Question

Electricity is distributed from electrical substations to neighborhoods at 15,000 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. a. How many turns does the primary coil on the transformer have if the secondary coil has 100 turns? b. No energy is lost in an ideal transformer, so the output power P out from the secondary coil equals the input power Pin to the primary coil. Suppose a neighborhood transformer delivers 250 A at 120 V. What is the current in the 15,000 V line from the substation?

Answer 1

a. 12,500 turns

The transformer equation states that

$\frac{N_p}{V_p}=\frac{N_s}{V_s}$

where

Np is the number of turns in the primary coil

Ns is the number of turns in the secondary coil

Vp is the voltage in the primary coil

Vs is the voltage in the secondary coil

For the transformer in the problem,

Vp = 15,000 V

Vs = 120 V

Ns = 100

So we can find Np by rearranging the equation:

$N_p = V_p \frac{N_s}{V_s}=(15,000 V)\frac{100}{120 V}=12,500$

b.  2 A

For an ideal transformer, the output power is equal to the input power:

$P_i = P_o\\V_p I_p = V_s I_s$

where

$V_p = 15,000 V$ is the voltage in the primary coil

$I_p$ is the current in the primary coil

$V_s = 120 V$ is the voltage in the secondary coil

$I_s = 250 A$ is the current in the secondary coil

Solvign the formula for Ip, we find:

$I_p = \frac{V_s I_s}{V_p}=\frac{(120 V)(250 A)}{15,000 V}=2 A$