Circuits - AP Physics 2

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Question

Consider the three circuits shown. In each circuit, the voltage source is the same, and all resistors have the same value. If each resistor represents a light bulb, which of the three circuits will produce the brightest light?

Answer

To answer this question, we first need to determine what we're looking for that will allow any given light bulb to produce bright light. The answer is power. The more energy that is delivered to the light bulb within a given amount of time, the brighter the light will be. Let's go ahead and look at the equation for power in a circuit.

$Power= \frac{Energy}{Time}=i\Delta V$

And since we're told in the question stem that the voltage source is identical in each circuit, we're looking for the circuit that has the largest current.

In order to find which circuit has the largest current, we'll need to invoke Ohm's law.

$\Delta V=iR$

$i=\frac{\Delta V}{R}$

What this shows is that a higher current will occur in the circuit that has the lowest total resistance. Thus, we'll need to determine what the total resistance is in each of the three circuits.

In circuit A, we can see that there is only a single resistor. Thus, we can give this circuit's total resistance a value of .

In circuit B, we have two resistors that are connected in series. Remember that when resistors are connected in this way, the overall resistance of the circuit increases. We find the total resistance by summing all the resistors connected in series.

Therefore, we can give circuit B a total resistance of .

Now, let's look at circuit C. We can see that there are two resistors connected in parallel, and each of these are connected in series with a third resistor. To solve for the total resistance of this circuit, we first need to determine the equivalent resistance of the two resistors connected in parallel. Once we find that value, then we can take into account the third resistor connected in series.

To solve for the resistance of the two resistors connected in parallel, we have to remember that they add inversely.

$\frac{1}{R_{Parallel}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}$

$R_{Parallel}=\frac{R}{2}$

Now that we've found the equivalent resistance for the two resistors connected in parallel, we can consider the third resistor connected in series.

$R_{Total}=\frac{R}{2}+R=\frac{3R}{2}$

Thus, we can give circuit C a value of $\frac{3R}{2}$ .

Now that we've found the total resistance for each circuit, we can obtain our answer. Since circuit A has the lowest total resistance, it will also have the greatest current. Consequently, it will also have the greatest power delivered to its resistor (the light bulb), thus causing the light coming from that bulb to be the brightest.

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit

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

Combine all of our voltage sources:

Plug in our values:

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Use Ohm's law:

We also know that:

$I_1+I_2=I_{Total}$

Substitute:

$I_2(1+\frac{R_2}{R_1})=I_{Total}$

Solve for :

$\frac{I_{Total}}{(1+\frac{R_2}{R_1})}=I_2$

We plug in our values.

$\frac{3.43A}{(1+\frac{2\Omega }{1\Omega })}=I_2$

Use Ohm's law:

$V=1.14A*2\Omega$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Using Ohm's law:

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_3(1+\frac{R_3}{R_4})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_3}{R_4})}=I_3$

Plug in our values:

$\frac{3.43 A}{(1+\frac{3 \Omega}{1\Omega})}=I_3$

Use Ohm's law:

$V=.856A*3\Omega$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R3}=V_{R4}$

Use Ohm's law:

We also know that:

$I_3+I_4=I_{Total}$

Substitute:

$I_4(1+\frac{R_4}{R_3})=I_{Total}$

Solve for :

$\frac{I_{Total}}{(1+\frac{R_4}{R_3})}=I_4$

$\frac{3.43amps}{(1+\frac{1\Omega}{3\Omega})}=I_4$

Plug in our values, we get

Use:

$V=2.57A*3\Omega$

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Question

You are researching new materials for usage in spacecraft electronics.

You have a new material, known as "Type G."

You carve out a cylinder of the material. It is tall, with a radius of .

You put electrodes on each face of the cylinder.

You determine the resistance to be .

What is the resistivity of "Type G?"

Answer

We will use the relationship

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

Where $\rho$ is the resistivity, is the resistance, is the surface area of the face the current is coming in or out of, and is the length from one face to the other.

Remember that the area of a circle is

$A=\pi\cdot r^2$

Combining our equations we get

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

We then need to plug in our values

$\rho=9.53\frac{\pi\cdot .015^2}{.025}$

$2.70\times 10^{-3}\Omega\cdot meters=\rho$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R6}=V_{R5}$

Use Ohm's law:

We also know that:

$I_5+I_6=I_{Total}$

Substitute:

$I_5(1+\frac{R_5}{R_6})=I_{Total}$

Solve for :

$\frac{3.43A}{(1+\frac{2\Omega}{3\Omega})}=I_5$

Plug in our values, we get

Use:

$V=2.06A*2\Omega$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the voltage drop across .

Answer

First, we need to find the total resistance of the circuit. In order to find the total resistance of the circuit, we need to combine all of the parallel resistors first, then add them together as resistors in series.

