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AP Physics 1 : Current and Voltage

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Example Questions

Example Question #11 : Current And Voltage

Give the SI units for voltage.

Possible Answers:

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

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

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

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

Correct answer:

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

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 $\displaystyle V$ is defined as:

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

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

Therefore, voltage can be written as

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

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Example Question #132 : Electricity

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

Possible Answers:

$7.78\times10^{4}A$ $\displaystyle 7.78\times10^{4}A$

$5.6\times10^{8}A$ $\displaystyle 5.6\times10^{8}A$

$1.2\times10^{6}A$ $\displaystyle 1.2\times10^{6}A$

$7.2\times10^{3}A$ $\displaystyle 7.2\times10^{3}A$

Correct answer:

$7.78\times10^{4}A$ $\displaystyle 7.78\times10^{4}A$

Explanation:

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

Time must first be converted to seconds.

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

Solving for current,

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

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Example Question #91 : Circuits

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

Possible Answers:

8hr $\displaystyle 8hr$

$7.8\times10^{6}hr$ $\displaystyle 7.8\times10^{6}hr$

$1.5\times10^{6}hr$ $\displaystyle 1.5\times10^{6}hr$

433hr $\displaystyle 433hr$

Correct answer:

433hr $\displaystyle 433hr$

Explanation:

For this problem

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

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Example Question #131 : Electricity

What quantity of charge passes through a clock in 30min $\displaystyle 30min$ if its current is $\displaystyle 5A$?

Possible Answers:

$\displaystyle 9C$

9000C $\displaystyle 9000C$

$1.8\times10^{3}C$ $\displaystyle 1.8\times10^{3}C$

700C $\displaystyle 700C$

Correct answer:

9000C $\displaystyle 9000C$

Explanation:

Time must first be converted to seconds.

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

Solving for charge,

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

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Example Question #1361 : Ap Physics 1

A circuit with one battery and one resistor with resistance $1 \Omega$ $\displaystyle 1 \Omega$ has current I_0 $\displaystyle I_0$ 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 $3\Omega$ $\displaystyle 3\Omega$. Determine the current flowing through the original resistor.

Possible Answers:

$\frac{I_0}{2}$ $\displaystyle \frac{I_0}{2}$

$\frac{3I_0}{2}$ $\displaystyle \frac{3I_0}{2}$

$\frac{I_0}{3}$ $\displaystyle \frac{I_0}{3}$

I_0 $\displaystyle I_0$

Correct answer:

$\frac{I_0}{2}$ $\displaystyle \frac{I_0}{2}$

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 $\frac{I_0}{2}$ $\displaystyle \frac{I_0}{2}$ flowing through the original resistor.

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Example Question #135 : Electricity

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

Possible Answers:

1.125V $\displaystyle 1.125V$

18V $\displaystyle 18V$

$\displaystyle 0V$

5.625V $\displaystyle 5.625V$

Correct answer:

5.625V $\displaystyle 5.625V$

Explanation:

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

$\Delta V=IR$ $\displaystyle \Delta V=IR$.

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

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

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

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

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Example Question #12 : Current And Voltage

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

Possible Answers:

$\displaystyle 8V$

$\displaystyle 4V$

40V $\displaystyle 40V$

10V $\displaystyle 10V$

Correct answer:

$\displaystyle 4V$

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.

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

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Example Question #211 : Electricity And Waves

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

$R1=3k\Omega$ $\displaystyle R1=3k\Omega$

$R2=R3=4k\Omega$ $\displaystyle R2=R3=4k\Omega$

$V_{SOURCE}=10V$ $\displaystyle V_{SOURCE}=10V$

Circuit

What is the current passing though R3?

Possible Answers:

2.5A $\displaystyle 2.5A$

$\displaystyle 2A$

1.25A $\displaystyle 1.25A$

$\displaystyle 1A$

$\displaystyle 4A$

Correct answer:

$\displaystyle 1A$

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:

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

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

$R_{equivalent}=5k\Omega$ $\displaystyle R_{equivalent}=5k\Omega$

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

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

That means that $\displaystyle 2A$ will pass through R1, and will split when it reaches the junction between R2 and R3 based on the ratio of the two resistance:

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

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

$\displaystyle 1A$ of current passes through the resistor R3.

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Example Question #211 : Electricity And Waves

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

Possible Answers:

They would all have equal resistance

Wire D: 1000 V $\displaystyle 1000 V$ and .1 A $\displaystyle .1 A$

Wire B: 1 V $\displaystyle 1 V$ and 100 A $\displaystyle 100 A$

Wire A: 10 V $\displaystyle 10 V$ and 10 A $\displaystyle 10 A$

Wire C: 100 V $\displaystyle 100 V$ and 1 A $\displaystyle 1 A$

Correct answer:

Wire B: 1 V $\displaystyle 1 V$ and 100 A $\displaystyle 100 A$

Explanation:

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

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

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

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

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Example Question #11 : Current And Voltage

Series curcuit

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

Possible Answers:

$\displaystyle 1A$

25A $\displaystyle 25A$

10A $\displaystyle 10A$

$\displaystyle 5A$

Correct answer:

$\displaystyle 1A$

Explanation:

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

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

For this problem

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

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

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

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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 #11 : Current And Voltage

Example Question #132 : Electricity

Example Question #91 : Circuits

Example Question #131 : Electricity

Example Question #1361 : Ap Physics 1

Example Question #135 : Electricity

Example Question #12 : Current And Voltage

Example Question #211 : Electricity And Waves

Example Question #211 : Electricity And Waves

Example Question #11 : Current And Voltage

All AP Physics 1 Resources

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