Transcript Series Circuits - PHS Regents Physics

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