Combine with :

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{1}+\frac{1}{2}=\frac{1}{R_{1,2}}$

$R_{1,2}=\frac{2}{3}\Omega$

Combine with :

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{3}+\frac{1}{1}=\frac{1}{R_{3,4}}$

$R_{3,4}=\frac{3}{4}\Omega$

Combine with :

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

$\frac{1}{2}+\frac{1}{3}=\frac{1}{R_{5,6}}$

$R_{5,6}=\frac{6}{5}\Omega$

Then, add the combined resistors, which are now all in series:

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{Total}=\frac{2}{3}+\frac{3}{4}+\frac{6}{5}=2.62\Omega$

Then, we will need to use Ohm's law to determine the total current of the circuit.

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

Combine all of our voltage sources:

Plug in our values:

$\frac{9V}{2.62 \Omega}=3.43A$

We know that the voltage drop across parallel resistors must be the same, so:

$V_{R6}=V_{R5}$

Use Ohm's law

We also know that:

$I_5+I_6=I_{Total}$

Substitute:

$I_6(1+\frac{R_6}{R_5})=I_{Total}$

Solve for :

$\frac{I_{Total}}{(1+\frac{R_6}{R_5})}=I_6$

$\frac{3.43A}{(1+\frac{3\Omega}{2\Omega})}=I_6$

Plug in our values, we get

Use

$V=1.37a*3\Omega$

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Question

$R_1=R_4=1\Omega$

$R_2=R_5=2\Omega$

$R_3=R_6= 3\Omega$

Determine the voltage drop across

Answer

The first step is to find the total resistance of the circuit.

In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.

Combining with , with , with .

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R_{1,2}}$

$\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{R_{3,4}}$

$\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_{5,6}}$

Then, combining $R_{1,2}$ with $R_{3,4}$ and $R_{5,6}$ :

$R_{1,2}+R_{3,4}+R_{5,6}=R_{Total}$

$R_{1,2}=.667 \Omega$

$R_{3,4}=.75 \Omega$

$R_{5,6}=1.2 \Omega$

$R_{Total}= 2.62\Omega$

Ohms is used law to determine the total current of the circuit

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

Combing all voltage sources for the total voltage.

Plugging in given values,

$\frac{9V}{2.62\Omega}=I$

The voltage drop across parallel resistors must be the same, so:

$V_{R1}=V_{R2}$

Using ohms law:

It is also true that:

$I_1+I_2=I_{Total}$

Using Subsitution

$I_1(1+\frac{R_1}{R_2})=I_{Total}$

Solving for :

$\frac{I_{Total}}{(1+\frac{R_1}{R_2})}=I_1$

$\frac{3.43}{(1+\frac{1}{2})}=I_1$

Plugging back into ohms law in order to find the voltage drop.

.

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Question

identical resistors are placed in parallel. They are placed in a circuit with a battery. If the current through the battery is , determine the current through each resistor.

Answer

Since each resistor is in parallel, the voltage drop across each will be .

Since each resistor is identical, they all have the same resistance.

Using

$I_1+I_2+I_3=I_{total}=3A$

and

It is determined that

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Question

What is the current flowing through ?

Answer

, , and are in parallel, so we add them by using:

$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{R_{1,2,3}}}$

We find that $R_{1,2,3} = \frac{15}{13}\Omega$

$R_{1,2,3}$ , , and are in series. So we use:

$R_{1,2,3}+R_4 +R_5 = R_{total}$

$\frac{15}{13}\Omega+1\Omega +1\Omega = R_{total}$

$R_{total}=\frac{67}{13}\Omega$

We will then find the total current of the circuit. This will also be the current of because this resistor is not in parallel with any others.

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

$I=\frac{15V}{(67/13)\Omega}$

$I=\frac{195}{67} amps$

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Question

identical resistors are placed in parallel. They are placed in a circuit with a battery. If the current through the battery is , determine the resistance of each resistor.

Answer

Since each resistor is in parallel, the voltage drop across each will be .

Since each resistor is identical, they all have the same resistance.

Using

$I_1+I_2+I_3=I_{total}=3A$

and

It is determined that

Using

$R=3\Omega$ for all three resistors

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Question

$R_1=10\Omega$

$R_2=20\Omega$

What is the current through the battery in the above circuit?

Answer

First, find the total resistance of the circuit. Since the resistors are in parallel, use the following formula:

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

Plug in known values.

$\frac{1}{10\Omega}+\frac{1}{20\Omega}=\frac{1}{R_{total}}$

$R_{total}=6.66\Omega$

Next, use Ohm's law to find current.

Plug in known values.

$I=\frac{25V}{6.66\Omega}=3.75A$

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Question

If the maximum amount of charge held by a capacitor at a voltage of 12V is 36C, what is the capacitance of this capacitor?

Answer

In this question, we're told the maximum amount of charge that a capacitor can hold at a given voltage. We're then asked to determine the capacitance. To do this, we'll need to use the expression for capacitance.

$C=\frac{Q}{V}$

Plug in the values given to us in the question stem:

$C=\frac{36 C}{12 V}=3F$

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Question

You have 4 capacitors, , , , and , arranged as shown in the diagram below.

