Circuit Symbols:
Battery
Resistor
Light-bulb
Switch
Wire
Three general types of circuits:
Closed Circuit - There is a
complete loop with wires
going from one side of the
battery through a
resistor(s) to the other
side of the battery.
Only Working
Open Circuit - There is not
Circuit
a complete loop.
Short Circuit - There is a
complete loop, but it does
not contain any resistors.
There are two ways to put
resistors into a circuit.
1. Resistors can be in series
OR
2. Resistors can be in parallel
Resistors in Series is like a trip to Costco
Resistors in Series
(like a trip to Costco)
Resistors are considered to be in series if the
current must go through all of the resistors in
order.
The current (amps) through all resistors in
series is the same.
The voltage across resistors in series may be
different
The rate of electron flow (or current) is
determined by which resistor?
The resistor with the largest amount of ohms.
Combining (adding) Resistors
R1
R2
R3
Amps
Series Resistors
Itotal = I1 = I2 = I3
Req = Rtotal = R1 + R2 + R3
Voltage is calculated with
Ohm’s Law
Q
Resistors in Parallel is like
a trip to Vons
Resistors in Parallel
(like a trip to Vons)
 Resistors are considered to be in parallel if the current
is shared between multiple resistors.
 The current (amps) through all resistors in parallel may
be different.
 The voltage across all parallel resistors is the same.
 Will a resistor with a large resistance have more or
less current through it then a resistor with a small
resistance?
The resistor with a large resistance will have a
smaller current then the resistor with the smaller
resistance.
Combining (adding) Resistors
Parallel Resistors
Current is calculated with
Ohm’s Law
1
1
1
1
 

Rtotal R1 R2 R3
Vtotal = V1 = V2 = V3
Example 1: A circuit has three
8.0 W, 5.0 W and a 12 W resistors
in series along with a 24 V
battery.
Draw the circuit.
Calculate the total resistance of the circuit.
Calculate the total current through the circuit.
 What is the current through each resistor?
Calculate the voltage across each resistor.
Example 2: A circuit has three
resistors: 6.0 W, 4.0 W and a 12 W
resistors in parallel along with a
24 V battery.
Draw the circuit.
Calculate the total resistance of the circuit.
Calculate the total current through the circuit.
 What is the voltage across each resistor?
Calculate the current across each resistor.
Electrical Outlets
Electrical outlets provide electric potential
(or the voltage) for any appliance plugged
in to it.
In the United States ALL outlets provide
120 V (in Europe it is 240 V)
Light bulbs are made to be the
only appliance plugged into a
socket.
The power rating of a light bulb (25 W or
100 W…) is as if that bulb was the only
bulb plugged in to a 120 V power source.
The resistance of a light bulb is calculated
by knowing the power rating and the
voltage (120 V)
Current and actual voltage used by a light
bulb depends on the circuit.
Example 3: What will the power output be if
an American-made 45 W light bulb is
plugged in to a 310 V power source?
Using 120 V, calculate the resistance of the light
bulb.
V2
1202
P
 45W 
 R  320W
R
R
Using the resistance and the voltage of the new
source, calculate the new power
V2
3102
P
P
 P  3.0 x102W
R
320
As more identical resistors R are added to the
parallel circuit shown, the total resistance
between points P and Q …
R
1. Increases
2. Remains the same
Q
…
3. Decreases
P
As more identical resistors R are added to the
parallel circuit shown, the total resistance between
points P and Q …
1. Increases
2. Remains the same
3. decreases
R
P
Q
…
Q
When one bulb is unscrewed, the other bulb will
remain lit in which circuit…
1. I
2. II
3. Both
4. Neither
Circuit II
Circuit I
When one bulb is unscrewed, the other bulb will remain
lit in which circuit…
1. I
2. II
3. both
4. neither
Circuit I
Circuit II
A 25W bulb and a 100W bulb are connected in
series. Which bulb will glow brighter?
25W
100W
120V
The Light Bulbs are really Resistors
A) Calculate the resistance for each resistor shown.
B) Calculate the total resistance of the circuit.
C) Calculate the current through each resistor.
D) Calculate the power used by each resistor.
E) Calculate the voltage across each resistor.
25W
100W
120V
Part A.
25W Bulb
100W Bulb
V2
P
R
V2
R
P
120
R
25
1202
R
100
R  576W
R 144W
2
144 W
576 W
120V
B) Calculate the total circuit resistance Rtotal
Rtotal = R1 + R2
= 576 + 144
= 720 W
Series Resistors
720 W
 .167 am ps
 .167 am ps
120V
C) Calculate the total circuit current (I)
V 120volts
I 
 .167 am ps
720 W
R
144 W
576 W
 .167 am ps
120V
D) Calculate the Power used by each resistor.
25 W Bulb
100 W Bulb
P = I2R
P = I2R
= .1672 x 576
= 16 watts
= .1672 x 144
= 4 watts
96 volts
24 volts
144 W
576 W
 .167 am ps
120V
E) Calculate the Voltage across each resistor.
25W Bulb
100W Bulb
V = IR
V = IR
= .167 x 576
= 96 volts
= .167 x 144
= 24 volts
120 volts
E) Consider the Percent Power Needed to
Light Each Bulb
100 W Bulb
4 watts
 100  4 percent
100watts
25 W Bulb
16 watts
 100  64 percent
25watts
Q
The circuit below consists of two identical light bulbs
burning with equal brightness and a single 12V battery.
When the switch is closed, the brightness of bulb A…
A
1. Increases
2. Decreases
3. Remains unchanged
The circuit below consists of two identical light bulbs
burning with equal brightness and a single 12V battery.
When the switch is closed, the brightness of bulb A…
1. Increases
2. decreases 3. remains unchanged
When the switch is closed,
bulb B goes out because all of
the current goes through the
wire parallel to the bulb. Thus,
the total resistance of the circuit
decreases, the current through
bulb increases, and it burns
brighter.
A
Q
Which bird is in trouble when the switch is closed?
1) Bird 1
2) Bird 2
3) Neither
4) Both
1
2
Which bird is in trouble when the switch is closed?
1) Bird 1 2) bird 2 3) neither
1
2
4) both
Charge flows through a light bulb. Suppose a wire
is connected across the bulb as shown. When the wire
is connected…
1. All the charge continues to
flow through the bulb, and
the bulb stays lit.
2. Half the charge flows
through the wire, the other
half continues through the
bulb.
3. Essentially all the charge
flows through the wire and
the bulb goes out.
4. None of these.
Q
Analyze the circuit:
A) Calculate Rtotal
B) Calculate the current through each resistor.
C) Calculate the voltage through each resistor.
16W
Parallel:
120V
1
R1 23
R123 

