1. Electric Circuits

2. Ohm’s Law
The relationship between the potential difference in a circuit and the resulting current.

3. 20.1 Electromotive Force and Current
The electric current is the amount of charge per unit time that passes
through a surface that is perpendicular to the motion of the charges.

4. 20.1 Electromotive Force and Current
The electric current is the amount of charge per unit time that passes
through a surface that is perpendicular to the motion of the charges. By
convention, we use the flow of positive charge or “electron holes”.
One coulomb per second equals one ampere (A).

5. To the extent that a wire or an electrical
20.2 Ohm’s Law
To the extent that a wire or an electrical
device offers resistance to electrical flow,
it is called a resistor.

6. Kirchoff’s Junction Law
The sum of the currents entering a junction are equal to the sum of the currents leaving.

7. What current flows through the 4th wire and which way does it go?
1 A into the junction
1 A out of the junction
9 A into the junction
9 A out of the junction

9. In which direction does the potential decrease?
Clockwise
Counterclockwise
Potential is the same everywhere in the circuit.

10. Ohm’s Law in the Cell
The resistance between the walls of a biological cell is 1.00 x 1010 Ω.
What is the current when the potential difference between the walls is 95 mV?
If the current is composed of Na+ ions (q = +e), how much charge moved between the cell walls in 0.2 s?

11. Ohm’s Law in the Cell
Suppose that the resistance between the walls of a biological cell is 1.00 x 1010 Ω.
What is the current when the potential difference between the walls is 95 mV? x A
If the current is composed of Na+ ions (q = +e), how much charge moved between the cell walls in 0.2 s? 1.9 x C

12. Power in a circuit
Through which resistor is the most power dissipated?
A. a B. b C. b,d D. d

13. Power rating of a household applicance
Commercial and residential electrical systems are set up so that each individual appliance operates at a potential difference of 120 V.
Power Rating or Wattage is the power that the appliance will dissipate at a potential difference of 120 V (e.g. 100 W bulb, 1000 W space heater).
Power consumption will differ if operated at any other voltage.
Energy is often expressed as kilowatt-hours, for metering purposes.

14. kW-h How much energy in a kilowatt-hour? 1000 Joules 0.28 Joules
3.6 million Joules
60,000 Joules

15. ∆Vbat = |∆Vcircuit |= IRs
Resistors in Series
Series wiring means that the resistors are connected so that there is the same current through each resistor.
The potential difference through each resistor is :
|∆VR |= IR
The equivalent resistor (Rs ) is made by adding up all resistors in series in the circuit.
∆Vbat = |∆Vcircuit |= IRs

16. V1 (voltage drop across the resistor1) = IR1
20.6 Series Wiring
By understanding that the potential difference across any circuit resistor will be a voltage DROP, while the potential difference back through the battery (charge escalator) will be a GAIN, one can replace the ΔV with “V” , and dispense with the absolute value signs.
V1 (voltage drop across the resistor1) = IR1
V2 (voltage drop across the resistor2) = IR2
Vbat = Vcir = IReq

17. Resistors in a Series Circuit
20.6 Series Wiring
Resistors in a Series Circuit
A 6.00 Ω resistor and a 3.00 Ω resistor are connected in series with
a 12.0 V battery. Assuming the battery contributes no resistance to
the circuit, find (a) the current, (b) the power dissipated in each resistor,
and (c) the total power delivered to the resistors by the battery.

18. 20.6 Series Wiring
(a)
(b)
(c)

19. Series Resistors What is the value of R (3rd resistor in circuit)?
a. 50 Ω b. 25Ω c. 10 Ω d. 0

20. Parallel wiring means that the devices are
connected in such a way that the same
voltage is applied across each device.
When two resistors are connected in
parallel, each receives current from the
battery as if the other was not present.
Therefore the two resistors connected in
parallel draw more current than does either
resistor alone.

21. 20.7 Parallel Wiring

22. The two parallel pipe sections are equivalent to a single pipe of the
20.7 Parallel Wiring
The two parallel pipe sections are equivalent to a single pipe of the
same length and same total cross sectional area.

23. parallel resistors Rp = (R1R2 ) / (R1 + R2)
20.7 Parallel Wiring
parallel resistors
note that for 2 parallel resistors this equation is equal to :
Rp = (R1R2 ) / (R1 + R2)
Note that the potential difference of the battery is the same as the potential difference across each resistor in parallel.

24. Example 10 Main and Remote Stereo Speakers
20.7 Parallel Wiring
Example 10 Main and Remote Stereo Speakers
Most receivers allow the user to connect to “remote” speakers in addition
to the main speakers. At the instant represented in the picture, the voltage
across the speakers is 6.00 V. Determine (a) the equivalent resistance
of the two speakers, (b) the total current supplied by the receiver, (c) the
current in each speaker, and (d) the power dissipated in each speaker.