Their capacitances are as follows:

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

If you have a 6V battery connected to the circuit, what's the total energy stored in the capacitors?

Answer

The equation for energy stored in a capacitor is

$W=\frac{Q^2}{2C}=\frac{CV^2}{2}=\frac{QV}{2}$

We can find the capacitance by adding the capacitors together, and we have the voltage, so we'll use the second equation, $W=\frac{1}{2}CV^2$ .

When adding capacitors, remember how to add in series and parallel.

$(Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$(Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

Capacitors and are in series, and are in parallel, and and are in parallel.

$C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$C_{AB}=(\frac{1}{1}+\frac{1}{3})^{-1}$

$C_{AB}= \frac{3}{4}$

$C_{ABC}&=C_{AB}+C_C$

$C_{ABC}=\frac{3}{4}+2$

$C_{ABC}= \frac{11}{4}$

$C_{tot}&=C_{ABC}+C_{D}$

$C_{total}=\frac{11}{4}+4$

$C_{total}=\frac{27}{4}$

Now that we have the total capacitance, we can use the earlier equation to find the energy.

$W=\frac{1}{2}CV^2$

$W= \frac{1}{2}(\frac{27}{4})(6)^2$

The total energy stored is 121.5J.

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Question

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is not connected to a battery. If some external agent pulls the capacitor apart such that D doubles, did the internal energy, U, stored in the capacitor increase, decrease or stay the same?

Answer

Equations required:

$C=\frac{\varepsilon_{o}A}{D}$

$U=\frac{1}{2}\frac{Q^{2}}{C}$

We see from the first equation the when D is doubled, C will be halved (since and $\varepsilon_o$ are constant). From the third equation, we seed that when C is halved, the potential energy, U will double.

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Question

Four arrangements of capacitors are pictured. Each has an equivalent capacitance. Rank these four arrangements from highest equivalent capacitance to lowest. Assume that all capacitors are identical.

Answer

Let's go through all of them and find the equivalent capacitance.

(A) This is just

$C_{eq}=C$

(B) There are two capacitors in series, so this is

$C_{eq}=\left ( \frac{1}{C}+\frac{1}{C}\right )^{-1}=\left ( \frac{2}{C}\right )^{-1}=\frac{1}{2}C$

(C) These capacitors are in parallel, so

$C_{eq}=C+C+C=3C$

(D) These are a combination of series and parallel. Two are in series and they are in parallel with a third,

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

So, ranking them we get

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Question

Consider the circuit:

If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

Answer

In parallel branches of a circuit, the voltage drops are all the same. Therefore, we know that the voltage drop across C3 is also 5V.

We can then use the following equation to calculate the total stored energy:

$U=\frac{1}{2}CV^2$

Since the voltage is the same for both capacitors, we can simply add the two capacitances to do one calculation for energy:

$U = \frac{1}{2}(8F)(5V)^2 = 100J$

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Question

If , and , how much energy is stored in ?

Answer

In this circuit, the voltage source, and , and are all in parallel, meaning they share the same voltage.

To find the energy, we can use the formula

$U=\frac{1}2CV^2$ , with being the energy, being the capacitance, and being the voltage drop across that capacitor.

To use the formula we need the voltage across .

Another hint we can use is that and having the same charge since they're in series. First let's find the equivalent capacitance:

$C_{12}= \frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{4}{5}F$

Now, we can use the formula

to calculate charge in the capacitor.

$Q=CV=(\frac{4}{5})(5)=4 C$

Now that we know a charge of exists in both capacitors, we can use the formula again to find the voltage in only .

$Q_{1}=C_{1}\cdot V_{1}$

$V_{1}=\frac{4}{4}=1V$

Finally, we plug this into the first equation to calculate energy.

$U=\frac{1}{2}CV^2=\frac{1}{2}(4)(1)^2=2J$

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Question

A $15\mu F$ capacitor is connected to a battery. Once the capacitor is fully charged, how much energy is stored?

Answer

To find the amount of energy stored in a capactior, we use the equation

$U=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}QV=\frac{1}{2}CV^2$ .

We're given the capacitance (), and the voltage (), so we'll use the third equation.

$U=\frac{1}{2}(15\mu F)(9V)$

$U= 6.075\cdot 10^{-4}J$

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Question

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the internal energy, U, stored in the capacitor increase, decrease or stay the same?

Answer

Relevant equations:

$C=\frac{\varepsilon_{o}A}{D}$

$U=\frac{1}{2}\frac{Q^{2}}{C}$

We must note for this problem that the voltage, V, is kept constant by the battery. Looking at the second equation, if C changes (by changing D) then Q must change when V is held constant. This means that our formula for U must be altered. How can we make claims about U if Q and C are both changing? We need a constant variable in the equation for U so that we can make a direct relationship between D and U. If we plug in the second equation into the third we arrive at:

$U=\frac{1}{2}CV^{2}$

Now, we know that V is held constant by the battery, so when C decreases (because D doubled) we see that U actually decreases here. So it matters whether or not the capacitor is hooked up to a battery.

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