Transcript Slide 1

Thomas Oberst
Cornell University
Electrical Circuits
Thomas Oberst
How you should be thinking
about electric circuits:
Voltage: a force that
pushes and pulls the
current through the
circuit (in this picture it
would be equivalent to
gravity)
Cornell University
Thomas Oberst
How you should be thinking
about electric circuits:
Resistance: friction that
impedes flow of current
through the circuit
(rocks in the river)
Cornell University
Thomas Oberst
How you should be thinking
about electric circuits:
Current: the actual
“substance” that is
flowing through the
wires of the circuit
(electrons!)
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Thomas Oberst
Cornell University
Sample Circuit:
11W
11W
12V
11W
11W
11W
•Measured resistance of each light bulb to be 11 Ohms
•Will apply a fixed voltage of 12V across the terminals
Thomas Oberst
Cornell University
When I connect the 12V, which light
bulb will light the brightest?
Which light bulb will light the
dimmest?
Rate the bulbs in order from brightest to
dimmest?
Thomas Oberst
Cornell University
Let’s find out if you’re guess is correct:
What is the resistance of each branch of the
circuit?
Branch 1 (bulbs 2 & 5):
R = 11 + 11 = 22 W
Branch 2 (bulbs 1, 3, & 4):
R = 11 + 1/(1/11 + 1/11)
= 11 + 1/(2/11) = 11 + 11/2 = 11 + 5.5
= 16.5 W
Thomas Oberst
Cornell University
What is the current flowing through each
branch of the circuit?
Branch 1 (bulbs 2 & 5):
V = IR
I = V/R = 12/22 = 0.54 A
Branch 2 (bulbs 1, 3, & 4):
I = V/R = 12/16.5 = 0.73 A
Thomas Oberst
Cornell University
Therefore, what must be the TOTAL current
flowing through the circuit?
I = I1 + I2 = 0.54 + 0.73 = 1.27 A
However, if I actually measure the current
using an Ammeter, I get…
I = 0.15 A
Thomas Oberst
Cornell University
That would mean that the light bulbs each have a
resistance of:
Vtot = Itot Rtot
12 = 0.15 [ 1/(1/(R branch 1) + 1/(R branch 2)) ]
= 0.15 [ 1/(1/(R+R) + 1/(R + 1/(1/R + 1/R))) ]
= 0.15 [ 1/(1/2R + 1/(R + 1/(2/R))) ]
= 0.15 [ 1/(1/2R + 1/(R + R/2)) ]
= 0.15 [ 1/(1/2R + 1/(3R/2)) ]
= 0.15 [ 1/(1/2R + 2/3R)]
= 0.15 [6R/7]
Or, R = (12 x 7)/(0.15 x 6) = 93 W
Thomas Oberst
Cornell University
What the heck is going on?
Thomas Oberst
Cornell University
When I measured the resistance of the light
bulbs, what temperature were they at?
Did the temperature of the bulbs change when
I turned on the 12V power supply?
Is resistance dependant on temperature?
Thomas Oberst
Cornell University
In fact, resistance is related to temperature by
the equation:
R(T) = R(T0) [1+ a DT ]
•Where DT = T – T0 is the difference between two
temperatures
•And a is the “temperature coefficient of resistivity”
and depends on the material that the light bulb
filaments are made of.
•All regular light bulbs have tungsten filaments, with
a = 0.0045 oC-1
Thomas Oberst
Cornell University
Can we use this information to actually calculate the
temperature of the light bulbs?
Recall that I measured R = 11W at room
temperature and I calculated that R must be
93W when the bulbs are lit (we will assume,
for simplicity, that all the bulbs are the same
temperature, which is not true because they
are lit at different brightnesses).
Thomas Oberst
R(T) = R(T0) [ 1 + a DT]
93 = 11 [ 1 + 0.0045 (Thigh – 23)]
Thigh = [ (93/11 – 1) / 0.0045 ] + 23
Thigh = 1679 oC
This is the temperature of
the filament, not of the
glass bulb surrounding the
filament!
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Thomas Oberst
Cornell University
Now that we know the correct resistance (93W) of each bulb,
back to figuring out why #1 was the brightest…
Thomas Oberst
Cornell University
What is the resistance of each branch of the
circuit?
Branch 1 (bulbs 2 & 5):
11 + 11
R = 93
93 = 22
186WW
Branch 2 (bulbs 1, 3, & 4):
R = 93
11 + 1/(1/11
1/(1/93 + 1/11)
1/93)
= 93
11 + 1/(2/11)
1/(2/93) = 11
93 ++ 11/2
93/2 == 11
93 ++ 5.5
46.5
= 149.5
16.5 WW
Thomas Oberst
Cornell University
What is the current flowing through each
branch of the circuit?
Branch 1 (bulbs 2 & 5):
V = IR
I = V/R = 12/22
12/186==0.54
0.065A
A
Branch 2 (bulbs 1, 3, & 4):
I = V/R = 12/139.5
12/16.5 ==0.73
0.086
AA
Thomas Oberst
Cornell University
Therefore, what must be the TOTAL current
flowing through the circuit?
I = I1 + I2 = 0.065
0.54 ++0.73
0.086
= 1.27
= 0.15
AA
However, if I actually measure the current
using an Ammeter, I get…
I = 0.15 A
Thomas Oberst
Cornell University
So we know the total current, and the current
flowing through each branch…
Now what is the current flowing through each
light bulb?
#2?
#5?
#1?
#3?
#4?
0.065 A
0.065 A
0.086 A
0.043 A
0.043 A
Thomas Oberst
Cornell University
What happens if I remove bulb #4?
Is the current going through all four bulbs the
same? What is it?
0.065 A
Is the total current still the same?
What is it?
2 x 0.065 = 0.13 A
Thomas Oberst
Cornell University
What happens if I remove bulb #5 (after replacing bulb
#4)?
Why did branch 1 go out?
Why didn’t branch 2 get brighter?!?!
(i.e., what happened to the current that was
previously going through branch 1?)
Thomas Oberst
Cornell University
To understand the answer, you must realize
that we are holding the total voltage fixed at
12V, and by taking out bulbs #2 and #5 I
changed the total resistance.
Thomas Oberst
Cornell University
So with V fixed and R being changed, I HAS
to change (recall V=IR). So it does not remain
0.15 A as it has been up until now.
Therefore, the current from branch 1 doesn’t
need to “go” anywhere. The total current of
the whole circuit has simply dropped.
(By an amount exactly equal to the current
that used to be in branch 1).
Thomas Oberst
Cornell University
The
End