 # My Chapter 18 Lecture Outline.

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My Chapter 18 Lecture Outline

Chapter 18: Electric Current and Circuits
EMF Current & Drift Velocity Resistance & Resistivity Kirchhoff’s Rules Series & Parallel Circuit Elements Applications of Kichhoff’s Rules Power & Energy Ammeters & Voltmeters RC Circuits

§18.1 Electric Current A metal wire.
Assume electrons flow to the right. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 18, Questions 1 and 14. Current is a measure of the amount of charge that passes though an area perpendicular to the flow of charge. Current units: 1C/sec = 1 amp

A current will flow until there is no potential difference.
The direction of current flow in a wire is opposite the flow of the electrons. (In the previous drawing the current is to the left.)

Example: If a current of 80
Example: If a current of 80.0 mA exists in a metal wire, how many electrons flow past a given cross-section of the wire in 10.0 minutes?

§18.2 EMF and Circuits An ideal battery maintains a constant potential difference. This potential difference is called the battery’s EMF(). Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 18, Question 10. The work done by an ideal battery in pumping a charge q is W = q.

The circuit symbol for a battery (EMF source) is
+ At high potential At low potential The circuit symbol for a battery (EMF source) is Batteries do work by converting chemical energy into electrical energy. A battery dies when it can no longer sustain its chemical reactions and so can do no more work to move charges.

§18.3 Microscopic View of Current in a Metal
Electrons in a metal might have a speed of ~106 m/s, but since the direction of travel is random, an electron has vdrift = 0.

Only when the ends of a wire are at different potentials (E  0) will there be a net flow of electrons along the wire (vdrift  0). Typically, vdrift < 1 mm/sec.

The current in the wire is:
Calculate the number of charges (Ne) that pass through the shaded region in a time t: l Symbol definitions: n is the density of charge carriers and l is the length of the shaded region (with volume Al). The current in the wire is:

Example (text problem 18. 19): A copper wire of cross-sectional area 1
Example (text problem 18.19): A copper wire of cross-sectional area 1.00 mm2 has a constant current of 2.0 A flowing along its length. What is the drift speed of the conduction electrons? Copper has 1.101029 electrons/m3. To simplify the calculation I used the given information in the printed problem to find the density of electrons in copper and supply this number as a given.

§18.4 Resistance and Resistivity
A material is considered ohmic if VI, where Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 18, Questions 2, 3, 4, 5, and 16. The proportionality constant R is called resistance and is measured in ohms (; and 1  = 1 V/A).

The resistance of a conductor is:
where  is the resistivity of the material, L is the length of the conductor, and A is its cross sectional area. With R a material is considered a conductor if  is “small” and an insulator if  is “large”.

The resistivity of a material depends on its temperature:
where 0 is the resistivity at the temperature T0, and  is the temperature coefficient of resistivity. A material is called a superconductor if  = 0.

Example (text problem 18. 28): The resistance of a conductor is 19
Example (text problem 18.28): The resistance of a conductor is 19.8  at 15.0 C and 25.0  at 85.0 C. What is the temperature coefficient of resistivity? Values of R are given at different temperatures, not values of . But the two quantities are related. Multiply both sides of equation (2) by L/A and use equation (1) to get:

Solve equation (3) for  and evaluate using the given quantities:
Example continued: Solve equation (3) for  and evaluate using the given quantities:

§18.5 Kirchhoff’s Rules Junction rule: The current that flows into a junction is the same as the current that flows out. (Charge is conserved) Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 18, Questions 6, 7, and 17. A junction is a place where two or more wires (or other components) meet. Loop rule: The sum of the voltage dropped around a closed loop is zero. (Energy is conserved.)

For a resistor: If you cross a resistor in the direction of the current flow, the voltage drops by an amount IR (write as IR). There is a voltage rise if you cross the other way (write as +IR). B A If the current flows from A to B, then the potential decreases from A to B. The potential difference between A and B is < 0 (V = IR) . I

For batteries (or other sources of EMF): If you move from the positive to the negative terminal the potential drops by  (write as ). The potential rises if you cross in the other direction (write as +). + At high potential At low potential

A current will only flow around a closed loop.
B VAB is the terminal voltage. Applying the loop rule:

In a circuit, if the current always flows in the same direction it is called a direct current (DC) circuit.

§18.6 Series and Parallel Circuits
Resistors: The current through the two resistors is the same. It is not “used up” as it flows around the circuit! Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 18, Questions 8, 9, 18, and 19. These resistors are in series. Apply Kirchhoff’s loop rule:

The pair of resistors R1 and R2 can be replaced with a single equivalent resistor provided that Req = R1 + R2. In general, for resistors in series

Current only flows around closed loops
Current only flows around closed loops. When the current reaches point A it splits into two currents. R1 and R2 do not have the same current through them, they are in parallel. Apply Kirchhoff’s loop rule: The potential drop across each resistor is the same.

