Download Electricity Test Review kaw (with answers)

ID: _____
Name: __________________________________ Block: ______
Electricity Test Review:
1)
Name these units and give the symbol for the variable:
________________ ________Charge
________________ ________Current
________________ ________Potential difference
2)
________________ ________Resistance
________________ ________Power
________________ ________Force
Sketch a graph of current to voltage for Ohm’s law: Sketch a graph of electric force to distance for Coulomb’s law:
Ohm’s law is V = IR. The current I is directly proportional to the voltage V because the resistance R is constant in a circuit. So it is a linear graph going
up as V increases.
In the Coulomb’s law, the electrical force is inversely proportional to the square of the distance. So the graph is a curved line going down rapidly as
distance increases.
3)
Calculate the equivalent resistance and current for the two circuits to the left below:
In the parallel circuit, more branches of resistors make the current much easier to flow through the wires. It means
that as branches are added in parallel, the resistance of the circuit is getting less. The equivalent resistance
which is the total resistance of the circuit is reciprocal sum of resistances. 1/Req = 1/R1 + 1/R2 + 1/R3
1/Req = 1/12 + 1/12 + 1/12 = 3/12 = ¼. Then Req =4 Ohms
The current is V/R, then 6/4 = 1.5 A
In the series circuit, more resistances (bridges) cause congestion of current. That means that as more resistors are added, the equivalent resistance
increases. Req = R1 + R2 + R3. Req = 1 + 2 + 3 = 6 Ohms. Current is V/R. 6/6 = 1 A
4)
What is the potential difference for the circuit to the left? Explain how you know this:
The potential difference is another name of voltage. In the series circuit, the total voltage is the sum of the
voltages applied to each resistor. The voltage is 10 + 10 + 10 which is 30 V.
5)
What is the current reading in the ammeter in the
circuit to the right?
The ammeter is an instrument measuring the current of the circuit. This circuit has 3 branches which
are arranged in parallel. 1/Req = 1/R1 + 1/R2 + 1/R3.
1/Req = 1/3 + 1/3 + 1/3 = 3/3 = 1/1. The total resistance is 1 Ohm. The current between the two ends
of the voltage source is 9/1 (V/R) which is 9 amperes. The 9 amperes of the current is distributed evenly into 3 branches. So the ammeter reads 3
amperes of the current in the schematic diagram.
6)
Demonstrate what happens to the electric force between two charges is you quadruple the distance between them:
This question is asking the relation between the force and distance in the Coulomb’s law. So let’s ignore other variables than these two variables. You
can write the Coulomb’s law this way: F = something / d2. As distance increases by a factor of 4, then the force will be something over 16d2. That
means the changed force is 1/16 the first force.
7)
Demonstrate what happens to the electric force between two charges if you increase the charge on the first by a factor of 3 and on the
second by a factor of 4:
This question is asking the relation between the force and charges. So let’s ignore other variables than these two variables. You can write the Colomb’s
law this way: F = something x q1(charge 1) x q2(charge 2). So F = something x 3 q1 x 4 q2 which is 12 times something x q1 x q2. That means the
changed force is 12 times the first force.
8)
Demonstrate what happens to the electrical force between two charges if you decrease the distance to 1/3 of the original distance:
This question is asking the relation between the force and distance. So let’s ignore other variables than these two variables. You can write the
Coulomb’s law this way: F = something / d2. As distance increases by a factor of 1/3, then the force will be something / (1/9)d2. That means the
changed force is 9 times the first force.
9)
Which of these causes an increase in electrical force between two charges?
a)
b)
c)
d)
Increasing the distance
Decreasing the distance
Increasing one of the charges
Increasing both of the charges
10) If the force between 2 charges is 3.5 x 10-7 N and the distance between them is decreased by half, what is the new value for the electrical
force between them?
F= 3.5 x 10-7 N. Since the distance is decreased by half, the force is increases by a factor of 4. The new force is 3.5 x 4 x 10-7 = 1.4 x 10-6 N.
11) Calculate the power in a 25 ohm resistor if 9.0V is applied across it. What is the current in this resistor?
I = V/R
9/25 = .36 A
V
I
R
12) A 1500 W microwave operates on household potential difference (120V). How much current flows through the microwave? What is its
resistance?
Power = Current x voltage. So Current = Power / Voltage. 1500/120 = 12.5 A.
Voltage = Current x Resistance. Resistance = Voltage / Current. 120/12.5 = 9.6 Ohms
13) 3 resistors are wired in parallel so that there is one in each branch. They are each 15 ohms. What is the equivalent resistance of these
resistors in this configuration?
This is a parallel circuit. 1/Req = 1/R1 + 1/R2 + 1/R3. 1/Req = 1/15 + 1/15 + 1/15 which is 3/15 which is 1/5. The equivalent resistance is 5 Ohms.
14) What is the equivalent resistance of the 3 resistors in the previous question if they are wired in series instead?
In the series circuit, you add the resistances. 15 + 15 + 15 = 45 Ohms.
15) Two identical charges of 3.2 nC are separated by a distance of 6.58 cm. What is the force between them?
F = k q1 q2 / d2. F = 9 x 109 x 3.2 x 10-9 x 3.2 x 10-9 / (0.0658)2.
F = 2.1 x 10-5 N
Some basic statements that you may want to transfer to a card:
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The flow of electrical current requires a difference in potential.
Electricity in a liquid depends on the presence of conducting ions.
Electricity in a metal depends on the presence of conduction electrons.
The greater the potential difference across a resistor, the greater the current.
Its current that kills you, not voltage.
A vending machine contains many solenoids that control when and which food is dispensed.
Human resistance falls between 1000 ohms (wet) and 100,000 ohms (dry).
The electric company sells you energy, not power.
Thick short wires carry more current than skinny long wire
A positive electrical force indicates repulsion.
Wool, hair and fur lose electrons to be positively charged.
As you add resistors to a series circuit, the equivalent resistance increases. When you add them to a parallel circuit, the equivalent
resistance decreases.