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AP Physics 1 : Circuits

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Example Questions

Example Question #2 : Current And Voltage

Consider the circuit:

Circuit_4

By how much does the current through the circuit decrease if R3 is removed?

$R1 = 5\Omega,\ R2 = 20\Omega,\ R3 = 25\Omega,\ R4 = 10\Omega$

Possible Answers:

0.87A

0.26A

0.48A

1.29A

0.11A

Correct answer:

0.48A

Explanation:

We are asked to compare two different scenarios, each involving the calculation of equivalent resistance, which will use the following formula:

$\frac{1}{R_{eq}}=\sum\frac{1}{R}$

Scenario 1: With R3 Present

$\frac{1}{R_{eq}} = \sum\frac{1}{R}= \frac{1}{5\Omega}+\frac{1}{20\Omega}+\frac{1}{25\Omega}+\frac{1}{10\Omega}$

$\frac{1}{R_{eq}} = \frac{39}{100}\Omega$

$R_{eq} = 2.56\Omega$

Now using Ohm's law:

V = IR

$I = \frac{V}{R}=\frac{12V}{2.56\Omega}=4.68A$

Scenario 2: Without R3

$\frac{1}{R_{eq}} = \sum\frac{1}{R}= \frac{1}{5\Omega}+\frac{1}{20\Omega}+\frac{1}{10\Omega}$

$\frac{1}{R_{eq}} = \frac{7}{20}\Omega$

$R_{eq} = 2.86\Omega$

Now using Ohm's law:

V = IR

$I = \frac{V}{R}=\frac{12V}{2.86\Omega}=4.20A$

Calculate the change in current:

$\Delta I = 4.68A-4.20A = 0.48A$

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Example Question #5 : Current And Voltage

Consider the circuit:

Circuit_4

How much current is flowing through R1?

$R1 = 2\Omega,\ R2 = 4\Omega,\ R3 = 4\Omega,\ R4 = 2\Omega$

Possible Answers:

12A

Correct answer:

Explanation:

Although it is possible to solve this problem by calculating an equivalent resistance, calculating a total current through the circuit, and then using Kirchoff's junction rule to find the current through R1, it is much easier to simply use Kirchoff's Loop rule. What this rule says is that through any closed loop in a circuit, all voltages must add up to zero. Written as an equation:

$\sum V = 0$

If we consider the closed loop path consisting of only the power supply and R1, we can use Ohm's law to calculate the current:

V = IR

12V = I(R1)

$I = \frac{12V}{2\Omega} = 6A$

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Example Question #126 : Electricity

Consider the circuit:

Circuit_4

The total current through the circuit is 25A and the current through R2 is . What is the value of R2?

Possible Answers:

$3.6\Omega$

$1.7\Omega$

$7.8\Omega$

$5.4\Omega$

$9.1\Omega$

Correct answer:

$1.7\Omega$

Explanation:

The total current through the circuit is unnecessary to solve this problem. We only need to know the current through the resistor, as well as Kirchoff's loop rule. The loop rule states that for any closed loop through a circuit, the voltages add up to zero.

$\Delta V = 0$

For this problem, we will consider the loop that consists solely of the voltage source and R2. From the rule, we know that 12V is lost through the resistor. Using Ohm's law, we can write:

V = IR

12V = I(R2)

Rearranging for R2, we get:

$R2 = \frac{12V}{I_2} = \frac{12V}{7A}= 1.7\Omega$

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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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AP Physics 1 : Circuits

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #2 : Current And Voltage

Example Question #5 : Current And Voltage

Example Question #126 : Electricity

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

All AP Physics 1 Resources

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