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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #1 : Series And Parallel

If we have 3 resistors in a series, with resistor 1 having a resistance of $5\Omega$ , resistor 2 having a resistance of $10\Omega$ , and resistor 3 having a resistance of $15\Omega$ , what is the equivalent resistance of the series?

Possible Answers:

$40\Omega$

$30\Omega$

$10\Omega$

$20\Omega$

Correct answer:

$30\Omega$

Explanation:

The total resistance of resistors in a series is the sum of their individual resistances. In this case,

$R_{tot}=R_1+R_2+...+R_n$

$R_{tot}=5\Omega +10\Omega + 15\Omega=30\Omega$

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Example Question #1 : Series And Parallel

You are presented with three resistors, each measure $3 \Omega$ . What is the difference between the total resistance of the resistors combined in series, and the total resistance of the resistors combined in parallel?

Possible Answers:

$0\Omega$

$8\Omega$

$6\Omega$

$9\Omega$

$1\Omega$

Correct answer:

$8\Omega$

Explanation:

Resistors in series:

R_s = R_1+R_2+R_3

$R_s=3\Omega+3\Omega+3\Omega$

$R_s=9\Omega$

Resistors in parallel:

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

$\frac{1}{R_p}=\frac{1}{3\Omega}+\frac{1}{3\Omega}+\frac{1}{3\Omega}$

$\frac{1}{R_p}=\frac{3}{3\Omega}$

$R_p=1\Omega$

$R_s-R_p=8\Omega$

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Example Question #1 : Series And Parallel

What is the total resistance of three resistors, $100\Omega$ , $10\Omega$ , and $1\Omega$ , in parallel?

Possible Answers:

$111\Omega$

$11.1\Omega$

$9.0\Omega$

$0.9\Omega$

$1.1\Omega$

Correct answer:

$0.9\Omega$

Explanation:

The equation for equivalent resistance for multiple resistors in parallel is:

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

Plug in known values and solve.

$\frac{1}{R_{eq}}=\frac{1}{100\Omega}+\frac{1}{10\Omega}+\frac{1}{1\Omega}$

$\frac{1}{R_{eq}}=\frac{111}{100\Omega}$

$R_{eq}=0.9\Omega$

Notice that for resistors in parallel, the total resistance is never greater than the resistance of the smallest element.

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Example Question #1 : Series And Parallel

Series parallel circuit jpeg

A circuit is created using a battery and 3 identical resistors, as shown in the figure. Each of the resistors has a resistance of $100\Omega$ . If resistor is removed from the circuit, what will be the effect on the current through resistor ?

Possible Answers:

The current through will decrease

The current through will increase by a factor of two

The current through will remain the same

The current through will increase by a factor of four

Cannot be determined without knowing the resistivity of the wire

Correct answer:

The current through will decrease

Explanation:

Since the resistors and form a parallel network, removing from the circuit increases the resistance of that part of the circuit. Because the new circuit is the series combination of and , the increased resistance leads to lower current in each of these resistors.

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Example Question #1 : Series And Parallel

Determine the total charge stored by a circuit with 2 identical parallel-plate capacitors in parallel with area 5m^2 , and a distance of $2 \mu m$ between the parallel plates. Assume the space between the parallel plates is a vacuum. The circuit shows a voltage difference of 10V.

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

Possible Answers:

$2.21*10^{-5}C$

$6.44*10^{-5}C$

$4.42*10^{-4}C$

$1.11*10^{-4}C$

Correct answer:

$4.42*10^{-4}C$

Explanation:

To determine total charge stored, we need to add up the capacitance of each capacitor(because they are capacitors in parallel) and multiply by the voltage difference. Recall that for capacitors,

$Q=C \Delta V$

For parallel plate capacitors:

$C=\frac{k\epsilon_oA}{d}$

Here, $\epsilon_o= 8.85*10^{-12}\frac{F}{m}$ , which is the permittivity of empty space, is the dielectric constant, which is since there is only vacuum present, A=5m^2 , which is the area of the parallel plates, and $d=2*10^{-6}m$ , which is the distant between the plates.

Plug in known values to solve for capacitance.

$C=\frac{1*8.85*10^{-12}\frac{F}{m}*5m^2}{2*10^{-6}m}= \frac{4.43*10^-11Fm}{2*10^{-6}m}$

$C=2.21*10^{-5}F$

Each of the two capacitors has capacitance $C_{capacitor}=2.21*10^{-5}F$

Since the capacitors are in parallel, the total capacitance is the sum of each individual capacitance. The total capacitance in the circuit $C_{circuit}$ is given by:

$C_{circuit}=2*C_{capacitance}=2*2.21*10^{-5}F= 4.42*10^{-5}F$

Plug this value into our first equation and solve for the total charge stored.

