Post on 14-Dec-2015
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Series Circuits:
Other examples:
Series circuits - ________________________________________ _________________________________________
Assume:
1. _____________________________________________________
2. _____________________________________________________
3. _____________________________________________________
4. _____________________________________________________
Wires have no potential drop (voltage) across them
Pos. current is out of the “high” voltage side
circuitelement ___
voltage source circuit
element ___
circuitelement ___
wire
have only 1 path for current
+________potential
________potential
-
wire
wirewire
All energy provided by source is used in the elements
No charge is “lost.” All current returns to the source.
high
low
I
I
1
2
could have switches, etc.
3
V V1
V2
I
I1 R1
R2
I2
_______________ Conservation: V =
_______________ Conservation: I =
_______________ (Total) Resistance: Req =
__________ Law applies to the total: V = and to each individual element: V1 =
V2 =
For a circuitwith 2 resistors:
Energy V1 + V2
Charge I1 = I2 Equivalent R1 + R2
Ohm’s I Req
I1R1
I2R2
Ohm’s Law
V = IR
I = V/R
Ex. Find all the voltages and currents in the circuit below:
20. V 40
120
V1 = I1 = R1 =
V2 = I2 = R2 =
V = I = Req =20. V
40
120
160 20/160 A
0.125 A
0.125 A5.0 V
15.0 V
0.125 A
Req = R1 + R2
V1 = I1R1
I = V/Req
I = I1 = I2
V2 = I2R2
V = 20. V
40
120
•V “divides up” ______________________________ as the R’s•This is because ___________ R requires _________ energy.•Series circuits are _______________________________.
in the same proportion
voltage dividers
Form the _________ of each resistance to Req = _______ ,and then multiply by the __________ voltage V
40
160 x 20 V = 5 V
120
160 x 20 V = 15 V
160
R1
Req
R2
Req
ratiototal
more more
Plot V vs. “distance around circuit.”
20
15
wire V dropped across the ______ resistor
wire
40
wire0
back to ____side of thebattery
______ drop across wires because we assume _________
distance around circuit
No R = 0
potentialdifference
(V)
____ side ofbattery
+
V dropped across the ______ R160 at the ___
side of thebattery
+
-
Important:
“I is ______________ everywhere in ___________ circuit” does
NOT mean that I is ___________ in _________________ circuit!
10. V R1=
R2 =
I=
I1=
I2=
10. V R2=
R3=
I=
I1=
I2=
I3 =
I is still the _______________ in all parts of the second circuit, but it is a ________________ I than the first one!
the same a series
the same every other
10/100 = .1 A
.1 A
.1 A
10/200 = .05 A
.05 A
.05 A
samedifferent
R1=
75 25
100 25
75 .05 A
Equivalent resistance: _________________________________
________________________________________________________
The total I =
20. V40
120
If you replace the resistors of a circuit with one resistor, the total I would be the same
Replacing this part of thecircuit with a single_______________ resistor:
Req = R1 + R2 =
=
40 + 120
equivalent
160
…gives you this circuit:
Req =160
20. V
VReq
= 20 V 160
= 0.125 A
This is the ____________ as before.same I
50 150
V = 20 V
70
20
16 8
V = 12 V
10
20
All _______________ circuits can be ___________________ in this way.
This can be done even if the ______________________ is not shown.
simplifiedseries
voltage source
15 5
Req =
_____ V = 20 V
V = 12 V
Req results in the _____________ as the _________________ circuit.
A.
B.
C.
D.
200
Req =
_____ 24
Req =
_____ 90
Req =
_____ 50
same I original
__________ Hookups:
V R1
R2
Original circuit:
To measure I1, the current through R1, _________________ the circuit and _____________ an ________________ next to R1
Meter
ammeterdisconnect
insert
V R1
R2
V R1
R2
A
Other possibilities: V R1
R2
V R1
R2
A A
___ is the same everywhere, so _________________________I
disconnect insert
anywhere gives same I.
To measure, V1, the voltage across R1, __________disconnect the circuit. Simply connect the ______________ across R1
voltmeterdo NOT
Other possibilities:
V R1
R2
V R1
R2
VOriginal circuit:
V R1
R2
V V R1
R2
V
Similarly, to measure the _________ voltage V or V2:
V R1
R2
V
V
total
V R1
R2