Circuit Components

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AP Physics 2 › Circuit Components

Questions 1 - 10

Lazy capacitor

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.

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

$n=2.21*10^{9}$

$n=5.85*10^{9}$

$n=5.15*10^{9}$

$n=3.33*10^{9}$

None of these

Explanation

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.

$V_{C1}=E*d$

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:

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

$E*\epsilon_0*A=q$

Convert to and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

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

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:

$\frac{3.54*10^{-10}}{1.6*10^{-19}}=n$

$n=2.21*10^{9}$

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

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

$6.75 \cdot 10^{-5} J$

$1.35 \cdot 10^{-5} J$

$1.215 \cdot 10^{-3} J$

None of the other answers is correct

Explanation

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.

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

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

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

The total resistance is less than any individual resistor.

The total resistance is higher than any individual resistor.

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

Nothing can be said for certain about the total resistance.

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

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.

Two parallel conducting plates separated by a distance are connected to a battery with voltage . If the distance between them is doubled and the battery stays connected, which of the following statements are correct?

The electric charge on the plates is halved

The potential difference between the plates is doubled

The capacitance is unchanged

The potential difference between the plates is halved

The electric charge on the plates is doubled

Explanation

The equation for capacitance of parallel conducting plates is:

$C=\frac{\epsilon_0A}{d}$

When the distance is doubled, the capacitance changes to:

$C'=\frac{\epsilon_0A}{d'}=\frac{\epsilon_0A}{2d}=\frac{1}{2}C$

The battery is still connected to the plates, the potential difference is unchanged. Because , and the capacitance is halved while the voltage stays the same, must necessarily drop to half to account for the change.

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

The total resistance is less than any individual resistor.

The total resistance is higher than any individual resistor.

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

Nothing can be said for certain about the total resistance.

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

Explanation

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

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

The electric charge on the plates is halved

The potential difference between the plates is doubled

The capacitance is unchanged

The potential difference between the plates is halved

The electric charge on the plates is doubled

Explanation

The equation for capacitance of parallel conducting plates is:

$C=\frac{\epsilon_0A}{d}$

When the distance is doubled, the capacitance changes to:

$C'=\frac{\epsilon_0A}{d'}=\frac{\epsilon_0A}{2d}=\frac{1}{2}C$

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?

$\frac{12}{13} \mu F$

$\frac{13}{12} \mu F$

$9 \mu F$

$\frac{1}{9} \mu F$

None of the other answers is correct

Explanation

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 $\frac{12}{13} \mu F$ .

Lazy capacitor

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.

$3.54*10^{-10}C$

$9.26*10^{-10}C$

$6.11*10^{-10}C$

$1.11*10^{-10}C$

$4.44*10^{-10}C$

Explanation

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.

$V_{C1}=E*d$

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:

$E*\epsilon_0*A=q$

Convert to and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

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

Lazy capacitor

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.

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

$n=2.21*10^{9}$

$n=5.85*10^{9}$

$n=5.15*10^{9}$

$n=3.33*10^{9}$

None of these

Explanation

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.

$V_{C1}=E*d$

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:

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

$E*\epsilon_0*A=q$

Convert to and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

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

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:

$\frac{3.54*10^{-10}}{1.6*10^{-19}}=n$

$n=2.21*10^{9}$

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

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

$6.75 \cdot 10^{-5} J$

$1.35 \cdot 10^{-5} J$

$1.215 \cdot 10^{-3} J$

None of the other answers is correct

Explanation

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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