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