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AP Physics 2 : AP Physics 2

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #41 : Circuit Components

The following is a picture of a circuit with 4 resistors, labeled , , , and .

4resistorcircuit

What is the total resistance of the circuit if the values of the resistors are as follows?

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

Possible Answers:

$11\Omega$

$10\Omega$

$1.2\Omega$

$5\Omega$

$6\Omega$

Correct answer:

$5\Omega$

Explanation:

Remember, the equations for find the total resistance of a circuit are as follows:

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

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

Let's complete this in pieces. First, we see that and are in series, so they just add together. Next, we see that and are in parallel, so we invert the added inverses. Finally, we see that ABC and are in series, so we add them together.

$R_{AB}&=R_A+R_B$

$R_{AB}=2+2$

$R_{AB}= 4$

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

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

$R_{ABC}=(\frac{1}{2})^{-1}=2$

$R_{ABCD}&=R_{ABC}+R_D$

$R_{ABCD}=2+3$

$R_{ABCD}=5$

Therefore, the total resistance of the circuit is $5\Omega$ .

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Example Question #2 : Resistors And Resistance

Parallel circuit 1 jpeg

$R_1=5\Omega$

$R_2=7\Omega$

What is the total resistance of the circuit above?

Possible Answers:

$0.35\Omega$

$6.0\Omega$

$12\Omega$

$35\Omega$

$2.92\Omega$

Correct answer:

$2.92\Omega$

Explanation:

Use our rule for adding resistors in parallel.

$\frac{1}{R1}+\frac{1}{R2}=\frac{1}{R_{total}}$

Plug in known values.

$\frac{1}{5\Omega}+\frac{1}{7\Omega}=\frac{1}{R_{total}}$

$R_{total}=2.92\Omega$

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Example Question #2 : Resistors And Resistance

Photo 4 1

$R_A=3\Omega$

$R_B=6\Omega$

$R_C=9\Omega$

What is the total resistance of the circuit shown above?

Possible Answers:

$18\Omega$

None of these

$\frac{18}{11}\Omega$

$11\Omega$

$\frac{11}{18}\Omega$

Correct answer:

$\frac{18}{11}\Omega$

Explanation:

The equations for adding resistances in parallel, as in the diagram is:

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

Solve.

$(R_{tot})^{-1}=\left(\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}$

$(R_{tot})^{-1}= \left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right)^{-1}$

$R_{tot}= \frac{18}{11}\Omega$

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Example Question #2 : Resistors And Resistance

Resistances:

$\begin{align*} A&= 1\Omega\\ B&= 2\Omega\\ C&= 3\Omega\\ D&= 4\Omega\\ E&= 5\Omega\\ F&=6\Omega \end{align*}$

If the circuit above is connected to a battery, what is the total power output of the system?

Possible Answers:

15.5000W

11.1375W

19.5467 W

16.1357 W

30.3030 W

Correct answer:

11.1375W

Explanation:

The rules for adding resistors are:

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

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

From the diagram, A, B, and C are in parallel, E and F are in parallel, and ABC , EF, and D are in series.

Find the resistance of A, B and C.

$R_{ABC}&=\left(\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}$

$R_{ABC}=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\right)^{-1}$

$R_{ABC}= \frac{6}{11}\Omega$

Find the resistance of E, and F.

$R_{EF}=\left(\frac{1}{R_E}+\frac{1}{R_F}\right)^{-1}$

$R_{EF}= \left(\frac{1}{5}+\frac{1}{6}\right)^{-1}$

$R_{EF}= \frac{30}{11}\Omega$

Find the total equivalent resistance by using the rule for adding (systems of) resistors in series.

$R_{ABCDEF}&=R_{ABC}+R_D+R_{EF}$

$R_{ABCDEF}= \frac{6}{11}+4+\frac{30}{11}$

$R_{ABCDEF}= \frac{80}{11}\Omega$

Therefore the total resistance is $\frac{80}{11}\Omega$

The power equation is:

P=VI

Since we know total resistance and potential difference, we can find current. Use Ohm's law.

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

$I= \frac{9}{\frac{80}{11}}$

$I= \frac{99}{11}$

I=9A

Now we can find the power.

