50

Since Vp/Vs = 50 > 1

Its a step down transformer.

B)Np/Ns = Vp/Vs

Ns = Np (Vs/Vp)

Ns = 1000 (8/400) = 20

Hence, Ns = 20

C)Ns/Np = Ip/Is

Ip = Is (Ns/Np)

Ip = 600 mA (20/1000) = 12 mA

Hence, Ip = 12 mA

D)P = VI

Lets calculate P for primary,

P = 400 x 12 x 10^-3 = 4.8 W

Hence, P = 4.8 W

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