Question

A step-up transformer has 22 turns on the primary coil and 800 turns on the secondary coil. If this transformer is to produce an output of 5300 V with a 16- mA current, what input current and voltage are needed?

Answer 1

`I_p=0.582\ A`

`V_p=145.75\ V`

Step-by-step explanation:

Given:

No. of turns in the primary coil,
`N_p=22`

No. of turns in the secondary coil,
`N_s=800`

output voltage of the transformer,
`V_s=5300\ V`

output current of the transformer,
`I_s=0.016\ A`

from the equation of transformer we have:

`I_s.N_s=I_p.N_p`

where:

`I_p` = primary current on the input side

`0.016* 800=I_p* 22`

`I_p=0.582\ A`

Also

`(V_s)/(N_s) =(V_p)/(N_p)`

where:

`V_p=` input voltage on the primary coil

`(5300)/(800) =(V_p)/(22)`

`V_p=145.75\ V`

Answer 2

581.82 mA

145.75 V

Step-by-step explanation:

N = Number of turns

I = Current

V = Voltage

p denotes primary (input)

s denotes secondary (output)

In a transformer the following relation is applied

`(I_p)/(I_s)=(N_s)/(N_p)\\\Rightarrow I_p=I_s(N_s)/(N_p)=16(800)/(22)\\\Rightarrow I_p=581.82\ mA`

The input current is 581.82 mA

`(V_p)/(V_s)=(N_p)/(N_s)\\\Rightarrow V_p=V_s(N_p)/(N_s)=5300(22)/(800)\\\Rightarrow V_p=145.75\ V`

The input voltage is 145.75 V