Electric Circuits - AP Physics C: Electricity and Magnetism

Card 0 of 231

Question

A parallel plate capacitor has a capacitance of . If the plates are apart, what is the area of the plates?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Answer

The relationship between capacitance, distance, and area is $C=\epsilon_0 \frac{A}{d}$ . We can rearrange this equation to solve for area.

$A=\frac{Cd}{\epsilon_0}$

Now, we can use the values given in teh question to solve.

$A=\frac{1F(110^{-2}m)}{8.85 10^{-12} \frac{F}{m}}=1.1*10^9m^2$

Compare your answer with the correct one above

Question

Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of $\pm \sigma$ on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area and are separated by distance . A material of dielectric constant $\kappa_{material}$ has been placed between the two plates.

Which of the following would not result in an increase in the measure of electric potential difference between the two plates?

Answer

The electric potential difference created between the plates of a parallel plate capactor is given by the equation:

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

The charge can be calculated by using the equation:

$Q=\sigma A$

The value of the capacitance is related to the dimensions of the capacitor with the equation:

$C=\kappa \frac{\varepsilon _{o}A}{d}$

Combining these equations yields:

$V=\frac{\sigma A}{\frac{\kappa \epsilon_0A}{d}}=\frac{\sigma d}{\varepsilon_{o} \kappa }$

The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.

Compare your answer with the correct one above

Question

You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of $700\mu F$ out of those metal sheets, and the sheets needed to be apart, what area do the two metal plates need to be?

$\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Answer

For parallel plate capacitors, the equation is $C=\frac{\epsilon_0A}{d}$ .

Solve for .

$A=\frac{dC}{\epsilon_0}$

Now we can plug in our given values.

$A=\frac{(0.510^{-3})(70010^{-6})}{8.8510^{-12}\frac{\text{C}^2}{\text{Nm}^2}}$

$A=3.9510^4\ \text{m}^2$

Compare your answer with the correct one above

Question

The plates of a parallel plate capacitor are apart and in area. A potential difference of is applied across the capacitor. Find the capacitance.

$\epsilon_0=8.85*10^{-12} \frac{F}{m}$

Answer

Capacitance is related to plate area and distance by the equation $C=\epsilon_0\frac{A}{d}$ .

Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.

$C=8.85* 10^{-12} \frac{F}{m}(\frac{5 m^2}{310^{-3}m})=8.85 10^{-12} \frac{F}{m}(1.6710^{3}m)=1.510^{-8}F$

Compare your answer with the correct one above

Question

The plates of a parallel plate capacitor are apart and in area. A potential difference of is applied across the capacitor. Compute the charge on each plate.

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Answer

Charge on a capacitor is given by the equation . We know that the voltage is , but we need to determine the capacitance based off of the area and distance between the plates.

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

$C=(8.8510^{-12}\frac{F}{m})(\frac{5m^2}{310^{-3}m})$

$C=(8.8510^{-12}\frac{F}{m})(1.6710^{3}m)=1.510^{-8}F$

We can plug this value into the equation for charge.

$Q=CV=(1.510^{-8}F)(1000V)$

$Q=1.5*10^{-5}C$

Compare your answer with the correct one above

Question

The plates of a parallel plate capacitor are apart. A potential difference of is applied across the capacitor. Compute the magnitude of the generated electric field.

$\epsilon_0=8.85*10^{-12}\frac{F}{m}$

Answer

The electric field given by a capacitor is given by the formula $E=\frac{Q}{\epsilon_0 A}$ . We do not have these variables, so we will have to adjust the equation.

$Q=CV\rightarrow E=\frac{CV}{\epsilon_0A}$

The capacitance can be determined by the area of the plates and the distance between them.

$C=\epsilon_0\frac{A}{d}\rightarrow E=\frac{\epsilon_0\frac{A}{d}V}{\epsilon_0A}$

This equation simplifies to $E=\frac{V}{d}$ , allowing us to solve using the values given in the question.

$E=\frac{1000V}{310^{-3}m}=3.3310^{5}\frac{N}{C}$

Compare your answer with the correct one above

Question

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

Answer

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

Compare your answer with the correct one above

Question

Two parallel conducting plates each have an area of and are separated by a distance, . One plate has a charge evenly distributed across it, and the other has a charge of . A proton (charge ) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?

Answer

Relevant equations:

$W = q\Delta V$

$W = \Delta KE$

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

$C = \frac{\epsilon _o A}{d}$

According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.

