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SPH4UW
Circuits and Ohm’s Law
Electric Terminology

Current: Moving Charges
Symbol: I
 Unit: Amp  Coulomb/second
 Count number of charges which pass point/sec


Power: Energy/Time
Symbol: P
 Unit: Watt  Joule/second = Volt Coulomb/sec
 P = IV

Physical Resistor

Resistance: Traveling through a resistor,
electrons bump into things which slows
them down. R = r L /A
A




r: Resistivity (E/J or Ωm)
L: Length
A: Area
L
Ohms Law I = V/R

Double potential difference  double current
Understanding
Two cylindrical resistors are made from the same
material. They are of equal length but one has twice
the diameter of the other.
1.
2.
3.
R1 > R2
R1 = R2
R1 < R2
R
rL
A
1
2
Simple Circuit
I
e
R
I

Practice…

Calculate I (current) when e = 24 Volts and
R=8W
Ohm’s Law: V =IR
V
I
R
24V

8W
 3A
Resistors in Series
One wire:

Effectively adding lengths:
 Req=r(L1+L2)/A

Since R  L, add resistance:
R
=
Req = R1 + R2
R
2R
Resistors in Series

Resistors connected end-to-end:

If current goes through one resistor,
it must go through other.
I1 = I2 = Ieq

Both have voltage drops:
V1 + V2 = Veq
Req =
R1
Veq = V1 + V2 = R1 + R2 R2
Ieq
Ieq
Req
Understanding
Compare I1 the current through R1,
with I10 the current through R10.
1.
I1 < I10
2.
I1 = I10
3.
I1 > I10
R1=1W
e
0
R10=10W
Note: I is the same
everywhere in this circuit!
ACT: Series Circuit
Compare V1 the voltage across R1, with V10 the voltage across R10.
V1 = I1 R1 = 1 x I
1. V1>V10
2. V1=V10
3. V1<V10
V10 = I10 R10 = 10 x I
Practice:
Resistors in Series
Calculate the voltage across each resistor if
the battery has potential V0 = 22 volts.
R1=1W
e0
R2=10W
Simplify (R1 and R2 in series):
• R12 = R1 + R2= 11 W
• V12 = V1 + V2= 22 V
• I12 = I1=I2=V12/R12 = 2 Amps
e0
R12
Expand:
•V1 =
•V2 =
I1R1= 2 x 1 = 2 Volts
I2R2= 2 x 10 = 20 Volts
Check: V1 + V2 = V12 ? Yes
R1=1W
e0
R2=10W
Resistors in Parallel

Two wires:
Effectively adding the Area
 Since R a 1/A add 1/R:

1
1
1


Req
R1
R2
R
R
=
R/2
Resistors in Parallel

Both ends of resistor are connected:

Current is split between two wires:
I1 + I2 = Ieq

Voltage is same across each:
V1 = V2 = Veq
I eq I1  12 1 1
1
R1


 
Req Veq
Veq
R1 R2
R2
Req
Understanding
What happens to the current through R2 when the
switch is closed?
1) Increases
2) Remains Same
3) Decreases
V2 = ε = I2R2
ACT: Parallel Circuit
What happens to the current through the battery?
(1) Increases
(2) Remains Same
Ibattery = I2 + I3
(3) Decreases
Practice: Resistors in Parallel
e
R2
R3
Determine the current through the battery.
Let E = 60 Volts, R2 = 20 W and R3=30 W.
Simplify: R2 and R3 are in parallel
1
1
1


