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

Answer:

(A) 18667 turns

(B) 1.7 A

Solution:

As per the question:

Voltage at which the electricity is distributed, $V_{p} = 1.6\times 10^{4}\ Hz$

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:

$\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}$

$N_{p} = \frac{V_{p}}{V_{s}}\times N_{s}$

$N_{p} = \frac{1.6\times 10^{4}}{120}\times 140 = 18,667\ turns$

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

$\frac{V_{p}}{V_{s}} = \frac{I_{s}}{I_{p}}$

$I_{p} = \frac{V_{s}}{V_{p}} \times {I_{s}}$

$I_{p} = \frac{120}{1.6\times 10^{4}} \times 230 = 1.7\ A$