All AP Physics 1 Resources
Example Questions
Example Question #1 : Resistivity
What is the resistance of a length of round copper wire with a radius of ?
Resistance and resistivity are related as follows:
Example Question #2 : Resistivity
Two students are performing a lab using lengths of wire as resistors. The two students have wires made of the exact same material, but Student B has a wire the has twice the radius of Student A's wire. If Student B wants his wire to have the same resistance as Student A's wire, how should Student B's wire length compare to Student A's wire?
Student B's wire should be the length of Student A's wire
Student B's wire should be four times as long as Student A's wire
Student B's wire should be twice as long as Student A's wire
Student B's wire should be the length of Student A's wire
Student B's wire should be the same length as Student A's wire
Student B's wire should be the length of Student A's wire
Resistance is proportional to length, and inversely proportional to cross-sectional area. Area depends on the square of the radius: , so Student B's wire has times the cross-sectional area of Student A's wire. In order to compensate for the increased area, Student B must make his wire the length of Student A's wire. This can be shown mathematically using the equation for resistance:
Example Question #2 : Resistivity
By how much will resistivity change if resistance and length are constant, and cross sectional area is doubled?
The resistivity will not change
The resistivity will double
The resistivity will be halved
The resistivity will be quadrupled
The resistivity will double
Recall the formula for resistance is given by
, where is the cross sectional area, is resistivity, and is length.
Solve for resistivity:
From this, we can tell that resistivity is proportional to cross sectional area by:
Since is doubled, and resistance and length are constant, resistivity will also be doubled.
Example Question #1 : Resistivity
Ratio is given by:
Resistivity of first resistor: Resistivity of second resistor
What is the ratio of resistivity of 2 resistors with identical resistances and area, where the first resistor is twice the length of the second resistor?
Resistivity is given by:
, where is the resistance, is the area, and is the length.
Since both have identical resistances and area, the first resistor will have half the resistivity since it has twice the length. Therefore the resistivity relation is
Example Question #1 : Resistivity
What is the resistance of a copper rod with resistivity of , diameter of , and length of ?
The equation for resistance is as follows: . Where is resistivity, is the length of the wire, and is the cross section of the wire which can be found using .
Example Question #11 : Resistivity
You have a very long wire connected to an electric station. Even though you are suppling 120V from the source, by the time it reaches the station, there is a loss of voltage. The wire is 100 meters long.
If reaches the power station, what is the resistivity of the wire? Assume a current of .
The voltage drop from the source to the station (the "load") indicates that there is an internal resistance in the wire. According to the voltage law, the total amount of voltage drop is equal to the total amount of voltage supplied. Since was supplied, and drops at the station, that means that drops along the wire.
Now that the voltage drop across the wire is known, Ohm's law will give the resistance of the wire:
The resistivity of the wire is equal to the resistance per unit length, therefore, in order to find resistivity you divide the total resistance by the length:
Example Question #111 : Circuits
Which of the following actions would decrease the resistance of a wire by a factor of ?
Halving the cross-sectional area of the wire
Tripling the length of the wire
Halving the length of the wire
Doubling the cross-sectional area of the wire
Tripling the cross-sectional area of the wire
Doubling the cross-sectional area of the wire
The equation for resistance is as follows:
Where is the resistance, is the resistivity of the material, is the length of the material and is the cross-sectional area of the material. Looking at this equation, by doubling the area we effectively reduce the resistance by a factor of two.
Example Question #1381 : Ap Physics 1
Find the resistivity of a cylindrical wire with resistance , length , and cross sectional area of .
There exist a formula that directly relates resistance and resistivity. The formula is .
is resistance, is resistivity, is length, and is cross sectional area. Solving for , we get . Plugging in our givens, we get .
Example Question #1 : Circuit Power
Consider the following circuit:
How much power is lost through R1?
In order to find the power loss in R1, we need to know the current flowing through R1. Since it is not in parallel with anything, all of the current flowing through the circuit will flow through R1. To find the current flowing through the circuit, we will need to first find the total equivalent resistance of the circuit.
To do this, we first we need to condense R3 and R4. They are in series, so we can simply add them to get:
Now we can condense R2 and R34. They are in parallel, so we will use the following equation:
The equivalent circuit now looks like:
Since everything is in parallel, we can simply add everything up:
Now that we have the total resistance of the circuit, we can use Ohm's law to find the current:
Rearranging for current, we get:
Now that we know the current flowing through R1, we can use the following equation to find the power loss:
Since we don't know the voltage drop across R1 (although we can calculate it), we can substitute Ohm's law into the equation:
Plugging in our values, we get:
Example Question #2 : Circuit Power
Consider the given circuit:
If and , what is the power loss through R2?
To calculate the power loss through R2 we need to either calculate the current flowing through it or the voltage drop across it. Calculating the current will be one less step, so we'll use that method.
We need to first calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add their resistance values.
Then we can use Ohm's law to calculate the current through the circuit:
Next, use the expression for power:
Substituting Ohm's law for the voltage across the resistor, we get: