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AP Physics 1 : Electricity and Waves

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Example Questions

Example Question #2 : Current And Voltage

Basic circuit

In this circuit above, $Z_1=4\Omega$ and $Z_2=12\Omega$ . The voltage drop across Z_1 is eight volts. What is the current across the circuit?

Possible Answers:

0.7A

0.5A

Correct answer:

Explanation:

The voltage drop is related to the current and resistance via Ohm's law:

V=IR

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

$I=\frac{8V}{4\Omega}$

I=2A

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Example Question #2 : Current And Voltage

Resistor bulb series circuit

A student has created the given circuit diagram. It consists of a battery, a resistor, and a light bulb. In one minute, 1.2C of charge flows through the resistor. How much charge flows through the light bulb in one minute?

Possible Answers:

60 C

0.024 C

2.4 C

0.6 C

1.2 C

Correct answer:

1.2 C

Explanation:

Since the bulb and the resistor are connected in series, the current is the same in each. Electric current is just the flow of charge through the circuit, so the same amount of charge flows through each in one minute.

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Example Question #5 : Current And Voltage

If a TV uses 90. kwh of energy over the course of 15min , and it has a voltage of 120V , how many coulombs passed through it during that time?

Possible Answers:

$670 \:coulombs$

$3.0\: coulombs$

$2700\: coulombs$

$270\: coulombs$

$67\: coulombs$

Correct answer:

$2700\: coulombs$

Explanation:

Because the TV uses 90 kWh , and it was used for .25 hrs , it must have used

90 kWh/.25 h = 360 kW .

$Power = Voltage \cdot Current$ so:

Current = 360 kW/120 V = 3A

$1A = 1\:coulomb/sec$ , and since the TV was used for $15 \:min(60) = 900\: sec$

$3 \:coulombs/sec \cdot 900\: sec = 2700 \:coulombs$

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Example Question #201 : Electricity And Waves

Give the SI units for voltage.

Possible Answers:

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

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

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

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

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 #132 : Electricity

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

Possible Answers:

$7.2\times10^{3}A$

$7.78\times10^{4}A$

$1.2\times10^{6}A$

$5.6\times10^{8}A$

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 #133 : Electricity

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

Possible Answers:

$1.5\times10^{6}hr$

$7.8\times10^{6}hr$

8hr

433hr

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 #134 : Electricity

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 #1361 : Ap Physics 1

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:

$\frac{I_0}{3}$

$\frac{3I_0}{2}$

$\frac{I_0}{2}$

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 #131 : Electricity

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 #101 : Circuits

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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AP Physics 1 : Electricity and Waves

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #2 : Current And Voltage

Example Question #2 : Current And Voltage

Example Question #5 : Current And Voltage

Example Question #201 : Electricity And Waves

Example Question #132 : Electricity

Example Question #133 : Electricity

Example Question #134 : Electricity

Example Question #1361 : Ap Physics 1

Example Question #131 : Electricity

Example Question #101 : Circuits

All AP Physics 1 Resources

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