http://iml.umkc.edu/physics/wrobel/phy250/homework.htm
Homework 8
chapter 33: 23, 33, 43
Problem 33.23
A person is working mear the secondary coil of a transformer as shown in Figure
33.25. The primary voltage is 120 V at 60 Hz. The capacitance Cs, which is the
stray capacitance between the hand and the secondary winding, is 20 pF. Assuming
the person has a body resistance to ground Rb = 50 kΩ, determine the rms voltage
across the body.
5,000 V
The secondary circuit consists of a “capacitor” with
capacitance Cs=20μF, a “resistor” with resistance Rb=50kΩ
and a source of electromotive force (the secondary coil of the
transformer). The source produces sinusoidal potential
difference with rms value of 5000 V and frequency of 60 Hz. At
this frequency the impedance of the “capacitor” is
ZC =
1
2πf ⋅ Cs
and the impedance of the “resistor” is equal to its resistance
ZR = R b .
Since the “elements” are connected in series, the equivalent
resistance of the system (of the two elements) is
Z = R 2b +
1
2πf ⋅ Cs
2
The rms current in the “resistor” (and the circuit) is therefore
I rms =
Vrms
=
Z
Vrms
1
R 2b +
2πf ⋅ Cs
.
From Ohm’s law, the rms voltage across the body (resistor) is
Vb, rms = I b, rms R b = R b
=
Vrms
⎛ 1 ⎞
⎟⎟
R 2b + ⎜⎜
⎝ 2πf ⋅ Cs ⎠
Vrms
⎛
⎞
1
⎟⎟
1 + ⎜⎜
2
f
C
R
π
⋅
⋅
⎝
s
b ⎠
2
=
= 1.88V
3
2
=
5kV
⎞
⎛
1
1 + ⎜⎜
⎟⎟
2
60
Hz
20
pF
50
k
π
⋅
⋅
⋅
Ω
⎠
⎝
2
Problem 33.33
diode
A diode is a device that allows current to
pass in only one direction (indicated by the
arrowhead). Find, in terms of Vrms and R,
the average power dissipated in the diode
circuit shown in Figure P33.35.
2R
R
R
R
diode
V
When the left side of the
diode
generator has a higher
2R
potential, the top diode
R
conducts (forward bias), and
the bottom diode does not
R
(reverse bias). The remaining
circuit of three resistors acts
like a single resistor. The
+
resistors R are connected in
V
series and the pair is connected
in parallel to the resistor 2R. Therefore the equivalent resistance
of the system is
1)
R +eq
1
1 ⎞
= ⎛⎜
+
⎟
⎝ 2R R + R ⎠
−1
=R
Knowing the rms value of the voltage we could find the
power in this circuit if the diode was not present. However, the
presence of the diode reduces the power to half its value
4
(because only half of the time current would flow in the circuit).
Therefore
2
1 Vrms
+
2) P = ⋅
2 R
2R
When the polarity of the emf is
R
reversed, the top diode is in
R
R
reverse bias and the bottom
diode
diode is in forward bias. This
time the equivalent resistance of
+
the system of resistors is
V
different
3)
−1
1
1 ⎞
7
⎛
−
R = R+⎜ +
⎟ = R
⎝ R R + 2R ⎠
4
With the same arguments as before, the power dissipated at this
polarity is
2
1 Vrms
−
4) P = ⋅
. R
2 175
The overall (time) average power is
2
2
Vrms
⎛ 4⎞ 11 Vrms
+
−
P= P +P =
⋅ ⎜1 + ⎟ = ⋅
2R ⎝ 7⎠ 14 R
5
Problem 33.43
A transmission line that has a resistance per unit length of 4.5⋅10-4 Ω/m is to
be used to transmit 5 MW over 400 miles (6.44⋅105 m). The output voltage of
the generator is 4.5 kV. (a) What is the line loss if a transformer is used to step
up the voltage to 500 kV? (b) What fraction of the input power is lost to the
line under these circumstances? (c) What difficulties would be encountered in
attempting to transmit the 5 MW at the generator voltage of 4.5 kV
500 kV
or
4.5 kV
λ = 4.5⋅10-4 Ω/m
5 MW
L = 6.44⋅105 m
a) In order to send out power of P = 5MW at 500 kV potential
difference, the current in the grid (transmission lines) must be
I rms, a =
P
Vrms, a
= 10A
The resistance of the two wires of the transmission lines
(connected is series) is
R = 2 ⋅ L ⋅ λ = 580Ω
Hence the loss of power in the lines (dissipated in the lines) is
Pa = I 2rms ⋅ R = (10A ) 580Ω = 58kW
2
b) Relative small fraction is lost in the lines
Pa 58kW
=
= 0.01 = 1%
P 5mW
6
c) If the power were send at 4.5 kV potential difference, the
current in the grid (transmission lines) should be much larger
I rms, c =
P
Vrms, c
=
5MW
= 1.1kA
4.5kV
But even if nothing were connected at the end of the lines, with
the potential difference of 4.5 kV, the current in the lines would
be
I rms, max =
Vrms, c 4.5kV
=
= 77A
58Ω
R
It would be impossible to send 5 MW of power into the lines.
7