Applying the junction rule at A: I = I1+ I2.
From the loop rules: Substituting for I1 and I2 in the junction rule:

The pair of resistors R1 and R2 can be replaced with a single equivalent resistor provided that
In general, for resistors in parallel

R2 and R3 are in parallel. Replace with an equivalent resistor R23.
Example (text problem 18.40): In the given circuit, what is the total resistance between points A and B? R3 = 24  R2 = 12  R1 = 15 A B Only part a is done here. R2 and R3 are in parallel. Replace with an equivalent resistor R23.

The circuit can now be redrawn:
Example continued: The circuit can now be redrawn: R23 = 8  R1 = 15 A B The resistors R23 and R1 are in series: B R123 = 23  A Is the equivalent circuit and the total resistance is 23 .

For capacitors in series the charge on the plates is the same.
For capacitors in series the charge on the plates is the same. Apply Kirchhoff’s loop rule:

The pair of capacitors C1 and C2 can be replaced with a single equivalent capacitor provided that
In general, for capacitors in series

Apply Kirchhoff’s loop rule:
Apply Kirchhoff’s loop rule: For capacitors in parallel the charge on the plates may be different. Here

The pair of capacitors C1 and C2 can be replaced with a single equivalent capacitor provided that Ceq= C1 + C2. In general, for capacitors in parallel

Example (text problem 18.49): Find the value of a single capacitor that replaces the three in the circuit below if C1 = C2 = C3 = 12 F. C2 C1 C3 B A C2 and C3 are in parallel Only part of problem is done here.

The circuit can be redrawn:
Example continued: The circuit can be redrawn: C1 C23 B A The remaining two capacitors are in series. C123 B A Is the final, equivalent circuit.

§18.8 Power and Energy in Circuits
The energy dissipation rate is: Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 18, Questions 11, 15, and 20. For an EMF source: For a resistor:

Resistance () Current (A) Power (W) 122 0.199 4.83 5.6 0.123 0.0847
Example: Use the results of the example starting on slide 35 to determine the power dissipated by the three resistors in that circuit. Resistance () Current (A) Power (W) 122 0.199 4.83 5.6 0.123 0.0847 75 0.0760 0.433

§18.9 Measuring Currents and Voltages
Current is measured with an ammeter. An ammeter is placed in series with a circuit component. R2 R1 A1 A3 A2 A1 measures the current through R1. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 18, Questions 1 and 14. An ammeter has a low internal resistance. A2 measures the current through R2. A3 measures the current drawn from the EMF.

The voltmeter measures the voltage drop across R1.
A voltmeter is used to measure the potential drop across a circuit element. It is placed in parallel with the component. A voltmeter has a large internal resistance. R2 R1 V The voltmeter measures the voltage drop across R1.

§18.10 RC Circuits C R Switch + Close the switch at t = 0 to start the flow of current. The capacitor is being charged. Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 18, Questions 12 and 13. Apply Kirchhoff’s loop rule:

The current I(t) that satisfies Kirchhoff’s loop rule is:
where  is the RC time constant and is a measure of the charge (and discharge) rate of a capacitor.

The voltage drop across the capacitor is:
The voltage drop across the resistor is: The charge on the capacitor is: Note: Kirchhoff’s loop rule must be satisfied for all times.

Plots of the voltage drop across the (charging) capacitor and current in the circuit.

Apply Kirchhoff’s loop rule:
While the capacitor is charging S2 is open. After the capacitor is fully charged S1 is opened at the same time S2 is closed: this removes the battery from the circuit. Current will now flow in the right hand loop only, discharging the capacitor. C R S1 S2 + I Apply Kirchhoff’s loop rule: The current in the circuit is But the voltage drop across the capacitor is now

The voltage drop across the discharging capacitor:

Example (text problem 18.85): A capacitor is charged to an initial voltage of V0 = 9.0 volts. The capacitor is then discharged through a resistor. The current is measured and is shown in the figure. The red lines on the graph are aids to find the e-folding time.

(a) Find C, R, and the total energy dissipated in the resistor.
Example continued: (a) Find C, R, and the total energy dissipated in the resistor. Use the graph to determine . I0 = 100 mA; the current is I0/e = 36.8 mA at t = 13 msec. Since  = V0 = 9.0 volts, R = 90  and C = 144 F. All of the energy stored in the capacitor is eventually dissipated by the resistor.

Example continued: (b) At what time is the energy in the capacitor half of the initial value? Want:

Example continued: Solve for t:

Summary Current & Drift Velocity Resistance & Resistivity Ohm’s Law
Kirchhoff’s Rules Series/Parallel Resistors/Capacitors Power Voltmeters & Ammeters RC Circuits

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