1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other:

$\frac{Q}{V}=\frac{\epsilon _o A}{d}$

$V = \frac{Qd}{\epsilon _o A}$

2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):

$W = \frac {eQd}{\epsilon _o A} = \Delta K$

Compare your answer with the correct one above

Question

A $5 \mu F$ and $10 \mu F$ capacitor are connected in series with a battery. What is the magnitude of the charge on one plate of the $5 \mu F$ capacitor?

Answer

Relevant equations:

$\frac{1}{C_{eq, series}} = \frac{1}{C_1} + \frac{1}{C_2}$

First, find the equivalent capacitance of the whole circuit:

$\frac{1}{C_{eq,series}}= \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$

$C_{eq, series} = \frac{10}{3} \mu F$

Use this equivalent capacitance to find the total charge:

$Q = CV = (\frac{10}{3} \mu F) * (12 V) = 40 \mu C$

Capacitors in series always have the same charge on each unit, so the charge on the $5 \mu F$ capacitor is also $40 \mu C$ .

Compare your answer with the correct one above

Question

Three resistors and a battery form the following circuit.

Calculate the equivalent resistance if the resistors have the following values.

$R_1=40\Omega$

$R_2=25\Omega$

$R_3=38\Omega$

Answer

First, calculate the equivalent resistance of and . Since these two resistors are arranged in series, we just take the sum of their values.

$R_{23}=R_{2}+R_{3}$

$R_{23}=25\Omega+38\Omega$

$R_{23}=63\Omega$

With resistors 2 and 3 combined together in a single value, the following circuit is formed.

Notice that and $R_{23}$ are arranged in parallel. To calculate the equivalent resistance of this parallel pair, we use the following equation.

$\frac{1}{R_{123}}=\frac{1}{R_1}+\frac{1}{R_{23}}$

Plug in the values, and solve for $R_{123}$ .

$\frac{1}{R_{123}}=\frac{1}{40\Omega}+\frac{1}{63\Omega}$

$R_{123}=24.5\Omega$

This is just like the circuit shown below.

Compare your answer with the correct one above

Question

Resistors are one of the most important basic components of a circuit. With very few exceptions, all circuits have at least one kind of resistor component. An ammeter is a device that measures current flowing through a circuit. Ammeters are always connected to a circuit in series.

Which of the following accurately explains why ammeters must be connected in series within a circuit, and never in parallel?

Answer

In order to give accurate readings of current, ammeters have very low resistances. If connected in parallel, the voltage pushing current through the circuit would push a very strong current through the ammeter and virtually no current through the circuit's regular path. This would not only lead to a bad reading of current, but more often than not a broken ammeter. This phenomenon is an indication of why resistors are so important: they limit the current such that a circuit does not exceed its current-carrying capacity.

Compare your answer with the correct one above

Question

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

Why was the potential of the battery measured differently when the wire was connected?

Answer

All wires have at least some internal resistance. The most likely explanation for this is that the wire is displaying slight resistance, and therefore caused the measured potential to be less than it was before.

Compare your answer with the correct one above

Question

A lamp has a bulb. If the house wiring provides to light up that bulb, how much current does the bulb draw?

Answer

The formula for power is ,and we are given the following values.

$P=60\ \text{W}$

$V=110\ \text{V}$

Solve for the current, .

$I=\frac{P}{V}=\frac{60\ \text{W}}{110\ \text{V}}=0.55\ \text{A}$

Compare your answer with the correct one above

Question

A particle accelerator with a radius of 500 meters can have up to $10^{15}$ protons circulating within it at once.

$q_{proton}=1.60*10^{-19}C$

How fast must the protons in the accelerator move in order to produce a current of 1A?

Answer

The current produced is the total charge that circulates the particle accelerator per unit time.

$I=\frac{dQ}{dt}$

We calculate this by the equation:

$I = \frac{nqv}{2\pi R}$

is the number of protons, is the charge per proton, is the velocity of each proton, and is the radius of the particle accelerator.

Using the given current, we then solve for the velocity:

$1A = \frac{(10^{15})(1.6010^{-19}C)v}{2\pi (500m)}$

$v=\frac{(1A)2\pi(500m)}{(10^{15})(1.6010^{-19}C)}= 1.96*10^7\frac{m}{s}$

Compare your answer with the correct one above

Question

A straight copper wire has a fixed voltage applied across its length. Which of the following changes would increase the power dissipated by this wire?

