PHY-2049 Current & Circuits February ‘08. A closed circuit Hot, Hot Hot.

PHY-2049

Current & CircuitsFebruary ‘08

A closed circuit

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Power in DC Circuit

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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….

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Let’s add resistors …….

Series CombinationsR1 R2

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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??

R1, I1

R2, I2

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What’s This???

In Fig. 28-39, 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 Figure P28.6. (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

Internal Resistance Rr

Emfi

Back to which is brighter?

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 remember

In 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.

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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 EQUATION The sum of the currents entering (or

leaving) a node in a circuit is ZERO

TWO resistors againi

R1 R2

V1 V2

V

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Resistors SERIESfor General

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or

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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

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START by assuming a DIRECTION for each Current

Let’s write the equations.

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!

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

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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

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eqq

solutionC

q

dt

dqR

/0

/0

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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

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

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