Chapter 18: Basic Electric Circuits

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AP Physics
Chapter 18
Basic Electric Circuits
Chapter 18: Basic Electric Circuits
18.1 Resistances in Series, Parallel, and
Series-Parallel
18.2 Multiloop Circuits and Kirchoff’s Rules
18.4 Ammeters and Voltmeters
Homework for Chapter 18
• Read Chapter 18
• HW 18.A: p.592-596: 7, 10, 11, 25, 27, 44, 45, 47, 72, 73.
Answer: Each appliance needs its own path so that it
does not affect the operation of other appliances. As
you add more appliances to a parallel circuit, the total
current increases.
18.1 Resistances in Series, Parallel,
and Series-Parallel
equivalent series resistance (Rs)
connected in series in a circuit:
Rs = R1 + R2 + R3 + … = iRi
- when resistors are
on Gold Sheet
• The equivalent series resistance is the algebraic sum of the individual
resistances.
• The equivalent series resistance is larger than that of the largest resistor in the
series combination.
• In a series circuit, the current is the same through all resistors:
I = I 1 = I2 = I 3 = …
•The total voltage is the sum of the voltages of the individual resistors:
V = V1 + V2 + V3 +… = i Vi = i I Ri
Resistors (Light Bulbs) in Series
a) When resistors are connected in series, the current through each of them is
the same. The sum of the voltage drops across each of them is equal to the
voltage of the battery.
b) The equivalent resistance Rs of the resistors in series.
equivalent parallel resistance (Rp)
connected in parallel in a circuit:
1
1
1
1
1
Rp = R1 + R2 + R3 + … = iRi
- when resistors are
on Gold Sheet
• The reciprocal of the equivalent parallel resistance is equal to the sum of the
reciprocals of the individual resistances.
• The equivalent parallel resistance is less than that of the smallest resistor in the
parallel combination.
• In a parallel circuit, the voltage is the same across all resistors:
V = V1 = V2 = V3 = …
•The total current is the sum of the currents of the individual resistors:
I = I1 + I2 + I3 +… = i Ii
Resistors (Light Bulbs) in Parallel
a) When resistors are connected in parallel, the voltage drop across each of the
resistors is the same. The current from the battery divides among the resistors.
b) The equivalent resistance Rp of the resistors in parallel.
• Series connections provide a way to increase total resistance.
• Parallel connections provide a way to decrease total resistance.
Example 18.1:
Solution:
Example 18.2:
Solution:
Series – Parallel Resistor Combinations
(5)
Use a V-I-R chart.
V
R1
R2
Total
I
R
Example 18.3:
Solution:
Example 18.4: For the circuit shown, complete a V-I-R table.
R1 = 5 
R2 = 7 
12 V
R3 = 2 
Check for Understanding
1. Which of the following is always the same for resistors in series?
a) voltage
b) current
c) power
d) energy
Answer: b
2. Which of the following is always the same for resistors in parallel?
a) voltage
b) current
c) power
d) energy
Answer: a
Check for Understanding
3. Are the voltages across resistors in series generally the same?
Answer: No
4. Could they be so in certain cases?
Answer: Yes, if all the resistors are equal
5. Are the currents in resistors in parallel generally the same?
Answer: No
6. Could they be so in certain cases?
Answer: Yes, if all the resistors are equal
Check for Understanding
7. Three resistors that have values of 5 , 2 , and 1  are connected in series
to a battery. Which gets the most power, and why?
Answer: The 5  resistor gets the most power because:
 P=IV
 all the resistors get the same current
 the voltage drop is the highest for the largest resistor (V=IR)
8. Three resistors that have values of 5 , 2 , and 1  are connected in parallel
to a battery. Which gets the most power, and why?
Answer: The 1 resistor gets the most power because:
 P=IV
 all the resistors have the same voltage
 it gets the most current
Answer: Five can safely be used. Each cooker draws 7.5 amps.
18.2 Multiloop Circuits and
Kirchoff’s Rules
Kirchhoff’s rules
- a general method for analyzing multiloop circuits
• developed by the German physicist Gustav Kirchhoff (1824-1887)
• rules are based on the conservation of charge and energy
junction or node
- a point in a multiloop circuit at which three or
more connecting wires are joined together.
• current either divides or merges at a junction
branch
- a circuit path between two junctions, and may contain
one or more elements.
A Multiloop Circuit
a) Always reduce a circuit as far as possible before analyzing it. This circuit can
be reduced to the one shown in b.
b) At junction A the current divides, and at B the current merges. The paths
between junctions A and B are called branches. How many branches does
this circuit have?
Answer: 3
current in = current out
(Junction Theorem)
current in is positive and current out is negative
Sign Convention for Kirchoff’s Rules
a) The voltage across a battery is taken to be
positive (+) if a battery is traversed from the
negative to the positive terminal.
