Review Exam 4

Chapter 18 Electric Currents

Resistance and Resistors The ratio of voltage to current is called the resistance: (18-2a) (18-2b)

18.4 Resistivity The constant ρ, the resistivity, is characteristic of the material.

18.4 Resistivity

For any given material, the resistivity changes with temperature: Since R 0 = Resistance at some reference temperature (T 0 ) α = temperature coefficient and R = resistance at some new temperature T

18.5 Electric Power

The unit of power is the watt, W. 1Watt = 1 Joule/second

Chapter 19 DC Circuits

Conservation Laws Conservation of Charge (I = Q/t) and I Conservation of Energy (PE = QV) and V Conservative Forces (pages ) ∑ of Work around a closed path = 0 Independent of Path For a closed loop ∑ of PE gained and lost or ∑ of V gained and lost = 0

R ε + - I V = IR Simple Circuit ε-V = 0 ε –IR =0 ε = IR

Resistors in Series From this we get the equivalent resistance (that single resistance that gives the same current in the circuit).

Resistors in Parallel This gives the reciprocal of the equivalent resistance: (19-4)

Repeat this Mantra The current is the same for elements connected in series The voltage is the same for elements connected in parallel

Multiple Loop Game Rules Draw picture Define number of loops Pick ARBITRARY directions for the currents Indentify branch points (where currents divide) Currents into branch points = currents out of branch points Circle each loop and ∑ V’s = 0

How to circle a loop Start at some ARBITRARY point Circle clockwise or counterclockwise (your choice) End at starting point

When you come to a Battery as you circle the loop ε How to Add V’s Direction you take - ε + ε When you come to a Resistor as you circle the loop you will travel with or against the current I - IR R + IR

19.5 Circuits Containing Capacitors in Series and in Parallel In this case, the total capacitance is the sum: (19-5)

19.5 Circuits Containing Capacitors in Series and in Parallel In this case, the reciprocals of the capacitances add to give the reciprocal of the equivalent capacitance: (19-6)

Ammeters – Very small resistance so voltage across it is very small In series

Voltmeters - Very large resistance, takes away little current. In Parallel

What would happen if I put an Ammeter in Parallel? A

ConcepTest 19.8Kirchhoff’s Rules ConcepTest 19.8 Kirchhoff’s Rules The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes

the point between the bulbs is at 12 VBut so is the point between the batteries When the switch is open, the point between the bulbs is at 12 V. But so is the point between the batteries. If there is no potential difference, then no current will flow once the switch is closed!! Thus, nothing changes. The lightbulbs in the circuit are identical. When the switch is closed, what happens? 1) both bulbs go out 2) intensity of both bulbs increases 3) intensity of both bulbs decreases 4) A gets brighter and B gets dimmer 5) nothing changes ConcepTest 19.8Kirchhoff’s Rules ConcepTest 19.8 Kirchhoff’s Rules 24 V Follow-up: Follow-up: What happens if the bottom battery is replaced by a 24 V battery?

ConcepTest 19.10Kirchhoff’s Rules ConcepTest More Kirchhoff’s Rules 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2 Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I = 0 5) 2 – I 1 – 3I 3 – 6 = 0

ConcepTest Kirchhoff’s Rules ConcepTest More Kirchhoff’s Rules 2 V 2  2 V 6 V 4 V 3  1  I1I1 I3I3 I2I2 Eqn. 3 is valid for the left loop Eqn. 3 is valid for the left loop: The left battery gives +2V, then there is a drop through a 1  resistor with current I 1 flowing. Then we go through the middle battery (but from + to – !), which gives –4V. Finally, there is a drop through a 2  resistor with current I 2. Which of the equations is valid for the circuit below? 1) 2 – I 1 – 2I 2 = 0 2) 2 – 2I 1 – 2I 2 – 4I 3 = 0 3) 2 – I 1 – 4 – 2I 2 = 0 4) I 3 – 4 – 2I = 0 5) 2 – I 1 – 3I 3 – 6 = 0