1 1
1
4
 

16 32 32 32
32
 8W
4
16W
32W
32W
120V
Series:
R123-4=8+16
R1234=24
16W
8W
Make
These are in parallel so their
voltage is the same along with
chart:
the total voltage
R
16W
120V
16W
32W
32W
All these numbers
will be the same.
1
16
2
16
3
32
4
32
234
8
1234
24
I
V
120
Make
These are in series so their
current is the same along with
chart:
the total current
R
16W
120V
8W
All these numbers
will be the same.
1
16
2
16
3
32
4
32
234
8
1234
24
I
V
120
Fill out the chart with V=IR
V = IR
V = IR
120 = I (24)
V = (5) (16)
I=5A
V = 80 V
V = IR
V = IR
V = (5) (8)
40 = I (16)
V = 40 V
I = 2.5 A
V = IR
40 = I (32)
I = 1.25 A
1
R
16
I
5
V
80
2
16
2.5
40
3
32
1.25
40
4
32
1.25
40
234
8
5
40
1234
24
5
120
Another way to do the problem (without
the chart)
I=V/R
120V
I=120v/24W
I=5 amps
24W
V=IR
V=(5)(16)
V=80volts
120V
16W
80volts
5amps
V=IR
V=(5)(8)
V=40volts
8W
40volts
120volts
I=V/R
=40volts/16 W
=2.5 amps
120V
I=V/R
=40volts/32 W
=1.25 amps
5 amps
5 amps
16W
16W
32W
80volts
32W
40
volts
When the series circuit shown is connected, Bulb A is
brighter than Bulb B. If the positions of the bulbs were
reversed…
1. Bulb A would again be brighter
2. Bulb B would be brighter
3. They would be equal brightness
When the series circuit shown is connected, Bulb A is
brighter than Bulb B. If the positions of the bulbs were
reversed…
1. Bulb A would again be brighter
2. Bulb B would be brighter
3. They would be the same
The bulbs are connected in series, so the same
current passes through both of them. Different
brightnesses indicate different filament resistances.
Bulb A is NOT brighter because it is “first in line”
for the current of the battery! After all, electrons
deliver the energy, and they flow from negative to
positive --- in the opposite direction!
Example: Find the voltage and
current for each resistor.
6W
3W
3W
6W
4W
12W
18 volts
2W
6W
3W
3W
6W
4W
12W
18 volts
2W
3W
3W
3W
4W
12W
18 volts
2W
3W
3W
3W
4W
12W
18 volts
2W
6W
3W
4W
12W
18 volts
2W
6W
3W
1
1
1
 