The voltage across a battery is taken to be
negative (-) if a battery is traversed from
the positive to the negative terminal.
b) The voltage across a resistor is taken to be
negative (-) if the resistance is traversed in
the direction of the assigned branch
currrent.
The voltage across a resistor is taken to be
positive (+) if the resistance is traversed in
the direction opposite that opposite that of
the assigned branch currrent.
Application of Kirchhoff’s Rules
Example 18.5:
Solution:
Example 18.6:
Solution:
Check for Understanding
1. A multiloop circuit has more than one
a)
b)
c)
d)
junction
branch
current
all of these
Answer: d
2. By our sign convention, if a resistor is traversed in the direction of the current
in it
a) the current is negative
b) the current is positive
c) the voltage is negative
d) the voltage is positive
Answer: c
Check for Understanding
3. If you traversed a battery from the negative to the positive terminal, is the
voltage positive or negative?
Answer: positive
Hint: the 60 W bulb has a higher
resistance than the 75 W bulb.
Answer: The greater the power, the brighter the bulb.
In the series circuit, the current is the same for both bulbs.
Since P = I2R, the greater the R, the greater the power. Therefore, the 60 W
bulb burns brighter.
In a parallel circuit, the voltage is the same for both bulbs.
Since P = V2/R, the greater the R, the lower the power. Therefore, the 75 W
bulb burns brighter.
Brightness of a Bulb
• The brightness of a bulb depends solely on the power dissipated by the bulb.
You can remember that from your own experience – when you go to the store to
buy a light bulb, you as for a 60 watt or 75 watt bulb. Watt is the unit of power.
• A 75 watt bulb is brighter than a 60 watt bulb, but be careful: a bulb’s power can
change depending on the current and voltage it’s hooked up to.
• Power is given by the equations P = IV, P = I2R, P = V2/R.
• The resistance of a bulb can be calculated by hooking it up to a 120V source:
For a 60 W bulb: I = P = 60 W = 0.5 A so, R = V = 120 V = 240 
V 120 V
I
0.5 A
For a 75 W bulb: I = 75 W = 0.625 A
so, R = 120 V = 192 
The resistance of a bulb is a property of the bulb itself, and never
changes.
Question: A light bulb is rated at 100 W in the United States, where the standard
wall outlet voltage is 120 V. If this bulb were plugged in in Europe, where standard
wall outlet voltage is 240 V, which of the following would be true?
a) The bulb would be ¼ as bright.
b) The bulb would be ½ as bright.
c) The bulb’s brightness would be the same.
d) The bulb would be twice as bright.
e) The bulb would be four times as bright.
Answer: e
The resistance does not change because it is a
property of the bulb itself. It will not vary no matter
what the bulb is hooked up to.
Since P = V2/R, if the voltage is doubled, the
power is quadrupled.
18.4 Ammeters and Voltmeters
ammeter
- a low resistance
device to measure current
The circuit symbol for an ammeter is
a circle with an A inside it.
• Current is the same for any resistors in
series with one another. So, if you’re
going to measure the current through a
resistor, the ammeter must
be in series
with the resistor
whose current you want to measure.
• The ammeter does not appreciably
affect the current measurement due to its
low resistance.
• In this circuit, ammeter A1 measures the current through resistor R1, while
ammeter A2 measures the current through resistor R2.
Question: Is there a way to figure out the current in the other three resistors
based only on the readings in these two ammeters?
Answer: The current through R5 must be the same as through R1, because both
resistors carry whatever current came directly from the battery. The current
through R3 and R4 can be determined from Kirchhoff’s junction rule: subtract the
current in R2 from the current in R1 and that’s what’s left over for the right-hand
branch of the circuit.
voltmeter
- a high resistance device used to measure voltage.
The symbol for a voltmeter is a circle with a
V inside it.
• Voltage is the same for any resistors
in parallel with each other. So, if you’re
going to measure the voltage across a
resistor, you must put the
voltmeter in parallel
resistor.
with the
• Since the voltmeter has high
resistance, it does not appreciably
affect the voltage across and element.
In the circuit, the meter labeled V2 measures voltage across the 100  resistor,
while the meter labeled V1 measures the potential difference between points A
and B (which is also the voltage across R1).
• Both ammeter and voltmeter functions can be combined
in a single multimeter.
Check for Understanding
1. A voltmeter has a
a) large capacitance
b) large resistance
c) small resistance
d) small capacitance
Answer: b
2. To measure the current through a circuit element, an ammeter is connected
a) in series with the element
b) in parallel with the element
c) between the high potential side of the element and ground
d) none of these
Answer: a
Check for Understanding
3. An ammeter has a
a) large resistance
b) large capacitance
c) small resistance
d) small capacitance
Answer: c
4. To measure the voltage across a circuit element, a voltmeter is connected
a) in series with the element
b) in parallel with the element
c) between the high potential side of the element and ground
d) none of these
Answer: b
HW 18.A: p.592-596: 7, 10, 11, 25, 27, 44, 45, 47, 72, 73.
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