CK-12-Physics - Intermediate

http://www.ck12.org Chapter 19. Electromagnetism

Vp

Vs=

Np

Ns→

12 , 000

300 , 000 =

100

Ns →Ns=

30 , 000 , 000
12 , 000 =^2 ,^500 →Ns=^2 ,^500 turns
b. If the current in the primary coil is 1,000 A, what is the current in the secondary coil?

Answer:

Since the output voltage is 25 times larger than the input voltage, the output current must be 25 times smaller than
the input current.

IpVp=IsVs→

Vs

Vp

=

Ip

Is

→ 25 =

1 , 000 A

Is

→Is= 40 A

Power stations actually use step-up transformers at the beginning of the power transfer process (as in the “Check
Your Understanding” above). A high-voltage transmission line then transfers the energy to local substations where
the voltage is reduced using the step-down transformers. From there, the energy is stepped-down using transformers
which can easily be seen on utility poles, such as the ones you may see near your house. SeeFigure19.11.

FIGURE 19.11

A typical residential transformer.

Figure19.12 shows a common AC adaptor (a transformer). Notice that the input is given as 120 VAC 60 Hz and the
input power as 8 W. The output is given as 5 VDC 500 mA. This transformer accomplishes two important tasks:

1. It converts an alternating voltage and current into a direct voltage and current.


2. It steps-down the voltage.

Check Your Understanding

a. What is the output power of the transformer?

Answer: The voltage isV= 7. 5 Vand the current isI= 200 mA= 200 × 10 −^3 A= 0. 200 A. Thus,P=IV=
( 0. 200 A)( 7. 5 V) = 1. 5 W

b. The input power is given as 6 W. According to the specifications on the transformer, there is four times more
power input than output. What has happened to the remaining energy?