Answer

Relevant equations:

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

Current and resistance are inversely proportional to one another, assuming voltage is fixed. Since , changes in current effect the power more than changes in resistance do. Thus, we need current to increase, meaning that resistance must decrease.

To decrease resistance, we could:

1. Change the material of the wire to one of lesser resistivity

2. Decrease the length of the wire

3. Increase the cross-sectional area of the wire

4. Decrease the temperature of the wire (very slight effect on resistance)

Compare your answer with the correct one above

Question

A battery is measured to have a potential of 5V. When connected to a wire with no resistors or other components, the voltage measured is 4.9V.

If the current through the wire is measured to be 2A, how much thermal energy is being lost per second as soon as the wire is connected to the battery?

Answer

First, we must know that the wire has some internal resistance . To calculate this, we need to know the potential drop through the wire, which must be the difference we saw from the initial voltage reading to the second. This value, 0.1V, we plug into Ohm's law to calculate the resistance of the wire.

$R=0.05\Omega$

The question asks for energy lost per second; this value is equivalent to the power.

Use our values to solve.

$P=(2A)^2(0.05\Omega)=0.2W$

Compare your answer with the correct one above

Question

A simple circuit contains two $12\Omega$ resistors in parallel, connected to a 20V source. What power is being provided by the source to the circuit?

Answer

The power supplied to the circuit can be calculated using the equation:

$P=IV=\frac{V^2}{R}$

To use this equation, we need to find the equivalent resistance of the circuit. Use the equation for equivalent resistance in parallel:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R_{eq}}=\frac{1}{12\Omega}+\frac{1}{12\Omega}=\frac{1}{6\Omega}$

$R_{eq}=6\Omega$

Now that we have the resistance and the voltage, we can solve for the power.

$P=\frac{V^2}{R}=\frac{(20V)^2}{6\Omega}\approx 67W$

Compare your answer with the correct one above

Question

Three resistors R1, R2, and R3 and the capacitor, C, are connected to an ideal battery V to complete the circuit as shown.

After the circuit has been connected for a very long time, the currents in each branch of the circuit are measured to be some values , , and , and the charge on the capacitor some value .

Which of the following equations is a correct expression regarding the voltage of the circuit after a long time?

Answer

Identifying the sum of the voltage drops and rises (Kirchoff's Loop Law) around the three possible loops of this circuit is the key to answering this question correctly. The following signs can be assigned to each of the circuit elements based on the direction of the currents given.

Use Ohm's law and the equation for capacitance to derive terms for the voltage across each element of the circuit.

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

There are three possible paths through the circuit, resulting in three correct equations that could be derived:

$0=+V-\frac{Q}{}C-R_{3}I_{3} -R_{1}I_{1}$

$0=-\frac{Q}{}C-R_{3}I_{3} +R_{2}I_{2}$

$0=+V-R_{2}I_{2} -R_{1}I_{1}$

Only one of the given answer options matches up correctly to these.

Compare your answer with the correct one above

Question

A simple circuit consists of a $2\ \text{M}\Omega$ resistor that's connected to a battery. How much power is dissipated by the resistor?

Answer

One of the formulas for power is $P=\frac{V^2}{R}$ . We are given the values of the voltage and resistance.

$V=9\ \text{V}$

$R=2\ \text{M}\Omega=210^6\ \Omega$

Using these values, we can solve for power.

$P=\frac{(9\ \text{V})^2}{210^6\ \Omega}$

$P=4*10^{-5}\ \text{W}$

Compare your answer with the correct one above

Question

$R_1=15\Omega,\ R_2=20\Omega,\ R_3=20\Omega$

What is the current delivered to the curcuit by the battery?

Answer

Using Ohm's law to solve for the value of the current from the battery requires calculation of the equivalent resistance of the circuit.

$V=IR_{eq}$

The resistors R2 and R3 are in parallel with one another. Once combined, their requivalent resistor (R23) is in series with R1.

The equivalent resistance is thus given by:

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

$R_{eq}=\left ( \frac{1}{20\Omega} + \frac{1}{20\Omega} \right )^{-1} + 15\Omega$

$R_{eq}=10\Omega + 15\Omega=25\Omega$

Use this value and the given voltage to solve for the current in the circuit:

$I = \frac{V}{R_{eq}}$

$I=\frac{50V}{25\Omega}=2.0A$

Compare your answer with the correct one above

Tap the card to reveal the answer