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Today 3/27
Circuits
Current
Potential (same as always)
HW:
3/27 “Circuits 2”
Due Monday 3/31
Rank the bulbs from brightest to
dimmest. Are any equally bright?
D
1
A
2
B
C
3
E
A=B=C >D=E
Play “Current Show”
powerpoint presentation
Rank the bulbs from brightest to
dimmest. Are any equally bright?
D
1
A
2
B
C
3
E
A=B=C >D=E
Rank the bulbs for current.
A=B=C>D=E
Rank the batteries for current.
2>1>3
Current is charge in
motion, = “how many
coulombs per second”
pass by
Rank the bulbs from brightest to
dimmest. Are any equally bright?
A
A gets all the current while B and C each
only get part so A is brightest.
B
C
At the junction the current divides 50/50
since both branches have the same
resistance.
Rank the bulbs for
current.
A>B=C
Rank the bulbs for More flow, more glow.
brightness.
A>B=C
Rank the bulbs from brightest to
dimmest. Are any equally bright?
Since branch D-E has mor R
than branch F (extra
something added in series) the
current does not divide 50/50.
Current favors the path of
least R. Beware of the words
“takes the path…” as it
implies none goes through
D-E.
Current for D = current for E
as all that goes through D also
goes through E. (See
“Current Show.”)
D
F
E
Rank the bulbs for
current.
F>D=E
Rank the bulbs for
brightness.
F>D=E
Networks
A
D
B
C
F
E
Which circuit has the least resistance and hence the greatest
current through the battery?
Added in...
Series
Series addition increases R,
less current
Added in...
Parallel
Parallel addition decreases R,
more current
Rank for resistance
A
B
C
Most R
B
C
D
Least R
A
D
E
E
Compared to A, B and
C have an extra clog
on an existing path.
Compared to A, D and
E have an extra path.
Conceptual Circuits (Current)
More flow-more glow
What goes around, comes around
Current divides at junctions
Resistance inhibits current
add something in series = more R
add something in parallel = less R
Think in terms of “networks.”
Voltage Model
Voltage: how much energy each coulomb of
charge gains (battery) or loses (bulb) in
going through an element. (as usual)
More glow, more voltage.
The Loop Rule:
What goes up, must come down! VA,A = 0,
voltage rises and drops must must cancel
around any loop.
Current Model Fails
A
Compare B to D
B
less
current
C
1/
more D
current
2
all
B gets all of the current
through A. D gets half of the
current through C.
But the current through C is
bigger
than
the
current
throug
1/
2
E A.
Can’t be sure which is
brighter!!!
Voltage Model
A
V
think
loops
V
VA
VB
B
Compare B to D
Compare the voltages
across B and D. More
volts more glow! Recall
current for B and D.
C
VC
D
VD
C brighter than D
E C has more voltage than D
Voltage Model
A
in series
The voltage for series
elements SPLITS
because it adds up to
B the battery voltage.
The voltage splits
equally for identical Rs.
Otherwise, the voltage
drop is greater across
the greater R.
Find Equivalent Resistance
5
6
12 V
3
2
6
8 4 8
12 V
12
7
12
9
Req = 6
?
Find current through the battery
2
5
12 V
I = 2 Amps
6
8
8
12 V
7
9
It is the same as the
current through Req.
Req = 6
6
Current through each element
0.5 Amp
5
12 V
2 Amps
0.5 Amp
2
I = 2 Amps
6
8
8
12 V
1 Amp
7
0.25 Amp
each
1 Amp 9
What is the current through each resistor?
What is the voltage across each resistor?
6
Voltage across each element
12 V
5
5V
1V
2
3V
6
8
8
I = 2 Amps
12 V
2 V each
7
7V
9V
9
What is the current through each resistor?
What is the voltage across each resistor?
6
Kirchhoff’s Rules, Loop
The sum of all voltages around any
closed loop is zero. (what goes up must
come down)
or VA,A = 0
!!must keep track of ups and downs!! (+/-)
Kirchhoff’s Rules, Junction
The sum of all currents at any junction
is zero. (what goes in must come out)
!!must keep track of ins and outs!! (+/-)