$Q_{circuit}=C_{circuit}* \Delta V$ , where $Q_{circuit}$ is the total charge stored in the capacitor. Since $\Delta V= 10V$

$Q_{circuit}=4.42*10^{-5}F* 10V= 4.42*10^{-4}C$

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Example Question #1 : Series And Parallel

What is the total current flowing through a system with 2 resistors in parallel with resistances of $2 \Omega$ and $5 \Omega$ , and a battery with voltage difference of 10V?

Possible Answers:

I=7A

I=10A

$I=\frac{10}{7}A$

I=7A

Correct answer:

I=7A

Explanation:

First we need to determine the overall resistance of the circuit before we know how much current is flowing through. Since the resistors are in parallel, their resistances will add reciprocally:

$\frac{1}{2\Omega}+\frac{1}{5\Omega}= \frac{1}{R}$

where is the total resistance of the circuit.

$\frac{7}{10\Omega}=\frac{1}{R}$

$\frac{10}{7}\Omega= R$

Now that we've solved for , we know that the current flowing through the circuit can be found using Ohm's law:

$I=\frac{V}{R}=\frac{10V}{\frac{10}{7}\Omega}=7A$

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Example Question #11 : Series And Parallel

Consider two circuits: one contains two resistors wired in series, each with a resistance of $20\Omega$ , while the other contains two resistors wired in parallel, one with a resistance of $50\Omega$ and the other with an unknown resistance. The circuits are completely independent, each having its own 120V battery, and each drawing a current of . What must the resistance of the unknown resistor be for the two circuits to have the same total resistance?

Possible Answers:

$50\Omega$

$160\Omega$

$400\Omega$

$40\Omega$

$200\Omega$

Correct answer:

$200\Omega$

Explanation:

The total resistance of a circuit in series can be described by the equation:

$R_{T} = R_{1} + R_{2} +...$

The series circuit in ths problem therefore has a resistance of:

$20\Omega + 20\Omega = 40\Omega$

The resistance of a circuit wired in parallel has a total resistance of:

$1/R_{T} = 1/R_{1} + 1/R_{2} + ...$

We are assuming that the two circuits have the same total resistance, so to find the resistance of the unknown resistor, we set up the following equation:

$1/40\Omega = 1/50\Omega + 1/R_{2}$

This, when solved, gives us a resitance of 200 ohms for our unknown element.

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Example Question #1331 : Ap Physics 1

Three 1.5V batteries are connected in series. What is their equivalent voltage?

Possible Answers:

4.5V

1.5V

3.0V

Correct answer:

4.5V

Explanation:

The equivalent voltage of batteries connected in series is the sum of the voltage of each battery, or

$V_{s}=\sum V_{n}=V_{1}+V_{2}+...$

In our problem,

$V_{s}=V_{1}+V_{2}+V_{3}=1.5+1.5+1.5=4.5V$

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Example Question #1332 : Ap Physics 1

Three 1.5V batteries are connected in parallel. What is their equivalent voltage?

Possible Answers:

4.5V

0.5V

1.5V

Correct answer:

1.5V

Explanation:

The equivalent voltage of batteries connected in parallel is equal to the voltage of 1 battery.

In this problem, V=1.5V

Note: Connecting batteries in parallel increases the capacity of the battery.

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Example Question #12 : Series And Parallel

Four resistors, $R_{1}=15\Omega$ , $R_{2}=10\Omega$ , $R_{3}=5\Omega$ and $R_{4}=20\Omega$ , are arranged as follows. What is the equivalent resistance of this setup? Screen shot 2015 09 08 at 12.13.13 pm

Possible Answers:

$5\Omega$

$15\Omega$

$7\Omega$

$31\Omega$

Correct answer:

$31\Omega$

Explanation:

To find the equivalent resistance of this system, we must first find the equivalent resistance of the resistors in parallel, then evaluate the resistors in parallel.

The parallel resistor equivalence is given by the following equation,

$\frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...$

In our problem,

$\frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}=\frac{2}{30}+\frac{3}{30}$

$\frac{1}{R_{p}}=\frac{5}{30}$

$R_{p}=\frac{30}{5}=6\Omega$

The parallel resistors can now be treated as one resistor with the resistance $R_{p}$ . To find the total resistance, we add the resistances of $R_{p}$ , $R_{3}$ and $R_{4}$ .

$R_{p}+R_{3}+R_{4}=6+5+20=31\Omega$

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #1 : Series And Parallel

Example Question #1 : Series And Parallel

Example Question #1 : Series And Parallel

Example Question #1 : Series And Parallel

Example Question #1 : Series And Parallel

Example Question #1 : Series And Parallel

Example Question #11 : Series And Parallel

Example Question #1331 : Ap Physics 1

Example Question #1332 : Ap Physics 1

Example Question #12 : Series And Parallel

All AP Physics 1 Resources

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