Example 21-2 A High-Voltage Transformer
Each of the 17 generators employed in the hydroelectric power plant at Hoover Dam in Colorado can generate up
to 133 MW of average power. This power is delivered at 8.00 kV rms to a transformer that connects to a long-distance
transmission line that operates at 5.00 * 102 kV rms. (a) What is the ratio of the number of windings in the secondary coil
of the transformer to the number in the primary coil? (b) What is the rms current in the high-voltage transmission line?
Set Up
In part (a) we’ll use Equation 21-10 to relate
the ratio Ns >Np (the number of windings in the
secondary coil divided by the number in the
primary coil) to the input and output rms voltages. In part (b) we’ll assume that the transmission line delivers its power to a resistor, so we
can relate the current, voltage, and power using
Equations 18-9 and 18-23.
Input and output voltages for
a transformer:
Vs,rms
Vs
Ns
=
=
Vp
Vp,rms
Np
Vp,rms = 8.00 kV
Vs,rms = 5.00 × 102 kV
(21-10)
Relationship among potential
difference, current, and
resistance:
(18-9)
V = iR
Power for a circuit element:
(18-23)
P = iV
Solve
(a) We know both the primary and secondary
rms voltages, so we can rearrange Equation
21-10 to solve for the ratio of the number of
windings.
From Equation 21-10,
Vs,rms
Ns
5.00 * 102 kV
=
=
= 62.5
Np
Vp,rms
8.00 kV
There are 62.5 times as many windings in the secondary coil as in
the primary coil (or, equivalently, 125 windings in the secondary for
every two windings of the primary).
(b) Since the resistance R is a constant, the current and voltage both have the same sin vt time
dependence. We use this to relate the average
power Paverage = 133 MW to the rms values of
the current and voltage in the transmission line.
If the voltage is V(t) = V0 sin vt, from Equation 18-9 the
current is
V1t2
i1t2 =
R
=
V0 sin vt
V0
=
sin vt = i 0 sin vt
R
R
In this expression i0 is the amplitude of the current. From
Equation 18-23 the instantaneous power is
P(t) = i(t)V(t) = (V0 sin vt)(i0 sin vt) = i0V0 sin2 vt
To find the average power Paverage, replace sin2 vt by its average
value 12 :
Paverage =
i0
V0
i 0 V0
ba
b = i rmsVrms
= a
2
22
22
The average power is the product of the rms current and the rms
voltage. Since i(t) has the same time dependence as V(t), the rms
value of current is equal to its maximum value divided by 12,
just as for the voltage. From this expression, the rms current in the
transmission line is
i s,rms =
Reflect
In the process of delivering power to the end
user, the voltage must ultimately be stepped
down to 120 V. This process is not normally
done with a single transformer. One reason is
that if only one transformer were used for this
stepping-down process, there would need to be
more than 4000 windings in the primary coil
for every one in the secondary.
Paverage
Vs,rms
=
133 MW
133 * 106 W
= 266 A
=
2
5.00 * 105 V
5.00 * 10 kV
Ratio of coil windings to step down 5.00 * 102 kV to 120 V:
Np
Ns
=
Vp
Vs
=
5.00 * 102 kV
= 4170
120 V
We found a current of 266 A in the transmission line. Although that sounds large, it is only
about a factor of 10 more than the current in
household wiring. In addition, the power loss
in the transmission wire is relatively low at this
current. A few hundred kilometers of transmission wire might have a resistance of 25 . The
loss in such a line would be about 1.8 MW.
Although in absolute terms this is a significant
power loss, it represents only 1.4% of the total
power transmitted.
The voltage between the two ends of the transmission line is
Vends = iR, where R = 25  is the resistance of the line. Use the
formula we derived in part (b) to find the power lost in the
transmission line:
Plost,average = i rmsVends,rms = i rms 1i rmsR2
= i 2rmsR = 1266 A2 2 125 2 = 1.8 * 106 W = 1.8 MW
Express this as a percentage of the power that is transmitted:
power lost
* 100%
power transmitted
1.8 MW
=
* 100% = 1.4%
133 MW
percentage =