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AP Physics 2 : Circuit Properties

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Example Questions

Example Question #71 : Circuits

A scientist is testing resistors made of the same material. Resistor has a cross sectional area of 3mm^2 and a length of 3mm^2 . Resistor has a cross sectional area of 6mm^2 and a length of 3mm^2 . How will the resistance of compare to that of ?

Possible Answers:

The resistance will quadruple

None of these

The resistance will double

The resistance will be cut in half

The resistance will stay the same

Correct answer:

The resistance will be cut in half

Explanation:

Use the equation for resistivity to visualize the proportions:

$R=\rho*\frac{l}{A}$

$\rho$ will stay constant as the material is the same. doubles and stays the same. Thus, the resistance, , will get cut in half.

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Example Question #71 : Circuit Properties

A resistor is made out of a material. The resistor has a cross-sectional area of 4mm^2 and a length of 8mm . It is found to have a resistance of $7\Omega$ . Determine the resistivity of the material.

Possible Answers:

$4.7\Omega\cdot mm$

$6.5\Omega\cdot mm$

$7.0\Omega\cdot mm$

$10.3\Omega\cdot mm$

$3.5\Omega\cdot mm$

Correct answer:

$3.5\Omega\cdot mm$

Explanation:

Using

$\rho=R\frac{A}{l}$

Plugging in values

$\rho=7*\frac{4}{8}$

$\rho=3.5\Omega\cdot mm$

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Example Question #71 : Circuits

A resistor is made out of a material. The resistor has a cross-sectional area of 4mm^2 and a length of 8mm . It is found to have a resistance of $7\Omega$ . A new resistor is built that has a length of 6mm and a cross sectional area of 3.3mm^2 . The new resistor is placed in series with a battery. Determine the current through the resistor.

Possible Answers:

.94A

None of these

3.25A

1.94A

1.25A

Correct answer:

.94A

Explanation:

Using

$\rho=R\frac{A}{l}$

Plugging in values

$\rho=7*\frac{4}{8}$

$\rho=3.5\Omega*mm$

Using

$R=\rho*\frac{l}{A}$

Plugging in values

$R=3.5*\frac{6}{3.3}$

$R=6.36\Omega$

Using

V=IR

Solving for

$I=\frac{V}{R}$

$I=\frac{6}{6.36}$

I=.94A

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Example Question #74 : Circuit Properties

A scientist is testing resistors made of the same material. Resistor has a cross sectional area of 4mm^2 and a length of 8mm . Resistor has a cross sectional area of 6mm^2 and a length of 3mm . How will the resistance of compare to that of ?

Possible Answers:

It is impossible to determine

None of these

The resistance of will be a quarter of that of

They will be the same

The resistance of will be a quarter of that of

Correct answer:

The resistance of will be a quarter of that of

Explanation:

Using

$R=\rho*\frac{l}{A}$

$\rho$ will stay constant as the material is the same. increases by a ratio of $\frac{3}{2}$ , and decreases by a ratio of $\frac{3}{8}$ . Thus, the resistance, , will decrease to a quarter of it's original value. $\left (\frac{\frac{3}{8}}{\frac{3}{2}}\right)$

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Example Question #881 : Ap Physics 2

A cubic resistor is placed in a circuit. If each edge measures 1cm long, the voltage of the battery is and the current through the resistor is , determine the resistivity.

Possible Answers:

$1.67\Omega\cdot cm$

$18\Omega\cdot cm$

$6\Omega\cdot m$

None of these

$6\Omega\cdot cm$

Correct answer:

$6\Omega\cdot cm$

Explanation:

Determining the resistance

V=IR

6=1*R

$R=6\Omega$

Using

$\rho=R\frac{A}{l}$

$\rho=6*\frac{1}{1}$

$\rho=6\Omega\cdot cm$

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Example Question #31 : Resistivity

A scientist carves out a cylinder of a new material she developed. It is 5.0 cm tall, with a radius of .80 cm . She puts electrodes on each face of the cylinder. She determines the resistance to be $.75\Omega$ . What is the resistivity of this new material?

Possible Answers:

$4.5*10^{-3} \Omega \cdot m$

$1.5 *10^{ -3} \Omega \cdot m$

$1.75 *10^{ -3} \Omega \cdot m$

None of these

$3.0*10^{-3} \Omega\cdot m$

Correct answer:

$3.0*10^{-3} \Omega\cdot m$

Explanation:

Using the relationship:

$\rho =R\frac{A}{l}$

Here, $\rho$ is the resistivity, is the resistance, is cross sectional area, and is the length.

The area of a circle is:

$A=\pi r^2$

Substitute:

$\rho =R\frac{\pi r^2}{l}$

Plug in given values and solve.

$\rho =.75\Omega \frac{\pi (.008m)^2}{.05m}$

$\rho=3.0 *10^{ -3} \Omega \cdot m$

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Example Question #72 : Circuit Properties

Consider the circuit:

Circuit_1

What is the rate of power consumption in the circuit if every resistor has a resistance of $3\Omega$ ?

