Chapter 16
Electrons in Motion: Circuits
In This Chapter
Getting to know current
Handling resistance
Working with power
Developing series circuits
Creating parallel circuits
T
his chapter is about what happens when you get electrons moving: You get electric current. Electric current runs through wires and has to contend with various items in those
wires — resistors, capacitors, and the like. In this chapter, you see how to handle such
items yourself.
Electrons in a Whirl: Current
If you have electrons moving through a wire, you have electrical current. Current, symbol I,
is defined as the amount of charge that passes by in 1 second. One Coulomb per second is
named 1 Ampere or just 1 Amp, symbol A. So this equation is how you find current:
q
I= t
Current is created when you have an electromotive force (EMF) providing a voltage across
a conductor. EMF can be supplied by batteries, generators, and similar items in physics
problems.
Q.
How many electrons per second pass by a
given point in a wire carrying 2.0 Amps?
A.
1.25 × 1019
3. Plug in the numbers to find the total
charge in 1 second:
q = (2.0)·(1.0) = 2.0 C
1. Use this equation:
q
I= t
2. Solve for q:
q = I·t
4. Divide by the charge per electron to find
the number of electrons:
Number =
2.0
= 1.25 # 1019
1.6 # 10- 19
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Part V: Zap: Electricity and Magnetism
1.
2.
How many electrons per second pass by a
given point in a wire carrying 6.0 Amps?
Solve It
What total charge passes by a given point
in a circuit carrying 12.0 Amps in 1 minute?
Solve It
Giving You Some Resistance: Ohm’s Law
When you have an EMF providing voltage across a wire, how much current flows?
The answer turns out to depend on how much resistance, symbol R, is in the wire.
This equation shows the relationship among voltage, current, and resistance:
V = I·R
Resistance is measured in ohms, symbol ω.
At the top of Figure 16-1, you see a battery with an EMF of 6.0 Volts. The current flows
from the positive terminal of the battery — corresponding to the longer of the two
lines in the battery symbol — to the negative terminal of the battery.
At the bottom of the figure is a resistor, R. How much current flows through the resistor? You know the answer from the equation V = I·R.
Chapter 16: Electrons in Motion: Circuits
6V
+
−
I
Figure 16-1:
A resistor
and a
battery in
a circuit.
R
Q.
A.
2. Solve for I:
If the resistor in Figure 16-1 is 3.0 ω, what
current flows through it, driven by the
6.0 V battery?
I= V
R
3. Plug in the numbers:
2.0 Amps
I = V = 6.0 = 2.0 A
R 3.0
1. Use this equation:
V = I·R
3.
If the resistor in Figure 16-1 is 1.5 ω, what
current flows through it, driven by the
6.0 V battery?
Solve It
4.
If the resistor in Figure 16-1 is 0.5 ω, what
current flows through it, driven by the
6.0 V battery?
Solve It
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Part V: Zap: Electricity and Magnetism
5.
What’s the voltage across a 3.0 ω resistor
with a current of 1.0 A going through it?
Solve It
6.
What’s the voltage across a 5.0 ω resistor
with a current of 1.5 A going through it?
Solve It
Powering It Up
Notice how light bulbs get hot? That happens because the filament in the light bulb
acts as a resistor — and resistors dissipate heat. How much heat does a resistor give
off? This equation is how to calculate the answer in terms of current or voltage:
2
P = I $ V = V = I2 $ R
R
Q.
If a 300.0 ω resistor has a current of
1.0 Amps going through it, how much
power does it turn into heat?
A.
300 Watts
1. Use this equation:
P = I2·R
2. Plug in the numbers:
P = I2·R = (1.02)·(300) = 300 Watts
Chapter 16: Electrons in Motion: Circuits
7.
8.
If a 500.0 ω resistor has a current of
2.0 Amps going through it, how much
power does it turn into heat?
Solve It
If a 2000.0 ω resistor has a current of
1.5 Amps going through it, how much
power does it turn into heat?
Solve It
One after the Other: Series Circuits
What if you have several resistors, one after another, in a circuit, as you see in
Figure 16-2?
6V
+
−
I
Figure 16-2:
Resistors in
series.
R1
R2
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Part V: Zap: Electricity and Magnetism
When you have resistors arranged this way, so that the current has to go through both
of them, the arrangement is called placing resistors in series. To find the net resistance,
you just add the two individual resistances:
R = R1 + R2
Q.
A.
3. Use this equation to find the current:
If R1 is 20 ω, and R2 is 40 ω, what is the current flowing in Figure 16-2, in which the
battery is 6.0 V?
V = I·R
4. Solve for I:
0.1 A
I = V/R
1. Use this equation:
5. Plug in the numbers:
R = R1 + R2
I = V/R = 6.0/60 = 0.1 A
2. Plug in the numbers:
R = R1 + R2 = 60 ω
9.
If R1 is 60 ω, and R2 is 100 ω, what is the
current flowing in Figure 16-2, in which
the battery is 6.0 V?
Solve It
10.
If R1 is 100 ω, and R2 is 200 ω, what is the
current flowing in Figure 16-2, in which
the battery is 6.0 V?
Solve It
Chapter 16: Electrons in Motion: Circuits
11.
12.
If R1 is 100 ω, and R2 is 200 ω, what is the
voltage across R1, in which the battery is
6.0 V?
Solve It
If R1 is 100 ω, and R2 is 200 ω, what is the
voltage across R2, in which the battery is
6.0 V?
Solve It
All for One: Parallel Circuits
Say you have resistors connected as you see in Figure 16-3, so that the current flowing
can go through either resistor.
6V
+
−
I
R1
Figure 16-3:
Resistors in
parallel.
