October 15, 2008

DC Circuits

This is the week that will have beenToday

Complete Resistance/Current with some problemsFriday

Examination #2: Potential Resistance & CurrentNext Week: DC CircuitsNext Topic: MagnetismNext Quiz: One week from Friday

The same old closed circuit

Note change in notation: VE

IER

ERIP

A

LR

A

IJ

IRE

22

The figure below gives the electrical potential V(x) along a copper wire carrying a uniform current, from a point at higher potential (x=0m) to a point at a lower potential (x=3m). The wire

has a radius of 2.45 mm. What is the current in the wire?

copper

12 volts 0 volts

What does the graph tell us??

*The length of the wire is 3 meters.*The potential difference across the

wire is 12 volts.*The wire is uniform.

Let’s get rid of the mm radius and convert it to area in square meters:A=r2 = 3.14159 x 2.452 x 10-6 m2

orA=1.9 x 10-5 m 2

Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters

We have all we need….

ma 49.41067.2

1012

R

Vi

:Law sOhm' From

67.2 109.1

0.3m-ohm 1069.1

3

6

5

8

ohms

volts

mx

mx

A

LR

When the potential difference across a certain conductor is doubled, the current is observed to increase by a factor of three. What can you conclude about the conductor?

A. It is a perfect conductor.B. It does not obey Ohm's

Law.C. It is a semiconductor.D. It obeys Ohm's Law.

Two conductors of the same length and radius are connected across the same potential difference. One conductor has twice the resistance of the other. To which conductor is more power delivered?

A. conductor with lower resistanceB. conductor with higher resistanceC. Equal amount of power delivered

to both conductors.

An electric utility company supplies a customer's house from the main power lines (120 V) with two copper wires, each of which is 50.0 m long and has a resistance of 0.108 Ω per 300 m.

(a) Find the potential difference at the customer's house for a load current of 110 A. 116 V

(b) For this load current, find the power delivered to the customer. 12.8 kW

(c) Find the rate at which internal energy is produced in the copper wires.436 W

A toaster is rated at 780 W when connected to a 240-V source.

What current does the toaster carry?3.25 AWhat is its resistance?73.8

This material will NOT be on the exam

DC CIRCUITSLet’s add resistors …….

Series Combinations

iiRseriesR

general

RRR

iRiRiRVVV

and

iRV

iRV

)(

:21

2121

22

11

R1 R2

i i

V1 V2V

SERIES Resistors

The rod in the figure is made of two materials. The figure is not drawn to scale. Each conductor has a square cross section 3.00 mm on a side. The first material has a resistivity of 4.00 × 10–3 Ω · m and is 25.0 cm long, while the second material has a resistivity of 6.00 × 10–3 Ω · m and is 40.0 cm long. What is the resistance between the ends of the rod?

Parallel Combination??

i iRR

general

RRR

so

R

V

R

V

R

Viii

iRV

11

111

..

21

2121

R1, I1

R2, I2

V

What’s This???In this Figure, find the equivalent resistance between points (a) F and H and [2.5]  (b) F and G. [3.13]  

(a) Find the equivalent resistance between points a and b in the Figure.

(b) A potential difference of 34.0 V is applied between points a and b. Calculate the current in each resistor.

Power Source in a Circuit

The ideal battery does work on charges moving them (inside) from a lower potential to one that is V higher.

A REAL Power Sourceis NOT an ideal battery

V

E or Emf is an idealized device that does an amount of work E to move a unit charge from one side to another.

By the way …. this is called a circuit!

Internal Resistance

A Physical (Real) Battery

Rr

Emfi

Internal Resistance

Back to which is brighter? (R1=R2)

Back to Potential

Represents a charge in space

Change in potential as one circuitsthis complete circuit is ZERO!

Consider a “circuit”.

This trip around the circuit is the same as a path through space.

THE CHANGE IN POTENTIAL FROM “a” AROUND THE CIRCUIT AND BACK TO “a” is ZERO!!