25. 20.7 Parallel Wiring
(a)
(b)

26. 20.7 Parallel Wiring
(c)
(d)

27. Find a) the total current supplied by
Circuits Wired Partially in Series and Partially in Parallel Example 12
Find a) the total current supplied by
the the battery and b) the voltage drop
between points A and B.

28. Find a) the total current supplied by
Circuits Wired Partially in Series and Partially in Parallel Example 12
Find a) the total current supplied by
the the battery and b) the voltage drop
between points A and B.

29. Circuits Wired Partially in Series and Partially in Parallel Example 12
Itot = V/Rp = 24V/240Ω = .10A
now go back to the 1st circuit in part c to calculate the voltage drop across A-B:
VAB = IRAB = (.10A)(130 Ω) = 13V

30. Your Understanding
What is the ratio of the power supplied by the battery in parallel circuit A to the power supplied by the battery in series circuit B?
¼ c. 2 e. 1
b d. 2 A. B.

31. Analyze the identical bulbs
Which bulbs will light
All bulbs
A, B
A only
none

32. What happens to A?
With switch closed, A is______________ it was when switch was open.
Brighter than
Less bright than
Equal to what
closed

33. Complex Circuit
The current through the 8 Ω resistor is .590 A. Determine the following:
Current in the 20 Ω resistor
Current in the 9 Ω resistor
Battery voltage

34. Complex Circuit
The current through the 8 Ω resistor is .590 Determine the following:
Current in the 20 Ω resistor: A
Current in the 9 Ω resistor: A
Battery voltage: 22.4 V

35. 20.11 The Measurement of Current and Voltage
An ammeter must be inserted into
a circuit so that the current passes
directly through it.
The resistance of the ammeter changes the current through the circuit.
The ideal ammeter has a very low resistance

36. 20.11 The Measurement of Current and Voltage
To measure the voltage between two points, in a circuit, a voltmeter is connected in parallel, between the points.
A voltmeter takes some current away from the circuit it measures.
The ideal voltmeter has a very large resistance so it diverts a negligible current.

37. The Battery
A battery consists of chemicals, called electrolytes, sandwiched in between 2 electrodes, or terminals, made of different metals.
Chemical reactions do positive work and separate charge. The electric field does the same amount of negative work, which translates as a potential difference between positive and negative terminals
A physical separator keeps the charge from going back through the battery. 37

38. ε and ∆V
For an ideal battery potential difference between the positive and negative terminals, ∆V, equals the chemical work done per unit charge.
∆V = Wch /q = ε
ε is the emf of the battery
Due to the internal resistance of a real battery, ∆V is often slightly less than the emf.
A capacitor can store charge, but has no way to do the work to separate the charge. It has a potential difference, but no ε.

39. How do we know there is a current?

40. 20.1 Electromotive Force and Current
If the charges move around the circuit in the same direction at all times,
the current is said to be direct current (dc).
If the charges move first one way and then the opposite way, the current is
said to be alternating current (ac).

41. EOC #3
A fax machine uses A of current in its normal mode of operation, but only A in the standby mode. The machine uses a potential difference of 120 V.
In one minute (60 s!), how much more charge passes through the machine in the normal mode than in the standby mode?
How much more energy is used?

42. EOC #3
A fax machine uses A of current in its normal mode of operation, but only A in the standby mode. The machine uses a potential difference of 120 V.
In one minute (60 s!), how much more charge passes through the machine in the normal mode than in the standby mode? C/s x 60 s = 2.6 C
How much more energy is used? 2.58 C x 120 J/C = 310 J

43. 20.1 Electromotive Force and Current
Conventional current is the hypothetical flow of positive charges (electron holes) that would have the same effect in the circuit as the movement of negative charges that actually does occur.

44. The resistance (R) is defined as the
20.2 Ohm’s Law
The resistance (R) is defined as the
ratio of the voltage V applied across
a piece of material to the current I through
the material.

45. To the extent that a wire or an electrical
20.2 Ohm’s Law
To the extent that a wire or an electrical
device offers resistance to electrical flow,
it is called a resistor.

46. The filament in a light bulb is a resistor in the form
20.2 Ohm’s Law
Example 2 A Flashlight
The filament in a light bulb is a resistor in the form
of a thin piece of wire. The wire becomes hot enough
to emit light because of the current in it. The flashlight
uses two 1.5-V batteries to provide a current of
0.40 A in the filament. Determine the resistance of
the glowing filament.