P=VI

$P= (9)(\frac{99}{80})$

$P= \frac{891}{80}W$

This fraction is equivalent to 11.1375 W

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Example Question #4 : Resistors And Resistance

Resistances:

$\begin{align*} A&= 3\Omega\\ B&= 1\Omega\\ C&= 2.4 \Omega\\ D&= 5\Omega\\ E&= 1.5\Omega\\ F&= 7\Omega \end{align*}$

What is the total resistance of the system in the diagram above?

Possible Answers:

$2.110\Omega$

$1.566\Omega$

$6.807\Omega$

$5.346\Omega$

$15.412\Omega$

Correct answer:

$6.807\Omega$

Explanation:

The rules for adding resistors are:

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

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

Note that these equations are the opposite of the equations used for adding capacitors.

From the diagram, A, B, and C are in parallel, E and F are in parallel, and ABC , EF, and D are in series.

Find the equivalent resistances for each system of resistors.

$R_{ABC}=\left(\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}\right)^{-1}$

$R_{ABC}=\left(\frac{1}{3}+\frac{1}{1}+\frac{1}{2.4}\right)^{-1}$

$R_{ABC}= \frac{4}{7}\Omega$

Find the equivalent resistance of the other system of parallel resistors.

$R_{EF}&=\left(\frac{1}{R_E}+\frac{1}{R_F}\right)^{-1}$

$R_{EF}= \left(\frac{1}{1.5}+\frac{1}{7}\right)^{-1}$

$R_{EF}= \frac{21}{17}\Omega$

The three systems of resistors are in series. Find the equivalent resistance using the appropriate equation.

$R_{ABCDEF}=R_{ABC}+R_D+R_{EF}$

$R_{ABCDEF}= \frac{4}{7}+5+\frac{21}{17}$

$R_{ABCDEF}= \frac{810}{119}\Omega$

The total resistance is about $6.807\Omega$

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Example Question #2 : Resistors And Resistance

You have 3 resistors in parallel with each other. What can you say for certain about the total resistance of the circuit?

Possible Answers:

The total resistance is equal to the arithmetic mean of the resistors.

The total resistance is somwhere between the highest resistor and the lowest resistor.

The total resistance is less than any individual resistor.

The total resistance is higher than any individual resistor.

Nothing can be said for certain about the total resistance.

Correct answer:

The total resistance is less than any individual resistor.

Explanation:

Because the resistors are in parallel, we can use the equation for finding the total resistance.

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

Using this equation, any positive numbers we plug into the equation for the resistances will yield a number that is less than any of the resistors individually. Using this property allows for many more resistances to be achieved besides the individual resistors one may have.

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Example Question #1 : Resistors And Resistance

Resistors

The circuit in the figure has a battery providing a potential. The resistors have resistances of $R_{1}=3\Omega$ , $R_{2}=6\Omega$ , $R_{3}=8\Omega$ . Find the voltage drop across $R_{3}$ .

Possible Answers:

$\Delta V_{3}=5V$

$\Delta V_{3}=2.35V$

$\Delta V_{3}=1.32V$

$\Delta V_{3}=4V$

Correct answer:

$\Delta V_{3}=4V$

Explanation:

First off, we need to find the current coming out of the battery. If we consider $R_{1}$ and $R_{2}$ in parallel, we can find the equivalent resistance of those two. The circuit will look like

Resistors2

Where

$R_{12}=\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}}\right)^{-1}$

Notice in this figure above that $R_{3}$ is in series with $R_{12}$ . Finding the total equivalent reistance of the circuit will allow us to find the total current coming out of the battery.

$R_{eq}=R_{12}+R_{3}=\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}}\right)^{-1}+R_{3}$

$R_{eq}=\left ( \frac{1}{3\Omega}+\frac{1}{6\Omega}\right)^{-1}+8\Omega=10\Omega$

The total current coming out of the battery is found using Ohm's Law,

$I_{battery}=\frac{\varepsilon }{R_{eq}}=\frac{5V}{10\Omega}=0.5A$

Notice that if $R_{12}$ is in series with $R_{3}$ than the current coming out of the battery is the same current traveling through $R_{3}$ since for a series circuit

$I_{battery}=I_{12}=I_{3}$

This means that the voltage drop across $R_{3}$ is just

$\Delta V_{3}=I_{3}R_{3}=\left ( 0.5A\right )\left ( 8\Omega\right )=4V$

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Example Question #1 : Resistors And Resistance

You have two resistors in series, with the pair being in parallel with another resistor. What can be said for certain about the total resistance?