R2
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Part V: Zap: Electricity and Magnetism
This arrangement is called placing resistors in parallel. Unlike resistors in series, in
which the current has to go through the resistors one after another, in this arrangement part of the current goes through one resistor and part through the other.
What’s the total resistance of two resistors in parallel? This equation shows you:
1= 1 + 1
R R1 R 2
Q.
If R1 is 20 ω, and R2 is 60 ω, what is the current flowing in Figure 16-3, in which the
battery is 6.0 V?
A.
0.4 A
3. Find R, the total resistance:
R = 1 = 15 ω
0.066
4. Use this equation:
V = I·R
1. Use this equation:
5. Solve for I:
1= 1 + 1
R R1 R 2
I = V/R
2. Plug in the numbers:
6. Plug in the numbers:
1 = 1 + 1 = 1 + 1 = 0.066
R R 1 R 2 20 60
13.
If R1 is 30 ω, and R2 is 90 ω, what is the current flowing in Figure 16-3, in which the
battery is 6.0 V?
Solve It
I = V/R = 6.0/15 = 0.4 A
14.
If R1 is 45 ω, and R2 is 120 ω, what is the current flowing in Figure 16-3, where in which
the battery is 6.0 V?
Solve It
Chapter 16: Electrons in Motion: Circuits
15.
If R1 is 39 ω, and R2 is 93 ω, what is the current flowing in Figure 16-3, in which the
battery is 6.0 V?
Solve It
16.
If R1 is 42 ω, and R2 is 56 ω, what is the current flowing in Figure 16-3, in which the
battery is 6.0 V?
Solve It
The Whole Story: Kirchhoff’s Rules
The circuit in Figure 16-4 is a whopper. What’s I1? What’s I2? Kirchhoff’s laws make
everything clear.
+
−
R1
I3
6V
−
+
+
−
A
R2
I2
I1
12V
−
Figure 16-4:
A circuit.
+
R3
+
−
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Part V: Zap: Electricity and Magnetism
Following are Kirchhoff’s laws for a circuit:
Junction rule. The junction rule says that the total current going into any point
in a circuit must equal the total current going out of that point. So ΣI = 0 at any
point.
Loop rule. The loop rule says that around any closed loop in a circuit, the sum
of potential rises (as from a battery) must equal the sum of the potential drops
(as from resistors). So ΣV = 0 around any loop.
An easy way of summing the voltages around a loop is to draw in the current arrows
(it’s no problem if you draw a current arrow backward; the current will just come out
negative) and label each resistor with a + where the current goes in and a – where it
comes out. Then go around the loop (either direction is fine), adding a +V if you
encounter a battery’s positive terminal first or a –V if you encounter its negative terminal first, and a +IR or –IR for each resistor, depending on whether you encounter
the + or – signs you added to each resistor first. Then set that whole expression to
zero, such as –6V + IR1 +12V +IR2 = 0.
Q.
A.
If R1 is 2.0 ω, R2 is 4.0 ω, and R2 is 6.0 ω,
what are the three currents in Figure 16-4,
in which the batteries are 6.0 V and 12.0 V?
5. If you substitute the top equation for I3 in
the second equation, you get
+6 – 2·(I1 + I2) – 4I2 = 0
+12 – 6 + 4I2 – 6I1 = 0
I1 = 1.36 Amps, I2 = 0.55 Amps, I3 =
1.91 Amps
1. The junction rule says that ΣI = 0 at any
point, so use point A at left in the figure. I1
and I2 flow into it, and I3 flows out of it, so:
I1 + I2 = I3
2. The loop rule says that ΣV = 0. Figure 16-4
has three loops: the two internal loops
and the external overall loop. Because
you have three unknowns — I1, I2, and
I3 — all you need are three equations,
and the ΣI = 0 rule has already given you
one. So go around the two internal loops
to get two more equations. From the top
loop, you get
So:
+6 – 2I1 – 6I2 = 0
+12 – 6 + 4I2 – 6I1 = 0
6. You can find I1 in terms of I2 by using the
first equation:
I1 = 3 – 3I2
7. Then you can substitute this value of I1 in
the second equation to get
+12 – 6 + 4I2 – 6 (3 – 3I2) = 0
So:
+6 – 2I3 – 4I2 = 0
3. From the bottom loop, you get
+12 – 6 + 4I2 – 6I1 = 0
4. Now you have three equations in three
unknowns:
I1 + I2 = I3
+6 – 2I3 – 4I2 = 0
+12 – 6 + 4I2 – 6I1 = 0
–12 + 22I2 = 0
And:
I2 = 12/22 = 6/11 = 0.55 Amp
Chapter 16: Electrons in Motion: Circuits
9. Now you can find I1. Start with:
8. You now have one of the currents:
I2 = 6/11 Amp. Plug that into:
I1 + I2 = I3
6
11
Which means that:
+6 – 2I3 – 4I2 = 0
I1 = I3 – I2
So:
Giving you:
I 1 = 21 - 6 = 15 = 1.36 Amps
11 11 11
+6 – 2I3 – 4 (6/11) = 0
Or, dividing by 2:
+3 – I3 – 12/11 = 0
So:
I3 = 21/11 = 1.91 Amps
17.
If R1 is 3 ω, R2 is 6 ω, and R2 is 9 ω, what
are the three currents in Figure 16-4, in
which the batteries are 6.0 V and 12.0 V?
Solve It
18.
If R1 is 5 ω, R2 is 10 ω, and R2 is 15 ω, what
are the three currents in Figure 16-4, in
which the batteries are 6.0 V and 12.0 V?
Solve It
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