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AP Physics 1 : Current and Voltage

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Example Question #131 : Electricity

Give the SI units for voltage.

Possible Answers:

$V=\frac{kg\cdot m^3}{C\cdot s^2}$

$V=\frac{kg\cdot m^2}{s^2\cdot C}$

$V=\frac{kg\cdot m}{C\cdot s^2}$

$V=\frac{kg\cdot m\cdot C}{s^2}$

Correct answer:

$V=\frac{kg\cdot m^2}{s^2\cdot C}$

Explanation:

We have to know what SI units are for this problem. SI units are units that cannot be broken down into smaller component parts. Voltage is defined as:

$V=\frac{J}{C}$ , where is joules and is Coulombs. Coulombs are already in SI units. However, joules is composed of other terms.

$J=N\cdot m=\left(\frac{kg\cdot m}{s^2}\right)\cdot m=\frac{kg\cdot m^2}{s^2}$

Therefore, voltage can be written as

$V=\frac{kg\cdot m^2}{s^2\cdot C}$

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Example Question #12 : Current And Voltage

$5.6\times 10^{8}C$ of charge pass through a television during a 2hr movie. How much is being used by the television?

Possible Answers:

$7.78\times10^{4}A$

$5.6\times10^{8}A$

$1.2\times10^{6}A$

$7.2\times10^{3}A$

Correct answer:

$7.78\times10^{4}A$

Explanation:

Current and charge are related by the equation $I=\frac{Q}{t}$ where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

Time must first be converted to seconds.

$2hr\times \frac{60min}{1hr}\times \frac{60sec}{1min}=7200 sec$

Solving for current,

$I=\frac{Q}{t}=\frac{5.6\times 10^{8}}{7200}=7.78\times10^{4}A$

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Example Question #13 : Current And Voltage

Who long (in hours) does it take a computer to use $7.8\times10^{6}C$ of charge $\left (I=5A \right )$ ?

Possible Answers:

433hr

$7.8\times10^{6}hr$

8hr

$1.5\times10^{6}hr$

Correct answer:

433hr

Explanation:

Current and charge are related by the equation $I=\frac{Q}{t}$ where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

For this problem

$t=\frac{Q}{I}=\frac{7.8\times10^{6}}{5}=1.5\times10^{6}sec\times \frac{1min}{60sec}\times \frac{1hr}{60min}=433hr$

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Example Question #11 : Current And Voltage

What quantity of charge passes through a clock in 30min if its current is ?

Possible Answers:

$1.8\times10^{3}C$

700C

9000C

Correct answer:

9000C

Explanation:

Current and charge are related by the equation $I=\frac{Q}{t}$ where is the current in Amperes (), is the charge in Coulombs (), and is the time in seconds ().

Time must first be converted to seconds.

$30min\times \frac{60sec}{1min}=1800 sec$

Solving for charge,

$Q=It=5\cdot1800=9000C$ .

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Example Question #15 : Current And Voltage

A circuit with one battery and one resistor with resistance $1 \Omega$ has current I_0 flowing through it. It is modified to add two more identical resistors in series of the first resistor and 1 more resistor in parallel to the other three with resistance $3\Omega$ . Determine the current flowing through the original resistor.

Possible Answers:

$\frac{I_0}{2}$

$\frac{3I_0}{2}$

$\frac{I_0}{3}$

I_0

Correct answer:

$\frac{I_0}{2}$

Explanation:

To find our answer, we have to know how each of the new additions changes the circuit. The addition of the two resistors in series do not affect the current because current is equivalent for resistors in series. However, since there is a resistor in parallel with equivalent resistance to the sum of the 3 resistors in series, current will split evenly amongst the two pathways, thus leading to a current of $\frac{I_0}{2}$ flowing through the original resistor.

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Example Question #16 : Current And Voltage

Consider a circuit composed of a 18V battery and three resistors in series. The three resistors are $R_1 = 3 \Omega, R_2 = 5 \Omega, R_3 = 8 \Omega$ . Calculate the voltage drop $(\Delta V)$ across R_2 .

