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AP Physics C: Mechanics : Electric Circuits

Study concepts, example questions & explanations for AP Physics C: Mechanics

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2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #211 : Ap Physics C

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?

Possible Answers:

The wire is causing electrons to be lost to the air, which lowers the potential measured

The difference in potential measured is small enough to be disregarded

The wire connected to the battery must be connected to something else where this potential is being lost

Once the current is distributed across the wire, there is less charge per unit length of the circuit and so the voltage is lower

The wire has a slight internal resistance and caused a potential drop

Correct answer:

The wire has a slight internal resistance and caused a potential drop

Explanation:

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.

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Example Question #61 : Electricity And Magnetism Exam

Three resistors and a battery form the following circuit.

Circuit_1

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

$R_1=40\Omega$

$R_2=25\Omega$

$R_3=38\Omega$

Possible Answers:

$R_{123}=134.5\Omega$

$R_{123}=92.3\Omega$

$R_{123}=45.2\Omega$

$R_{123}=24.5\Omega$

$R_{123}=2.5\Omega$

Correct answer:

$R_{123}=24.5\Omega$

Explanation:

First, calculate the equivalent resistance of R_2 and R_3 . 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.

Circuit_2

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.

Circuit_3

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Example Question #11 : Electric Circuits

A lamp has a 60W bulb. If the house wiring provides 110V to light up that bulb, how much current does the bulb draw?

Possible Answers:

$1.37\ \text{A}$

$0.03\ \text{A}$

$5.83\ \text{A}$

$0.72\ \text{A}$

$0.55\ \text{A}$

Correct answer:

$0.55\ \text{A}$

Explanation:

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

$P=60\ \text{W}$

$V=110\ \text{V}$

Solve for the current, .

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

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Example Question #12 : Electric Circuits

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?

Possible Answers:

$1.95*10^6\frac{m}{s}$

$1.96*10^7\frac{m}{s}$

$2.04*10^8\frac{m}{s}$

$3.00*10^8\frac{m}{s}$

$9.21*10^{7}\frac{m}{s}$

Correct answer:

$1.96*10^7\frac{m}{s}$

Explanation:

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:

$1A = \frac{(10^{15})(1.60*10^{-19}C)v}{2\pi (500m)}$

$v=\frac{(1A)2\pi(500m)}{(10^{15})(1.60*10^{-19}C)}= 1.96*10^7\frac{m}{s}$

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Example Question #1 : Interpreting Circuit Diagrams

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

Ps0_capacitorcircuit

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 I_1 , , 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?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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.

Ps0_capacitorcircuit_2

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

V=IR

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

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.

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Example Question #1 : Using Ohm's Law

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?

Possible Answers:

$5.5*10^{-6}W$

$4*10^{-5}W$

$2*10^{-2}W$

$7.2*10^{-2}W$

$8*10^{-4}W$

Correct answer:

$4*10^{-5}W$

Explanation:

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

$V=9\ \text{V}$

$R=2\ \text{M}\Omega=2*10^6\ \Omega$

Using these values, we can solve for power.

$P=\frac{(9\ \text{V})^2}{2*10^6\ \Omega}$

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

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Example Question #71 : Electricity And Magnetism Exam

A cicuit with a voltage source of 40V has a $15\Omega$ resistor and a $5\Omega$ resistor in series, followed by two $5\Omega$ resistors in parallel. After the parallel branches rejoin, there is a $10\Omega$ resistor before the circuit is closed at the voltage source.

What is the current supplied by the source?

Possible Answers:

1.5A

0.82A

1.2A

1.0A

Correct answer:

1.2A

Explanation:

The current supplied by the source can be calculated using a derivation of Ohm's law:

$I=\frac{V}{R}$

Start by finding the equivalent resistance of the circuit.

Sum the first two resistors in series:

$R_{12}=15\Omega+5\Omega=20\Omega$

Calculate the equivalent resistance of the two resistors in parallel:

$\frac{1}{R_{34}}=\frac{1}{5\Omega}+\frac{1}{5\Omega}=\frac{2}{5\Omega}$

$R_{34}=\frac{5}{2}\Omega=2.5\Omega$

Now, all of the reistances can be viewed as being in series.

$R_{eq}=R_{12}+R_{34}+R_{5}$

$R_{eq}=20\Omega+2.5\Omega+10\Omega=32.5\Omega$

Returning to our current calulation, we can find our final answer:

$I=\frac{V}{R}=\frac{40V}{32.5\Omega}=1.2A$

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Example Question #1 : Using Ohm's Law

Ps0_resistorcircuit

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

V=50V

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

Possible Answers:

4.6A

1.1A

0.91A

2.0A

Correct answer:

2.0A

Explanation:

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.

Ps0_resistorcircuit_2

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:

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

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

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Example Question #1 : Calculating Circuit Power

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?

Possible Answers:

Increasing the temperature of the wire

Decreasing the cross-sectional area of the wire

Changing the material of the wire to a metal of greater resistivity

None of these options result in increased power dissipation

Decreasing the length of the wire

Correct answer:

Decreasing the length of the wire

Explanation:

Relevant equations:

V = IR

P = IV = I^2 R

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

Current and resistance are inversely proportional to one another, assuming voltage is fixed. Since P = I^2 R , 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)

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Example Question #13 : Electric Circuits

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?

Possible Answers:

42W

67W

56W

20W

Correct answer:

67W

Explanation:

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.

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

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Electric Circuits

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #211 : Ap Physics C

Example Question #61 : Electricity And Magnetism Exam

Example Question #11 : Electric Circuits

Example Question #12 : Electric Circuits

Example Question #1 : Interpreting Circuit Diagrams

Example Question #1 : Using Ohm's Law

Example Question #71 : Electricity And Magnetism Exam

Example Question #1 : Using Ohm's Law

Example Question #1 : Calculating Circuit Power

Example Question #13 : Electric Circuits

All AP Physics C: Mechanics Resources

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