Possible Answers:

None of the answers are certain.

The total resistance is lower than any individual resistor.

The total resistance is the arithmetic mean of the resistors.

The total resistance is higher than any individual resistor.

The total resistance is somewhere in between the highest resistor and the lowest resistor (inclusive).

Correct answer:

None of the answers are certain.

Explanation:

Nothing can be said for certain about the total resistance. This is because, depending on the strength of the resistors, there can be practically any value for the total resistance. For instance, if our resistors were 1, 1, and 2, respectively, then after using the resistor equations (R¹ and R²), we find that the total resistance is 1 Ohm, which would eliminate the answers "arithmetic mean", "greater than any individual", and "less than any individual", which leaves "somewhere between the highest and lowest" and "none". However, we can use another example to disprove the first of those two.

Let the resistors equal 100, 100, and 1, respectively. The total resistance for that setup is $\frac{200}{201}$ , which is less than any individual resistor, not in between the highest and lowest, but we've already shown an example that contradicts this finding. Therefore, none of the answers are correct and can be said for certain.

$R^1\ (series)\ R_{tot}=R_1+R_2+...+R_n$

$R^2\ (parallel)\ R_{tot}=(\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_n})^{-1}$

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Example Question #1 : Resistors And Resistance

A circuit has two identical resistors in series. The resistors are then changed so they are in parallel. How will the current of the circuit change?

Possible Answers:

None of these.

It will double.

It will be cut in half.

It will be a quarter of it's original value.

It will be quadrupled.

Correct answer:

It will be quadrupled.

Explanation:

$R_{initial-total}=R_1+R_1=2R_1$

$\frac{1}{R_{final-total}}=\frac{1}{R_1}+\frac{1}{R_1}$

$\frac{1}{R_{final-total}}=\frac{2}{R_1}$

$R_{final-total}=\frac{R_1}{2}$

Putting the resistors in parallel makes the total resistance $\frac{1}{4}$ of the original value. Thus, the current is quadrupled.

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Example Question #321 : Electricity And Magnetism

Which of the following changes to a circuit will increase its overall resistance?

Possible Answers:

Decreasing the overall length of the circuit's conducting material

Adding a new resistor in parallel with another resistor in the circuit

Increasing the cross-sectional area of the wire that is used to carry the current

Decreasing the resistivity of the wire carrying the current

Adding a new resistor in series with another resistor in the circuit

Correct answer:

Adding a new resistor in series with another resistor in the circuit

Explanation:

In this question, we're asked to identify an answer choice that will increase the overall resistance of a circuit.

All conducting materials have an intrinsic resistance. This kind of resistance, called resistivity, is dependent on the type of material used to conduct the current. In addition to this, resistance is also dependent on the length of the conducting material as well as the cross-sectional area. A shorter length will lead to a smaller amount of resistance. Likewise, a larger cross-sectional area will result in decreased resistance.

Furthermore, it's important to recall that resistors add in series but add inversely in parallel. The reason for this is because resistors connected in series must all have the same current flowing through them. Because the sum total of the voltage drop of each of them must equal the overall voltage supplied by the external voltage source, the overall current (which will flow through each of them) is decreased. For a constant voltage, a decreased current means that the circuit as a whole must have a greater resistance.

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #41 : Circuit Components

Example Question #2 : Resistors And Resistance

Example Question #2 : Resistors And Resistance

Example Question #2 : Resistors And Resistance

Example Question #4 : Resistors And Resistance

Example Question #2 : Resistors And Resistance

Example Question #1 : Resistors And Resistance

Example Question #1 : Resistors And Resistance

Example Question #1 : Resistors And Resistance

Example Question #321 : Electricity And Magnetism

All AP Physics 2 Resources

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