Circuit Components - AP Physics 2

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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

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

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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Question

Consider the circuit:

What is the total equivalent capacitance?

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

Answer

Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:

$C_{eq}=C2+C3 = 7F+1F= 8F$

The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:

$\frac{1}{C_{tot}}=\sum\frac{1}{C} = \frac{1}{5}+\frac{1}{8} = \frac{13}{40}$

$C_{tot} = 3.08F$

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Question

In this parallel-plate capacitor, the distance between the plates is , the area of each plate is , and the voltage across them is .

Calculate the capacitance.

$\epsilon_0=8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}$

Answer

The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.

For a parallel-plate capacitor,

$C=\frac{\epsilon _{0}A}{d}$

Plugging in the numbers given results in

$C=\frac{8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}\cdot0.0015m^2}{0.03m}$

$C=4.43\cdot 10^{-13}F$

Be careful to convert the units to meters!

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Question

You have 3 capacitors in series. Their capcitance's are $4\mu F$ , $3\mu F$ , and $2\mu F$ . What is the total capacitance of the system?

Answer

To find the total capacitance of capacitors in series, you use the following equation:

$\frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...+\frac{1}{C_n}$ .

Our values for , , and are 4, 3, and 2. Now, we can plug in our values to find the answer.

$\frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\frac{1}{C_{tot}}=\frac{1}{4}+\frac{1}{3}+\frac{1}{2}$

$\frac{1}{C_{tot}}= \frac{13}{12}$

The answer we have is the inverse of this. Therefore, the total capacitance is $\frac{12}{13} \mu F$ .

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Question

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

Their capacitances are as follows:

$\begin{align} A&=3F \ B&=6F \ C&=4F \ D&=1F \end{align}$

What is the total capacitance of the circuit?

Answer

Remember, the equations for adding capacitances are as follows:

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

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

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

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

$C_{AB}=(\frac{1}{3}+\frac{1}{6})^{-1}$

$C_{AB}= 2$

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

$C_{ABC}=2+4$

$C_{ABC}= 6$

$C_{total}=C_{ABC}+C_{D}$

$C_{total}=6+1$

$C_{total}=7$

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Question

Three capacitors are in parallel with each other. What can be said for certain about the total capacitance?

Answer

Capacitors in parallel combine according to the following equation:

$C_{tot}=C_1+C_2+...+C_n$ .

Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.

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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 voltage difference between the plates increase, decrease or stay the same?

Answer

The fact that the system is still connected to the battery indicate a constant V so regardless what happens to the capacitor, V stays fixed.

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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 charge on each plate increase, decrease or stay the same?

Answer

Relevant equations:

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

Plug the second equation into the first:

$Q=\frac{\varepsilon _{o}AV}{D}$

Considering all the variables in the numerator are held fixed for this problem, we see that increasing D will decrease the charge stored on each plate.

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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 of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the charge on each plate increase, decrease or stay the same?

Answer

The charge has no where to go. Without the battery connected, the charge has no physical avenue on or off the plates.

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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, how did the capacitance change? Also, do we need to use the information provided about the battery?

Answer

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

As D increases, C will decrease

Does the battery matter? No. Capacitance is a "geometric" quantity. This means that it can be determined solely by physical parameters like the area and separation distance. The charge on the plates, voltage difference, electric field and any other quantity you could think of does not influence the capacitance other than A or D.

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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

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

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates at apart, determine the excess charge on the positive plate.

Answer

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=Ed$

Combine equations and solve for the electric field:

$\frac{V_1}{d}=E$

Convert mm to m and plug in values:

$\frac{V_1}{d}=E$

$4000\frac{N}{C}=E$

Using the electric field in a capacitor equation:

$E=\frac{q}{A\epsilon_0}$

Rearrange to solve for the charge:

$E\epsilon_0A=q$

Convert to and plug in values:

$(4000)(8.8610^{-12})(10010^{-4})=q$

$3.54*10^{-10}C=q$

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Question

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates are 0.1mm apart, determine the number of excess electrons on the negative plate.

$e=1.6*10^{-19}C$

Answer

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=Ed$

Combine equations and find the electric field:

$\frac{V_1}{d}=E$

Convert mm to m and plugg in values:

$\frac{V_1}{d}=E$

$4000\frac{N}{C}=E$

Use the electric field in a capacitor equation:

$E=\frac{q}{A\epsilon_0}$

$E\epsilon_0A=q$

Convert to and plug in values:

$(4000)(8.8610^{-12})(10010^{-4})=q$

$3.5410^{-10}=q$

The magnitude of total charge on the positive plate is equal to the total charge on the negative plate, so to find the number of excess elections:

$\frac{3.5410^{-10}}{1.610^{-19}}=n$

$n=2.2110^{9}$

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Question

Suppose I have a uniform electric field within a parallel plate capacitor with field strength of $E=41\frac{N}{C}$ .

Suppose the capacitor's plates are in length and in width, and the space between the plates is .

Determine the capacitance of this system given that the space in between is a vacuum, and that the permittivity of empty space $\epsilon_0$ is $8.85*10^{-12}\frac{F}{m}$ .

Answer

The formula for capacitance is given by:

$C=\frac{\kappa \epsilon_0 A}{d}$ ,

Where $\kappa$ is dielectric strength, is distance between plates, $\epsilon _0$ is permittivity of empty space, and is cross sectional area.

To determine , we do

$A=(.5.001)m^2=510^{-4}m^2$

$\kappa=1$ because between the plates is a vacuum.

Putting it all together,

$C=2.95*10^{-13}F$

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