To rememberIn a real circuit, we can neglect the

resistance of the wires compared to the resistors.We can therefore consider a wire in a circuit to be

an equipotential – the change in potential over its length is slight compared to that in a resistor

A resistor allows current to flow from a high potential to a lower potential.

The energy needed to do this is supplied by the battery.

VqW

NEW LAWS PASSED BY THIS SESSION OF THE FLORIDUH LEGISLATURE.LOOP EQUATION

The sum of the voltage drops (or rises) as one completely travels through a circuit loop is zero.

Sometimes known as Kirchoff’s loop equation.

NODE EQUATIONThe sum of the currents entering (or leaving)

a node in a circuit is ZERO

TWO resistors again

jj

21

21

RR

Resistors SERIESfor General

RRR

or

iRiRiRV

i

R1 R2

V1 V2

V

A single “real” resistor can be modeledas follows:

R

a b

V

position

ADD ENOUGH RESISTORS, MAKING THEM SMALLERAND YOU MODEL A CONTINUOUS VOLTAGE DROP.

We start at a point in the circuit and travel around until we get back to where we started.

If the potential rises … well it is a rise.If it falls it is a fall OR a negative rise.We can traverse the circuit adding each rise

or drop in potential.The sum of all the rises around the loop is

zero. A drop is a negative rise.The sum of all the drops around a circuit is

zero. A rise is a negative drop.Your choice … rises or drops. But you must

remain consistent.

Take a trip around this circuit.

Consider voltage DROPS:

-E +ir +iR = 0or

E=ir + iRrise

Circuit Reduction

i=E/Req

Multiple Batteries

Reduction

Computes i

Another Reduction Example

PARALLEL

1212

1

600

50

30

1

20

11

RR

START by assuming a DIRECTION for each Current

Let’s write the equations.

In the figure, all the resistors have a resistance of 4.0 and all the (ideal) batteries have an emf of 4.0 V. What is the current through resistor R?

The Unthinkable ….

RC Circuit Initially, no current

through the circuit Close switch at (a) and

current begins to flow until the capacitor is fully charged.

If capacitor is charged and switch is switched to (b) discharge will follow.

Close the Switch

I need to use E for E

Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)

Really Close the Switch

I need to use E for E

R

E

RC

q

dt

dq

or

EC

q

dt

dqR

C

qiRE

dt

dqi since

0

Equation Loop

Note RC = (Volts/Amp)(Coul/Volt) = Coul/(Coul/sec) = (1/sec)

This is a differential equation.

To solve we need what is called a particular solution as well as a general solution.

We often do this by creative “guessing” and then matching the guess to reality.

You may or may not have studied this topic … but you WILL!

RC

REaeCE

R

E

RC

q

dt

dq

CEq

R

E

RC

q

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at

p

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E/R0CEa

0for t

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KCE0

solution from and 0q 0,When t

and 0dq/dt charged,fully is device When the

:solution particularat Look

Solution General

at-

at-

Time Constant

RC

Result q=CE(1-e-t/RC)

q=CE(1-e-t/RC) and i=(CE/RC) e-t/RC

RCteR

Ei /

Discharging a Capacitor

qinitial=CE BIG SURPRISE! (Q=CV)i

iR+q/C=0

RCt

RCt

eRC

q

dt

dqi

eqq

solutionC

q

dt

dqR

/0

/0

0

In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase of thermal energy Eth in the resistor as a function of time t.

(a)What is the electric potential across the battery? (60)(b) If the resistance is doubled, what is the POWER dissipated by the circuit? (39)(c) Did you put your name on your paper? (1)

Looking at the graph, we see that theresistor dissipates 0.5 mJ in one second.

Therefore, the POWER =i2R=0.5 mW

ma 88.41088.4

1038.2Ω 21

mW 0.5

3

252

ampi

ampR

Pi

mVamp 10221104.88iRV

or iR reisitor theacross drop Voltage3-

If the resistance is doubled what is the power dissipated by the circuit?

mJRiP

ma

R

248.0

43.242

10102

R

Vi

mV 102V 42

2

3