R23 R2 R3
V23  V2  V3  60V
I 23 
V23
R23
1
1
1
1



R23 20W 30W 12W
R23  12W
60V
I 23 
12W
 5A
e
R23
Understanding
Johnny “Danger” Powers uses one power strip to plug in his
microwave, coffee pot, space heater, toaster, and guitar amplifier
all into one outlet.
25 A
10 A
Toaster
5A
Coffee Pot
10 A
Microwave
This is dangerous because…
(By the way, power strips are wired in parallel.)
1. The resistance of the kitchen circuit is too high.
2. The voltage across the kitchen circuit is too high.
3. The current in the kitchen circuit is too high.
Summary
• Resistors
–
–
–
–
Physical
Series
Parallel
Power
R = r L/A
Req = R1 + R2
1/Req = 1/R1 + 1/R2
P = IV
Summary
Series
Parallel
R1
R1
R2
R2
Wiring
Each resistor on the
same wire.
Each resistor on a
different wire.
Voltage
Different for each resistor.
Vtotal = V1 + V2
Same for each resistor.
Vtotal = V1 = V2
Current
Same for each resistor
Itotal = I1 = I2
Different for each resistor
Itotal = I1 + I2
Increases
Resistance R = R + R
eq
1
2
Decreases
1/Req = 1/R1 + 1/R2
Understanding
1
R
2
3
2R
R/2
Which configuration has the smallest resistance?
1
2
3
Which configuration has the largest resistance?
1
2
3
Parallel + Series Tests

Resistors R1 and R2 are in series if and
only if every loop that contains R1 also
contains R2

Resistors R1 and R2 are in parallel if and
only if you can make a loop that has ONLY
R1 and R2
Understanding
Determine the voltage and current in each
resistor
First we notice
the voltage drop
through these
two resistor
groupings total
10V.
R1=15W
Let’s combine to find REQ
R2=25W
e0=10V
1
1 1
1
  
REQ R1 R2 R3
1
1
1


15W 25W 17W
 6.04W

R3=17W
R4=5W
REQ
Let’s now add
R4 to this REQ
Rtot  REQ  R4
 6.04W  5W
 11.04W
Understanding
Let’s determine
the current, IT
R1=15W
R2=25W
e0=10V
e0=10V
RT=11.04W
R3=17W
R4=5W
R 4:
V  IR
V
10V
I 
 0.906 A
R 11.04W
I 4  0.91A
V4  I tot R4
This current will flow
through each of the
resistor groupings
R1, R2, R3,
experience the
same voltage
R 1:
 0.906 A 5W 
 4.5V
Since voltage drop has to total
10V, VEQ=10V-4.5V=5.5V
I1 
VEQ
R1
R 2:
I2 
VEQ
R2
R 3:
I3 
VEQ
R3

5.5V
15W
 0.36 A

5.5V
25W
 0.22 A

V1  5.5V
V2  5.5V
V2  5.5V
5.5V
17W
 0.32 A
The V-I-R Chart
Determine the Voltage, Current, and
Resistance
5Ω
12V
7Ω
10Ω
Step 1:
Fill out the table with known
resistors and the Total Voltage for
circuit
V
R1
R2
R3
Total
12
I
R
5
7
10
The V-I-R Chart
Determine the Voltage, Current, and
Resistance
5Ω
1
1
1


Req 7W 10W
RT  5W  4.117W
Req  4.117W
 9.117W
12V
7Ω
10Ω
Step 2:
Using resistor laws, determine
total resistance of circuit.
V
R1
R2
R3
Total
12
I
R
5
7
10
9.117
The V-I-R Chart
Determine the Voltage, Current, and
Resistance
5Ω
V  IR
V
R
12V

