Review Exam 4

Chapter 18

Electric Currents

Resistance and Resistors

The ratio of voltage to current is called the resistance:

(18-2a)

(18-2b)

18.4 Resistivity

The constant ρ, the resistivity, is characteristic of the material.

18.4 Resistivity

18.4 ResistivityFor any given material, the resistivity changes with temperature:

Since

00T T-T 1R R

R0 = Resistance at some reference temperature (T0)

α = temperature coefficient and R = resistance at some new temperature T

R ,T if negative If

R ,T if positive, If

18.5 Electric Power

18.5 Electric Power

The unit of power is the watt, W.

1Watt = 1 Joule/second

Chapter 19

DC Circuits

Conservation Laws• Conservation of Charge (I = Q/t) and I

• Conservation of Energy (PE = QV) and V

• Conservative Forces (pages 148-149)

• ∑ of Work around a closed path = 0

• Independent of Path

• For a closed loop ∑ of PE gained and lost or ∑ of V gained and lost = 0

Rε+

-

I

V = IR

Simple Circuit

ε-V = 0ε –IR =0ε = IR

Resistors in Series

From this we get the equivalent resistance (that single resistance that gives the same current in the circuit).

Resistors in Parallel

This gives the reciprocal of the equivalent resistance:

(19-4)

Repeat this Mantra

• The current is the same for elements connected in series

• The voltage is the same for elements connected in parallel

Multiple Loop Game Rules• Draw picture

• Define number of loops

• Pick ARBITRARY directions for the currents

• Indentify branch points (where currents divide)

• Currents into branch points = currents out of branch points

• Circle each loop and ∑ V’s = 0

How to circle a loop• Start at some ARBITRARY point

• Circle clockwise or counterclockwise (your choice)

• End at starting point

When you come toa Battery as you circlethe loop

ε

How to Add V’s

Direction you take - ε

+ ε

When you come toa Resistor as you circlethe loop you will travelwith or against the current

I

- IR

R

+ IR

19.5 Circuits Containing Capacitors in Series and in Parallel

In this case, the total capacitance is the sum:

(19-5)

19.5 Circuits Containing Capacitors in Series and in Parallel

In this case, the reciprocals of the capacitances add to give the reciprocal of the equivalent capacitance:

(19-6)

Ammeters – Very small resistance so voltage across it is very small

In series

Voltmeters - Very large resistance, takes away little current.

In Parallel

What would happen if I put an Ammeter in Parallel?

A

ConcepTest 19.8ConcepTest 19.8 Kirchhoff’s RulesKirchhoff’s Rules

The lightbulbs in the The lightbulbs in the

circuit are circuit are identicalidentical. .

When the switch is When the switch is

closed, what happens?closed, what happens?

1) both bulbs go out

2) intensity of both bulbs increases

3) intensity of both bulbs decreases

4) A gets brighter and B gets dimmer

5) nothing changes

When the switch is open, the point the point

between the bulbs is at 12 Vbetween the bulbs is at 12 V. But so But so

is the point between the batteriesis the point between the batteries. If

there is no potential difference, then

no current will flow once the switch is

closed!! Thus, nothing changes.

The lightbulbs in the The lightbulbs in the

circuit are circuit are identicalidentical. .

When the switch is When the switch is

closed, what happens?closed, what happens?

1) both bulbs go out

2) intensity of both bulbs increases

3) intensity of both bulbs decreases

4) A gets brighter and B gets dimmer

5) nothing changes

ConcepTest 19.8ConcepTest 19.8 Kirchhoff’s RulesKirchhoff’s Rules

24 VFollow-up:Follow-up: What happens if the bottom battery is replaced by a 24 V battery?

ConcepTest 19.10ConcepTest 19.10 More Kirchhoff’s RulesKirchhoff’s Rules

2 V

2

2 V 6 V

4 V

3 1

1

I1 I3

I2

Which of the equations is valid Which of the equations is valid

for the circuit below? for the circuit below?

1) 1) 2 – I2 – I11 – 2I – 2I22 = 0 = 0

2) 2) 2 – 2I2 – 2I11 – 2I – 2I2 2 – 4I– 4I3 3 = 0= 0

3) 3) 2 – I2 – I11 – 4 – 2I – 4 – 2I2 2 = 0= 0

4) 4) II33 – 4 – 2I – 4 – 2I2 2 + 6 + 6 = 0= 0

5) 5) 2 – I2 – I11 – 3I – 3I3 3 – 6– 6 = 0= 0

ConcepTest 19.10 ConcepTest 19.10 More Kirchhoff’s RulesKirchhoff’s Rules

2 V

2

2 V 6 V

4 V

3 1

1

I1 I3

I2

Eqn. 3 is valid for the left loopEqn. 3 is valid for the left loop:

The left battery gives +2V, then

there is a drop through a 1

resistor with current I1 flowing.

Then we go through the middle

battery (but from + to – !), which

gives –4V. Finally, there is a

drop through a 2 resistor with

current I2.

Which of the equations is valid Which of the equations is valid

for the circuit below? for the circuit below?

1) 1) 2 – I2 – I11 – 2I – 2I22 = 0 = 0

2) 2) 2 – 2I2 – 2I11 – 2I – 2I2 2 – 4I– 4I3 3 = 0= 0

3) 3) 2 – I2 – I11 – 4 – 2I – 4 – 2I2 2 = 0= 0

4) 4) II33 – 4 – 2I – 4 – 2I2 2 + 6 + 6 = 0= 0

5) 5) 2 – I2 – I11 – 3I – 3I3 3 – 6– 6 = 0= 0