Possible Answers:

24 W

18W

12 W

21 W

15 W

Correct answer:

18W

Explanation:

To calculate the power consumption of the circuit, we need to first reduce it to an equivalent circuit with a single resistor. Since each resistor has the same resistance, this solution will keep resistance calculations as multiples of until the circuit is fully reduced.

Start with the two branches in parallel. We can condense R3 and R4, then solve for the total resistance of R2, R3, and R4.

$R_{34}=R+R=2R$

$\frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{R_2}+ \frac{1}{R_{34}}=\frac{1}{R}+\frac{1}{2R} = \frac{3}{2R}$

$R_{eq} = \frac{2}{3}R$

The equivalent circuit now has three resistors in series (R1, Req, and R5), so we can simply add them all up:

$R_{tot} = R+\frac{2}{3}R+R=\frac{8}{3}R$

Plug in the value for R:

$R_{tot}=\frac{8}{3}R=\frac{8}{3}(3\Omega)=8\Omega$

Now we can use the equation for power:

P = IV

Substituting in Ohm's law for current, we get:
$P = \frac{V^2}{R} = \frac{(12V)^2}{8\Omega} = 18W$

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Example Question #73 : Circuit Properties

You have 4 resistors, , , , and , set up like this:

4resistorcircuit

Their resistance are as follows:

$\begin{align*} A&=1\Omega \\ B&=3\Omega \\ C&=4\Omega \\ D&= 2\Omega \end{align*}$

If the battery has 8V, what is the total power dissipated through the resistors?

Possible Answers:

16W

12W

20W

Correct answer:

16W

Explanation:

The equation for power is

P=VI

In order to get the power, we need the current. To find the current, we need to get the total resistance, and use Ohm's Law ( V = IR ).

To find the total resistance, remember the equations for adding resistors is this:

$(Series)\ R_{tot}&=R_1+R_2+...+R_n$

$(Parallel)\ \frac{1}{R_{tot}}&=\frac{1}{R_1}+\frac{1}{R_2}=...=\frac{1}{R_n}$

Resistors and are in series, resistors and are in parallel, and resistors ABC and are in series.

$R_{AB}&=1+3$

$R_{AB}= 4$

$R_{ABC}&=(\frac{1}{R_{AB}}+\frac{1}{R_C})^{-1}$

$R_{ABC}= (\frac{1}{4}+\frac{1}{4})^{-1}$

$R_{ABC}= 2$

$R_{total}&=R_{ABC}+R_{D}$

$R_{total}= 2+2$

$R_{total}= 4$

Now, we can find the current.

$I&=\frac{V}{R}$

$I=\frac{8}{4}$

I= 2

Finally, we can find the power.

P&=VI

P=(8)(2)

P= 16

Therefore, the power is 16W (watts).

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Example Question #2 : Circuit Power

Physics2set1q6

Calculate the power consumed across resistor .

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

Possible Answers:

3 W

10W

Correct answer:

3 W

Explanation:

Physics2set1q6

$I_{1}=2A$

$I_{2}=1A$

$R_{1}=4\Omega$

$R_{2}=3\Omega$

$C_{1}=2F$

To calculate power, we need two of the following three quantities: voltage, current, and resistance.

In this case, since we are lacking the voltage, let's try to find the current.

We can use Kirchoff's junction law to calculate current .

The current coming into the junction = the current coming out of the junction.

Let's take a look at the central junction to the right of resistor .

$I_{1}=I_{2}+I_{3}$

$2=I_{3}+1$

$I_{3}=1$

Now that we know and , we can calculate power across the resistor.

$P={I^2}R=3W$

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Example Question #1 : Circuit Power

Physics2set1q4a Physics2set1q4b

Elements A-D represent light bulbs.

Which of the following is true about these two circuits? Assume voltage sources have the same value and all the light bulbs are all identical.

Possible Answers:

Bulbs C and D have different brightnesses.

All bulbs will have the same brightness.

Bulbs A and B will be brighter than bulbs C and D.

Bulbs A and B have different brightnesses.

Bulb A will be as bright as bulb C, but bulbs B and D will have a different brightness.

Correct answer:

Bulbs A and B will be brighter than bulbs C and D.

Explanation:

Since bulbs A and B are in parallel, they will have the same voltage, and since the bulbs are identical in resistance, they will have the same current running through them and will be just as bright.

Let's say the voltage source as a value of and each bulb has a resistance of .

The current going through bulbs A and B is $\frac{V}{R}$ .

However, the current going through bulbs C and D is $\frac{V}{2R}$ .

The current going through bulbs C and D is half as much as the other two, so their brightness will be less.

So, bulbs A and B will be brigher than bulbs C and D.

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AP Physics 2 : Circuit Properties

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #71 : Circuits

Example Question #71 : Circuit Properties

Example Question #71 : Circuits

Example Question #74 : Circuit Properties

Example Question #881 : Ap Physics 2

Example Question #31 : Resistivity

Example Question #72 : Circuit Properties

Example Question #73 : Circuit Properties

Example Question #2 : Circuit Power

Example Question #1 : Circuit Power

All AP Physics 2 Resources

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