DC Circuits

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DC Circuits. October 15, 2008. This is the week that will have been. Today Complete Resistance/Current with some problems Friday Examination #2: Potential  Resistance & Current Next Week: DC Circuits Next Topic: Magnetism Next Quiz: One week from Friday. Good Luck on the Exam.

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DC Circuits

October 15, 2008

This is the week that will have been
• Today
• Complete Resistance/Current with some problems
• Friday
• Examination #2: Potential  Resistance & Current
• Next Week: DC Circuits
• Next Topic: Magnetism
• Next Quiz: One week from Friday

Good Luck on the Exam

copper

12 volts 0 volts

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?

What does the graph tell us??

*The length of the wire is 3 meters.

*The potential difference across the

wire is 12 m volts.

*The wire is uniform.

Let’s get rid of the mm radius and

convert it to area in square meters:

A=pr2 = 3.14159 x 2.452 x 10-6 m2

or

A=1.9 x 10-5 m 2

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

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?
• It is a perfect conductor.
• It does not obey Ohm's Law.
• It is a semiconductor.
• 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?
• conductor with lower resistance
• conductor with higher resistance
• 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

DC CIRCUITSLet’s add resistors …….

R1 R2

V1 V2

V

i

i

Series Combinations

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?
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]

?

• (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.

V

By the way …. this is called a circuit!

A REAL Power Sourceis NOT an ideal battery

Internal Resistance

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

A Physical (Real) Battery

Internal Resistance

Represents a charge in spaceBack to Potential

Change in potential as one circuits

this 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.
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
A single “real” resistor can be modeledas follows:

R

b

a

V

position

ADD ENOUGH RESISTORS, MAKING THEM SMALLER

AND 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.
Multiple Batteries

Watch the DIRECTION !!

Reduction

Computes i

In the figure, all the resistors have a resistance of 4.0 W and all the (ideal)

batteries have an emf of 4.0 V. What is the current through resistor R?

The Unthinkable ….

Resistors and Capacitors in the same circuit??

Is this cruel or what??

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.

How Fast ?

What do you

think will

happen when

we close the

swutch?

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

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!
Discharging a Capacitor

qinitial=CE BIG SURPRISE! (Q=CV)

i

iR+q/C=0

In Fig. (a), a R = 21, Ohma resistor is connected to a battery. Figure (b) shows the increase
• of thermal energy Eth in the resistor as a function of time t.
• What is the electric potential across the battery? (60)
• If the resistance is doubled, what is the POWER dissipated by the circuit? (39)
• Did you put your name on your paper? (1)

Looking at the graph, we see that the

resistor dissipates 0.5 mJ in one second.

Therefore, the POWER =i2R=0.5 mW