Question

Electricity is distributed from electrical substations to neighborhoods at 1.6×104 V . This is a 60 Hz oscillating (AC) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the 120 V that is delivered to your house. Part A How many turns does the primary coil on the transformer have if the secondary coil has 140 turns? Express your answer using two significant figures. n prim n p r i m = nothing turns SubmitRequest Answer Part B No energy is lost in an ideal transformer, so the output power P out from the secondary coil equals the input power P in to the primary coil. Suppose a neighborhood transformer delivers 230 A at 120 V . What is the current in the 1.6×104 V line from the substation? Express your answer using two significant figures. I i I i = nothing A SubmitRequest Answer Provide Feedback Next

Answer 1

(A) 18667 turns

(B) 1.7 A

Solution:

As per the question:

Voltage at which the electricity is distributed,
`V_(p) = 1.6* 10^(4)\ Hz`

Frequency of the oscillating voltage, f = 60 Hz

Step down voltage,
`V_(s) = 120\ V`

No. of turns in the secondary coil,
`N_(s) = 140\ turns`

Current in the secondary coil,
`I_(s) = 230\ A`

Now,

(A) To calculate the primary no. of turns, we use the relation:

`(V_(s))/(V_(p)) = (N_(s))/(N_(p))`

`N_(p) = (V_(p))/(V_(s))* N_(s)`

`N_(p) = (1.6* 10^(4))/(120)* 140 = 18,667\ turns`

(B) To calculate the current in the primary coil,
`I_(p)`, we use the relation:

`(V_(p))/(V_(s)) = (I_(s))/(I_(p))`

`I_(p) = (V_(s))/(V_(p)) * {I_(s)}`

`I_(p) = (120)/(1.6* 10^(4)) * 230 = 1.7\ A`