9.117W
 1.32 A
V  IR
I
12V
7Ω
Step 4:
Step 3:
Since initial current amount will
Using
Ohm’s
law, determine
the
also pass
through
resistor R1,
we
Current
of circuit.
can determine
its voltage drop.
10Ω
R1
R2
R3
Total
 1.32 A 5W 
 6.6V
V
6.6
I
1.32
12
1.32
R
5
7
10
9.117
The V-I-R Chart
Determine the Voltage, Current, and
Resistance
5Ω
12V
7Ω
Step 5:
Since R2 and R3 have the same
Voltage drop, we then must have
12V-6.6V=5.4V.
10Ω
R1
R2
R3
Total
V
6.6
5.4
5.4
12
I
1.32
1.32
R
5
7
10
9.117
The V-I-R Chart
Determine the Voltage, Current, and
Resistance
5Ω
12V
Step 6:
Use ohm’s law to find currents
7Ω
10Ω
R1
R2
R3
Total
V
6.6
5.4
5.4
12
I
1.32
0.77
0.54
1.32
R
5
7
10
9.117
Understanding
Let’s follow a
conventional current
path through R1
VT  e 0  IR1  IR4
 10V   0.366 A15W    0.91A 5W 
 10V  5.5V  4.5V
 0V
R1=15W
R2=25W
e0=10V
R3=17W
R4=5W
You can pick any path through the circuit and the total voltage
increases and decrease will balance
You can reverse the direction of the current and thus the signs,
(batteries increase the voltage, resistors drop the voltage) and obtain
the same results.
Let us calculate the Current and the Power
(used/generated) by the elements of the following circuit.
V2
P
R
V  IR
  2 A 2.5W 
 5V
What happens to
the Power delivery
and consumption if
another identical
bulb is place in
parallel or in series
with the first?
V
R
5V

2.5W
 2A
I
R  2.5W
e  5V
 5V 

2
2.5W
 10W
P  IV
  2 A 5V ) 
 10W
Let us calculate the Current and the Power
(used/generated) by the elements of the following circuit
when bulbs are in parallel.
2
  2 A 2.5W 
 5V
V  IR
V2
P
R
V  IR
R  2.5W

 5V 
2
2.5W
 10W
5V  I  2.5W 
I  2A
P
2A
R  2.5W
2A
V
R
5V

2.5W
 2A
I
e  5V
P  IV
4A
  2 A 5V ) 
V
R
 5V 

2
2.5W
 10W
R  2.5W
 10W
Because the bulbs
(resistors) are in parallel,
we use the parallel law to
determine total resistance
of the circuit.
V
R
5V

1.25W
 4A
I
e  5V
P  IV
  4 A 5V ) 
 20W
Let us calculate the Current and the Power
(used/generated) by the elements of the following circuit
V  IR
when bulbs are in series.
V2
  2 A 2.5W 
 5V
V
R
5V

2.5W
 2A
I
R  2.5W
e  5V

 5V 
P
 1A  2.5W 
V2
P
R
V  IR
2
2.5W
 10W
2.5V 


 2.5V
2
2.5W
 2.5W
P  IV
1
A
  2 A 5V ) 
 10W
Because the bulbs
(resistors) are in series,
we use the series law to
determine total resistance
of the circuit.
R
V
R
5V

5W
 1A
I
R  2.5W R  2.5W
e  5V
P  IV
 1A 5V ) 
 5W
Understanding
A voltmeter is a
device that’s used
to measure the
voltage between
two points in a
circuit. An ammeter
is used to measure
current. Determine
the readings on the
voltmeter and the
ammeter
v
a
J
2400V
A
b
Understanding
Let’s first combine the 2
parallel resistors
1
1
1


Req 600W 300W
Req  200W
Now we have 3
resistors in series
Rtot  50W  200W  150W
Req  400W
v
a
A
J
b
2400V
Thus we can determine the
current supplied by the battery
e
2400V
I 
 6A
R 400W
Understanding
At junction J this 6A
current splits. Since
the top resistor is
twice the bottom
resistor, half as
much current will
flow through it.
Therefore the current
through the top
resistor is 2A and the
bottom resistor is 4A
v
2A
J
4A
a
A
6A
6A
2400V
b
6A
6A
Thus the reading of
the ammeter is 4A
Understanding
The voltmeter will
read a voltage drop
of :
I3R3   6 A150W  900V
Between points a and
b
Therefore the
potential at point b
is 900V lower than
at point a
v
2A
J
4A
a
A
6A
6A
2400V
b
6A
6A