Chapter 19 Solutions 19.2. Set Up:
Solve:
*19.3. Set Up: The number of ions that enter gives the charge that enters the axon in the specified time.
Solve:
19.7. Set Up:
From Table 19.1, copper has
Solve:
*19.13. Set Up:
The volume of the wire remains the same when it is stretched.
so Solve:
Reflect: When the length increases the resistance increases, and when the area decreases the resistance
increases.
*19.15. Set Up: First use Ohm’s law to find the resistance at 20.0°C; then calculate the resistivity from the
resistance. Finally use the dependence of resistance on temperature to calculate the temperature coefficient of
resistance. Ohm’s law is R = V/I, R = ρL/A, R = R0[1 + α (T – T0)], and the radius is one-half the diameter.
Solve: (a) At 20.0°C, R = V/I = (15.0 V)/(18.5 A) = 0.811 Ω. Using R = ρL/A and solving for ρ gives ρ = RA/L =
Rπ(D/2)2/L = (0.811 Ω)π[(0.00500 m)/2]2/(1.50 m) = 1.06 ×
Ω m.
(b) At 92.0°C, R = V/I = (15.0 V)/(17.2 A) = 0.872 Ω. Using R = R0[1 + α (T – T0)] with T0 taken as 20.0°C, we
have 0.872 Ω = (0.811 Ω)[1 + α (92.0°C – 20.0°C)]. This gives α = 0.00105
Reflect: The results are typical of ordinary metals.
*19.27. Set Up: When the switch is open there is no current and the terminal voltage of the battery equals its
emf,
When the switch is closed, current I flows and the terminal voltage V of the battery is
current I is the same at all points of the circuit.
Solve:
with
and
The
so
and
Reflect: When current flows through the battery, the terminal voltage is less than the emf because of the voltage
across the internal resistance.
19.36. Set Up: Power is energy per unit time. The power delivered by a voltage source is
Solve: (a) (b) 19.37. Set Up:
Solve: (a)
(b) P increases when R decreases, so the 100 W bulb has less resistance.
Reflect:
The bulb with smaller R draws more current and this more than compensates for smaller R in
*19.39. Set Up:
and energy is the product of power and time.
Solve:
Reflect: The energy delivered depends not only on the voltage and current but also on the length of the pulse.
*19.49. Set Up: For a parallel connection the full line voltage appears across each resistor. The equivalent
resistance is
and
Solve: (a) (b) (c) For the
resistor,
For the
resistor,
*19.55. Set Up: We need to reduce the given resistance without removing it from the circuit. We can do this
by soldering a second resistor in parallel with the incorrect one. The equivalent resistance for resistors in parallel
is
Solve: We wish for the equivalent resistance to equal
Thus, the unknown resistance is given by
This simplifies to
Reflect: There would probably be no easy way to solve this problem if we had soldered a resistor that was
smaller than the desired value into the circuit.
19.60. Set Up: The circuit diagram is given in the figure below. The junction rule has been used to find the
magnitude and direction of the current in the upper branch of the circuit. There are no remaining unknown
currents.
Solve: (a) The junction rule gives that the current in R is 2.00 A to the left.
(b) The loop rule applied to loop (1) gives:
(c) The loop rule applied to loop (2) gives:
*19.61. Set Up: We can use the power consumption in the
resistor to find the current through it. The
circuit, unknown currents, and a circuit loop are all shown in the figure below.
Solve:
so
The loop rule for loop (1) gives
The ammeter reads 0.714 A.
19.67. Set Up: The capacitor discharges exponentially through the voltmeter. Since the potential difference
across the capacitor is directly proportional to the charge on the plates, the voltage across the plates decreases
exponentially with the same time constant as the charge. The reading of the voltmeter obeys the equation
where RC is the time constant.
Solve: (a) Solving for C and evaluating the result when t = 4.00 s gives
(b) Reflect: In most laboratory circuits, time constants are much shorter than this one.
19.68. Set Up: The time constant is
Solve: (a)
gives
required.
(b)
so
so
and
Three time constants are
*19.69. Set Up:
After a long time the current has dropped to zero, there is no potential drop across
the resistor and the full battery voltage is applied to the capacitor network. Just after the switch is closed the
charge and voltage across each capacitor is zero and the battery voltage equals the voltage drop across the
resistor. The time constant is
where C is the equivalent capacitance of the network.
Solve: (a) The 20.0 pF, 30.0 pF, and 40.0 pF capacitors are in series and their equivalent capacitance
is
given by
After a long time the voltage across the 10.0 pF capacitor is 50.0 V. The charge on this capacitor
is
The voltage across
is 50.0 V and
This is the charge on each of the capacitors in series, so
(b) Note that
(c) as it should.
(d) The 10.0 pF capacitor and
are in parallel so their equivalent is
Reflect: The charges on the capacitors start at zero and increase to their final values. The current starts at its
maximum value and then decays to zero.
19.74. Set Up: We know that the power used by the refrigerator is given by
and the corresponding
energy used by the refrigerator when operated for a time t is
Solve: When the refrigerator is operating the power used is
which is
In a 30-day month there are
Since the refrigerator operates 50% of the
time, the refrigerator will be in use for
over the 30-day period. Thus, the amount of energy used during
the month is
The total cost to run the refrigerator for the month is
Reflect: Since most household appliances operate at 120 V, the cost of operation of a given appliance is roughly
proportional to its current draw and the percent of time it is in use. For example, an appliance that uses
of
current and operates 5% of the time during the month would cost roughly
per month to
operate.
19.86. Set Up: For resistors in series, the currents are the same and the voltages add.
resistors in parallel, the currents add and the voltages are the same.
For
Solve: (a) Replacing series and parallel combinations of resistors by their equivalents gives the equivalent
networks as shown in the figure below. The equivalent resistance of the network is
(b) The voltage across the
resistor in Figure (a) above is
This is the voltage
across the
resistor in Figure (b) above. The current through the
resistor is 4.8 A. This is also the
current through the
resistor and the voltage across that resistor is 96.0 V. Therefore, the voltage between
points x and y is
This voltage is divided equally between the
resistors in
Figure (b), so the voltage between points x and y is
The voltmeter will read 57.6 V.
19.90. Set Up:
The charge is reduced to
of its maximum value when
Solve: (a) At
(b)