AP Physics 2 - Electricity and Magnetism

3 sets of parallel resistors

$R_1=R_4=1\Omega$

$R_2=R_5= 2\Omega$

$R_3=R_6 = 3\Omega$

Determine the total resistance of the given circuit.

$2.62\Omega$

$3\Omega$

$12\Omega$

$3.67\Omega$

Explanation

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:

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

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

A proton is traveling parallel to a wire in the same direction as the conventional current. The proton is traveling at $66\frac{m}{s}$ . The current in the wire is . The proton and the wire are apart. Determine the magnetic force on the proton.

$1.76*10^{-21}N$

None of these

$1.95*10^{-21}N$

$4.05*10^{-21}N$

$1.22*10^{-21}N$

Explanation

Finding the magnetic field at the location of the proton.

$B=\frac{\mu_0*I}{2*\pi*r}$

Converting to and plugging in values

$B=\frac{4*\pi*10^{-7}*7.5}{2*\pi*.009}$

$B=1.67*10^{-4}T$

Using

$F_{magnetic}=qvB$

$F_{magnetic}=1.6*10^{-19}*66*1.67*10^{-4}$

$F_{magnetic}=1.76*10^{-21}N$

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

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.

$1.76*10^{-21}N$

None of these

$1.95*10^{-21}N$

$4.05*10^{-21}N$

$1.22*10^{-21}N$

Explanation

Finding the magnetic field at the location of the proton.

$B=\frac{\mu_0*I}{2*\pi*r}$

Converting to and plugging in values

$B=\frac{4*\pi*10^{-7}*7.5}{2*\pi*.009}$

$B=1.67*10^{-4}T$

Using

$F_{magnetic}=qvB$

$F_{magnetic}=1.6*10^{-19}*66*1.67*10^{-4}$

$F_{magnetic}=1.76*10^{-21}N$

Which of the following will cause a superconductor to have 0 resistance?

Cooling it to below its critical temperature

Heating it to above its critical temperature

Increasing the applied voltage to beyond the critical voltage

Stretching the wire to a sufficiently small diameter

Increasing the applied pressure to above the critical pressure.

Explanation

A superconductor is a material that has 0 electrical resistance when cooled to below a certain temperature.

In general, materials have a decreasing resistance as they are cooled. With a superconductor, once the critical temperature is reached, the resistance abruptly goes to 0. Superconductivity is a quantum mechanical phenomenon.

You place a box with square faces of side length 4m in a uniform electric field of strength $39 \frac{N}{C}$ . There is no charge in the box. What is the total electric flux going through the box?

$0\frac{N\cdot m^2}{C}$

$624\frac{N\cdot m^2}{C}$

$104\frac{N\cdot m^2}{C}$

$3744\frac{N\cdot m^2}{C}$

$312\frac{N\cdot m^2}{C}$

Explanation

A way to think of flux is to count the number of electric field lines exiting a shape and subtract from it the number of field lines entering the shape. The only way for there to be more lines exiting the shape than entering the shape (and the only way to have any flux) is when the shape is enclosing charge.

In our problem, the box has no charge in it. Therefore, it has the same amount of field lines leaving it as it does entering it. This means that there is 0 net flux through the box.

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:

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

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

A circuit contains a battery and a $20\Omega$ resistor in series. Determine the magnitude of the magnetic field outside of the loop away from the wire.