Resistors and Resistance - AP Physics 2

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Question

Suppose a circuit consists of a battery connected to four resistors, all of which are connected in parallel. If a fifth resistor is added in parallel to this circuit, how will the current in the other four resistors change?

Answer

Since all the resistors are connected in parallel, they will all experience the same voltage drop, which is equal to the voltage of the battery. Furthermore, since each resistor has the same resistance, the current flowing through the four resistors will be equal.

The current flowing through each resistor can be shown by Ohm's law:

Also, the total current of the circuit can be found by noting:

$V=i_{Total}R_{Total}$

Moreover, the total resistance can be calculated by noting that resistors in parallel add inversely.

$\frac{1}{R_{Total}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}$

Since each resistor is the same, we can note that:

$\frac{1}{R_{Total}}=\frac{4}{R}$

$R_{Total}=\frac{R}{4}$

Plugging $R_{Total}$ into the previous equation, we see that:

$i_{Total}=\frac{V}{\frac{R}{4}}=\frac{4V}{R}$

So in the case of four resistors connected in parallel, the total current in the circuit is equal to $\frac{4V}{R}$ . And since the overall current is split evenly between the four resistors, we see that:

$i=\frac{(\frac{4V}{R})}{4}=\frac{V}{R}$

Now, if we consider the case of five resistors connected in parallel, we will see that the total resistance is:

$\frac{1}{R_{Total}^{'}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}$

$\frac{1}{R_{Total}^{'}}=\frac{5}{R}$

$R_{Total}^{'}=\frac{R}{5}$

Therefore, the overall current in the circuit when a fifth resistor is added in parallel is:

$i_{Total}^{'}=\frac{V}{R_{Total}^{'}}=\frac{V}{(\frac{R}{5})}=\frac{5V}{R}$

And once again, since each of the five resistors are connected in parallel, the total current will be split evenly between them, thus giving:

$i^{'}=\frac{(\frac{5V}{R})}{5}=\frac{V}{R}$

$i=i^{'}=\frac{V}{R}$

Therefore, we see that the current through each resistor is the same, both before and after the addition of the fifth resistor.

To summarize, the voltage drop across each resistor is the same since they are connected in parallel. Also, since each resistor has the same resistance, the current through each is the same. Adding a fifth resistor in parallel decreased the overall resistance of the circuit and increased the overall current flowing through the circuit. But because this new, greater current is divided evenly among all the resistors, including the new one that was added, the current through each of the resistors does not change.

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Question

$R_1=5\Omega$

$R_2=7\Omega$

What is the total resistance of the circuit above?

Answer

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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Question

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

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

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

Answer

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

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

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

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 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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Question

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?

Answer

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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Question

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?

Answer

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:

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

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

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

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

Now we can find the power.

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

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

This fraction is equivalent to

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Question

$R_A=3\Omega$

$R_B=6\Omega$

$R_C=9\Omega$

What is the total resistance of the circuit shown above?

Answer

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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Question

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

Answer

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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Question

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?

Answer

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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Question

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 $R_{3}$ .

Answer

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

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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Question

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?

Answer

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

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

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

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

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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Question

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

Answer

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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