Possible Answers:

1.125V

18V

5.625V

Correct answer:

5.625V

Explanation:

In order to find the voltage drop $(\Delta V)$ across R_2 , we will use Ohm's law:

$\Delta V=IR$ .

However, we first need to solve for the current (I) in the circuit. This will be calculated by applying Ohm's law to the entire circuit, using the total resistance $(R_{tot})$ for resistors in series.

$V=IR\Rightarrow I = \frac{V}{R_{tot}}=\frac{18V}{16 \Omega}=1.125 A$

Finally, we use this calculated current to find the voltage drop across R_2 .

$\Delta V_2=IR_2=(1.125A)5\Omega=5.625V$

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Example Question #1361 : Ap Physics 1

Determine the voltage drop across a resistor of $2\Omega$ experiencing a current flowing through it, if it is connected to a battery of 10V ?

Possible Answers:

40V

10V

Correct answer:

Explanation:

In circuits with resistors, the only thing necessary to determine voltage drop across a resistor is the current through it and the resistance, as given by Ohm's law.

$\Delta V=IR=2A*2\Omega=4V$

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Example Question #1361 : Ap Physics 1

You have the following circuit. The values for the components are:

$R1=3k\Omega$

$R2=R3=4k\Omega$

$V_{SOURCE}=10V$

Circuit

What is the current passing though R3?

Possible Answers:

2.5A

1.25A

Correct answer:

Explanation:

The first step is to find the equivalent resistance of the entire circuit. Note that R2 and R3 are in parallel, and they are in series with R1. Therefore, in order to find the equivalent resistance:

$(R2\left | \right |R3)+R1=\frac{R2*R3}{R2+R3}+R1=\frac{4k\Omega*4k\Omega}{4k\Omega+4k\Omega}+3k\Omega$

$R_{equivalent}=2k\Omega+3k\Omega$

$R_{equivalent}=5k\Omega$

Then, in order to find the total current, use Ohm's Law:

$V=IR\rightarrow I=\frac{V}{R}\rightarrow I=\frac{10}{5}A=2A$

That means that will pass through R1, and will split when it reaches the junction between R2 and R3 based on the ratio of the two resistance:

$I_{R3}=\frac{R2}{R2+R3}I_{source}$

$I_{R3}=\frac{1}{2}(2A)=1A$

of current passes through the resistor R3.

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Example Question #19 : Current And Voltage

Which of the following wires with a current running through them would have the least resistance?

Possible Answers:

Wire C: 100 V and 1 A

Wire D: 1000 V and .1 A

They would all have equal resistance

Wire B: 1 V and 100 A

Wire A: 10 V and 10 A

Correct answer:

Wire B: 1 V and 100 A

Explanation:

The correct answer is Wire B with 1 V and 100 A of current.

The formula for voltage, current and resistance is as follows:

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

Thus, the wire with the lowest ratio of voltage to current will have the least resistance.

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Example Question #101 : Circuits

Series curcuit

What is the current in the given circuit if $R_1=10\Omega$ , $R_2=15\Omega$ and V=25V ?

Possible Answers:

10A

25A

Correct answer:

Explanation:

To find the current we must first find the equivalent resistance. For resistors in series, the equivalent resistance is

$R_{eq}=R_1+R_2+...$

For this problem

$R_{eq}=R_1+R_2=10+15=25\Omega$

Now we use Ohm's law, $V=IR_{eq}$ , to find the current,

$V=IR_{eq}\Rightarrow I=\frac{V}{R_{eq}}=\frac{25}{25}=1A$

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AP Physics 1 : Current and Voltage

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #131 : Electricity

Example Question #12 : Current And Voltage

Example Question #13 : Current And Voltage

Example Question #11 : Current And Voltage

Example Question #15 : Current And Voltage

Example Question #16 : Current And Voltage

Example Question #1361 : Ap Physics 1

Example Question #1361 : Ap Physics 1

Example Question #19 : Current And Voltage

Example Question #101 : Circuits

All AP Physics 1 Resources

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