Download Circuits Test #2 - Review

Circuits Test #2 - Review
Vocabulary Matching:
__D__ 1. Resistance
A. The rate of flow of electricity through a circuit
__F__ 2. Voltage
B. A material that allows the flow of electricity
__A__ 3. Current
C. Electricity can flow through this circuit
__B__ 4. Conductor
D. Slows down the flow of electricity in a circuit
__E__ 5. Insulator
E. A material that resists electrical flow
__G__ 6. Schematic Diagram
F. Pushes electricity through a circuit
__H__ 7. Open Circuit
G. A shorthand way to draw an electrical circuit
__C__ 8. Closed Circuit
H. A circuit with a break in it, electricity will not flow
Parts of a Circuit:
__E__ 9. Battery
A. Slows down the flow of electricity in a circuit
__C__ 10. Wires
B. Can turn the electricity on and off in a circuit
__B__ 11. Switch
C. Connects the parts of the circuit together
__D__ 12. Light Bulb
D. Used to create radiant energy
__A__ 13. Resistor
E. Pushes electricity through the circuit
Ohm’s Law:
__E__ 14. Amperes
A. Unit of electrical resistance
__G__ 15. Volts
B. Symbol for electrical resistance
__A__ 16. Ohms
C. Symbol used to represent current in Ohm’s formula
__D__ 17. V
D. Symbol for electrical potential or voltage
__B__ 18. Ω
E. Unit of electrical current
__F__ 19. A
F. Symbol for electrical current
__C__ 20. I
G. Unit of electrical potential or voltage
Ohm’s Law Problems:
21. How much voltage is needed to produce 2 A through a 4 Ω light bulb?
V = IR = (2A) (4Ω) = 8V
22. A 9 volt battery produces 4.5 amps across a resistor. What is the resistance?
R = V/I = (9V)/(4.5A) = 2Ω
23. If a light in your house (120V) draws 0.5 amps, how much resistance is in the bulb?
R = V/I = (120V)/(0.5A) = 240Ω
24. If a light bulb uses 6V and has a resistance of 12Ω, what is the current it draws?
I = V/R = (6V)(12Ω) = 0.5A
Fill-in-the-Blanks:
25. Ohm’s Law states “For a given resistor at a particular temperature, the current is
directly proportional to the applied voltage.”
26. According to Ohm’s Law, resistance and current are inversely proportional.
27. In a series circuit there is only one path for the current to flow. The current is the
same at all points along the wire and the resistance increases as resistors are added.
28. In a parallel circuit there are multiple paths for the current to flow. Each resistor
provides a new path for electrons to flow and the voltage across each branch is equal to
the voltage of the source.
**29. As resistors are added to a series circuit, the resistance increases and the current
decreases.
**30. As resistors are added to a parallel circuit, the resistance decreases and the
current increases.
Draw the following diagrams using circuit symbols:
31. A circuit with a battery, a switch and 2 light bulbs in parallel.
32. A circuit with a battery, a switch and 2 resistors in series.
Circuit Diagrams: For each diagram, complete the table and fill in the blanks.
33.
VT = ___9___ V
RT = ___18___ Ω
IT = __0.5___ A
Series Circuit
Resistance
Current
Voltage
(Ω)
(A)
(V)
10
0.5
5
5
0.5
2.5
3
0.5
1.5
I = V/R = 9/18
34.
VT = ___6___ V
RT = ___2.0___ Ω
IT = ___3___ A
Parallel Circuit
Resistance
Current
Voltage
(Ω)
(A)
(V)
6
1
6
3
2
6
35.
1/RT = 1/6 + 1/3 = 0.1667 + 0.333 = 0.4997
RT = 1/0.4997
VT = ___12___ V
IT = ___1.33___ A
RT = ___9___ Ω
Series Circuit
Resistance
Current
Voltage
(Ω)
(A)
(V)
2
1.33
2.66
3
1.33
3.99
4
1.33
5.32
I = V/R = 12/9
36.
VT = __4.5___ V
RT = __0.55___ Ω
IT = __8.25___ A
Parallel Circuit
Resistance
Current
Voltage
(Ω)
(A)
(V)
2
2.25
4.5
3
1.5
4.5
1
4.5
4.5
1/RT = ½ + 1/3 + 1 = 0.5 + 0.333 + 1 = 1.833
37.
VT = ___6___ V
RT = ___2.0___ Ω
RT = 1/1.833
IT = ___3___ A
Parallel Circuit
Resistance
Current
Voltage
(Ω)
(A)
(V)
4
1.5
6
6
1
6
12
0.5
6
1/RT = ¼ + 1/6 + 1/12 = 0.25 + 0.1667 + 0.0833 = 0.5
RT = 1/0.5