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BW#1 Ch. 22: CQ2-6 P. 1, 3, 6, 8, 10, 12
Q22.1. Reason: You can use a compass placed beside the wire and observe the compass needle change direction when
the battery is disconnected and reconnected in the opposite direction. So something must be happening in the wire. You
can feel the heat and see the light emanating from the light bulb. You can use charged glass and plastic rods to find that
charge does not accumulate at the light bulb. Energy is transferred somehow from the battery to the bulb, so we conclude
that something carries that energy.
Assess: The compass indicates that the space around the wire is somehow changed when the circuit is connected, and we
also know that the light bulb is giving off energy that must come from somewhere. We conclude that charges flow through
the wires.
Q22.2. Reason: In order for a bulb to light, charged particles must move through the bulb and give up energy in the
process. We can look at the battery and determine the terminal with the higher electric potential (the positive end) and
hence the direction of the electric field through the external circuit. Knowing the direction of the electric field we can
establish which way positive and/or negative charges will travel. An ammeter will assure us that indeed charges are
moving through the circuit. But the ammeter will not tell us which direction the charges are moving, hence we cannot
distinguish their charge.
Assess: There are no simple observations and/or measurements you can make on this circuit to distinguish a current
composed of positive charge carriers from a current composed of negative charge carriers.
Q22.3. Reason: In order for electrons to move through a wire as a current, the wire must be connected to a source of
electric potential difference. This source of electric potential difference will establish an electric field in the circuit and it is
this electric field that causes electrons to travel through the circuit.
Assess: Frequently, we use misleading language when describing the current in a circuit. In fact, you may have used the
misleading statement “flow of current.” Current is defined as the motion of charges. It is the charges that flow, not the
current. Current is the flow of charges.
Q22.4. Reason: (a) Because there are no junctions or branching of the wires, they must carry the same current. This is
because of conservation of charge. Charges can’t appear or disappear. The current is the same everywhere in a single-loop
circuit, including in the filament itself.
(b) Because of the answer to part (a) the brightness of the bulb will stay the same. The current will be the same if the wires
are switched.
Assess: This question is asked because many students have difficulty with this concept. Concentrate on this idea until it
makes complete sense.
Q22.5. Reason: The circuit has no place to store electrons and no way to generate electrons. Also, for every electron
that enters the circuit, another electron leaves the circuit. Based on these two statements, we can conclude that I1 is equal
to I4. As the circuit shows, the current I1 splits into two currents (I2 and I3) which then combine to give I4. This allows us to
conclude that both I1 and I4 are greater than I2 and I3. Since the wires that carry the current I2 and I3 are identical (they each
have the same resistance), the current I1 will divide in half to give I2 and I3. This allows us to make the following
statements:
I2  I3 < I1  I4, and (I2  I3)  (1/2)(I1  I4)
Assess: We cannot create or destroy charges in the wire, neither can we store them in the junction. The rate at which
electrons flow into one or many wires must be exactly balanced by the rate at which they flow out of the other wires. We
are conserving the amount of charge and the flow of that charge (current).
Q22.6. Reason: Equation 22.1 gives the definition of current: I  Q/t.
The initial current is 4 A: I1  Q1/t1  4 A.
Now we double the charge and halve the time interval:
I2 
Q2 2Q1
 1
 4( I1 )  4(4 A)  16 A
t2
2 t1
Assess: Watch the dependencies. Had it been the other way, half the charge in twice the time, the final current would be
1 A.
P22.1. Prepare: We will find the total charge that flows through the hair dryer and then divide it by the electron charge
to find the number of electrons.
Solve: Equation 22.2 is Q  It. The amount of charge delivered is

60 s 
Q  (10.0 A)  5.0 min 
  3000 C
1 min 

The number of electrons that flow through the hair dryer is
Q
3000 C
N 
 1.9 1022
e 1.60 1019 C
Assess: This is an enormous amount of charge and is typical of such devices.
P22.3. Prepare: Current is defined in terms of the amount of charge Q passing through a cross section of a wire in a
time interval t : I  Q/t.
We will also need to know how many electrons are in a coulomb. In Chapter 20 we learned that the charge on one electron
is  e  16 1019 C, so there are 1/16 1019 electrons in one coulomb.
Solve: Solve the equation for Q.
 1 electron 
20
Q  I t  (10 A)(30 s)  30 C 
  19 10 electrons
19
1

6

10
C


Assess: It is hard to get our minds around such large numbers, but try we must. That many electrons really do go past a
point (cross section) of the wire in those conditions. The electrons generally aren’t traveling very fast, but a lot of them go
by to make up a 1.0 A current. And remember that when electrons are the charge carriers, they go in the opposite direction
from the direction of the current.
P22.6. Prepare: Electric current, charge, and time are related by I  Q/t.
Solve: The current during the lightning strike is I  Q/t  2.5 C/(2.0 104 s)  1.3 104 A.
Assess: As expected, this is a large amount of current.
P22.8. Prepare: The direction of the current in a material is opposite to the direction of motion of the negative charges
and is the same as the direction of motion of positive charges. That is, I  I   I   q /t  q /t  q  q /t
Solve: The charge due to positive ions moving to the right per second is
q  N (2e)  (5.0 1015 )(2 1.60 1019 C)  1.60 103 C
The charge due to negative ions moving to the left per second is
q  N (e)  (6.0 1015 s)(1.60 1019 C)   0.96 103 C
Thus, the current in the solution is
q  q 1.60 103 C  (0.96 10 3 C)

 2.56 103 A  2.6 mA
t
1s
Assess: This is a typical current for ionic solutions.
I
P22.10. Prepare: We will use Equation 22.2 to find the charge that leaves the battery.
Solve:
The total charge in the battery is
Q  I t  (90 A)(3600 s)  3.2  105 C
Assess: As expected, this is a large amount of charge.
P22.12. Prepare: For a junction, the law of conservation of charge requires Iin  I out. This says that the total current
into a junction must equal the total current out of that junction.
Solve: First let’s redraw the figure, label the junctions, and give the unknown currents a name.
Now write a junction equation for each junction:
Junction 1: 5 A  3 A  i1 or i1  2 A
Junction 2: i1  iB  1 A or iB  i1 – 1 A  1 A
Junction 3: 3 A  i2  2 A or i2  1 A
Junction 4: 1 A  i2  iC or iC  1 A  i2
Assess: Notice that the total current going into this arrangement and the total current going out is 5 A.