Question

Electricity is distributed from electrical substations to neighborhoods at 13000 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.
Required:
a. How many turns does the primary coil on the transformer have if the secondary coil has 130 turns?
b. No energy is lost in an ideal transformer, so the output power Pout from the secondary coil equals the input power Pin to the primary coil. Suppose a neighborhood transformer delivers 280 A at 120 V. What is the current in the 1.3×10^4 V line from the substation?

Answer 1

Answer:

a)  N₁ = 14083 turns,  b)   I₁ = 2.58 A

Explanation:

The relationship that describes the relationship between the primary and secondary of the transformer is

         $\frac{V_2}{N_2} = \frac{V_1}{N_1}$

a) They indicate that the secondary has N2 = 130 turns, the turns of the primary are

         N₁ = $N_2 \frac{V_1}{V_2}$

         N₁ = $130 \ \frac{13000}{120}$

         N₁ = 14083 turns

b) since there are no losses, the power of the neighboring transformer is

          P = V I

          P = 120 280

          P = 33600 W

this is the same power of the substation

          P = V₁  I₁

          I₁ = P / V₁

          I₁ = 33600/13000

          I₁ = 2.58 A