Rtotal R1 R2
1
1 1
 
Rtotal 4 12
Rtotal 3W
4W
12W
18 volts
2W
6W
3W
3W
1
1
1
 
Rtotal R1 R2
1
1 1
 
Rtotal 4 12
Rtotal 3W
18 volts
2W
6W
3W
3W
18 volts
2W
6W
3W
5W
18 volts
6W
3W
1
1 1
 
Rtotal 6 5
5W
Rtotal  2.73W
18 volts
3W
2.73W
18 volts
3W
2.73W
18 volts
5.73W
18 volts
Now, find the total current flowing
5.73W
V
I
R
18volts
I
5.73W
I  3.14 am ps
18 volts
9.42volts
3W
3W
V=IR
V=(3.14)(3W)
4W
V=9.42
12W
18 volts
6W
6W
2W
18-9.42
8.57volts
9.42volts
3W
3.14 amps
6W
3W
6W
4W
12W
18 volts
2W
18-9.42
8.57volts
9.42volts
3W
3.14 amps
6W
4W
12W
18 volts
2W
18-9.42
8.57volts
9.42volts
3W
3.14 amps
6W
4W
12W
18 volts
2W
18-9.42
8.57volts
9.42volts
6W
3W
3.14 amps
5W
18 volts
18-9.42
8.57volts
9.42volts
6W
3W
V
I
R
5W
8.57 volts
I
6W
I  1.43amps
18 volts
18-9.42
8.57volts
9.42volts
3W
3.14 amps
6W
1.43 amps
5W
1.71 amps
18 volts
18-9.42
8.57volts
9.42volts
6W
1.43 amps
3W
4W
3.14 amps
1.71
amps
V=IR
V=(1.71)(2)
V=3.42volts
12W
18 volts
3.42Volts
2W
1.71
amps
18-9.42
8.57volts
9.42volts
6W
5.15 volts
3W
4W
3.14 amps
1.71
amps
12W
18 volts
1.43 amps
3.42Volts
2W
1.71
amps
18-9.42
I=V/R
I=5.15volts/12W
9.42volts
I= 0.43 amps
8.57volts
6W
5.15 volts
3W
4W
3.14 amps
1.71
amps
12W
0.43 amps
18 volts
1.43 amps
3.42Volts
2W
1.71
amps
18-9.42
I=V/R
I=5.15volts/4W
9.42volts
I= 1.28 amps
Or…
8.57volts
1.71 amps – 0.43 =
6W 1.28 amps
5.15 volts
3W
4W
3.14 amps
1.71
amps
12W
0.43 amps
18 volts
3.42Volts
2W
1.71
amps
6W
3W
3W
6W
4W
12W
18 volts
2W
6W
3W
3W
6W
4W
12W
18 volts
2W
6W
3W
3W
6W
4W
12W
2W
18 volts
Q
Given: R1=1W; R2=2 W; R3=3 W. Rank the bulbs
according to their relative brightness
1.
R1 > R 2 > R 3
2.
R 1 > R 2 = R3
3.
R 1 = R 2 > R3
4.
R1 < R 2 < R 3
5.
R 1 = R 2 = R3
R1
R2 R3
15
Given: R1=1W; R2=2 W; R3=3 W. Rank the bulbs
according to their relative brightness
1.
R1 > R 2 > R 3
2.
R 1 > R 2 = R3
3.
R 1 = R 2 > R3
4.
R1 < R 2 < R 3
5.
R 1 = R 2 = R3
R1
R2 R3
2
V
P  IV  I R 
R
2
Q
If the four light
bulbs in the figure below
are identical, which
circuit puts out more
total light?
1. I
2. II
3. Same
Circuit II
If the four light
bulbs in the figure below
are identical, which
circuit puts out more
total light?
1. I
2. II
Circuit II
3. Same
The resistance of two light bulbs in parallel in
smaller than that of two bulbs in series. Thus the
current through the battery is greater for circuit I
than for circuit II.
Since the power dissipated is the product of
current and voltage, it follows that more is
dissipated in circuit I.