October 15, 2008 DC Circuits

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

The same old closed circuit

Note change in notation: V  E

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=  r 2 = x x m 2 or A=1.9 x m 2 Material is Copper so resistivity is (from table) = 1.69 x ohm meters

We have all we need….

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 resistance B. conductor with higher resistance C. 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 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 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 A What is its resistance? 73.8

This material will NOT be on the exam

DC CIRCUITS Let’s add resistors …….

Series Combinations R1 R2 ii V1 V2 V 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 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 Source is NOT an ideal battery V E or E mf 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

Back to which is brighter? (R 1 =R 2 )

Back to Potential Represents a charge in space 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

TWO resistors again i R 1 R 2 V 1 V 2 V

A single “real” resistor can be modeled as follows: R a b 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.

Take a trip around this circuit. Consider voltage DROPS : - E +ir +iR = 0 or E =ir + iR rise

Circuit Reduction i= E /R eq

Multiple Batteries

Reduction Computes i

Another Reduction Example PARALLEL

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 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!

Time Constant

Result q=C E (1-e -t/RC )

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

Discharging a Capacitor q initial =CE BIG SURPRISE! (Q=CV) i iR+q/C=0

In Fig. (a), a R = 21, Ohm a resistor is connected to a battery. Figure (b) shows the increase of thermal energy E th 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 the resistor dissipates 0.5 mJ in one second. Therefore, the POWER =i 2 R=0.5 mW

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