AP Physics 1 - Ohm's Law

What is the resistance of a resistor if the current going through it is and the voltage across it it is ?

$5\Omega$

$20\Omega$

$0.2\Omega$

$12\Omega$

Explanation

Use Ohm's law.

Plug in known values and solve for resistance.

$R=\frac{10V}{2A}=5\Omega$

Basic circuit

$V_{in}=12V$

is composted of two resistors in parallel, $3\Omega$ and $7\Omega$

is a single $6\Omega$ resistor.

In the circuit above, what is the current?

Explanation

To find the current, first find the total resistance of the circuit. Begin by simplifying , the two resistors in parallel as follows:

$\frac{1}{Z_1}=\frac{1}{3\Omega}+\frac{1}{7\Omega}$

$\frac{1}{Z_1}=\frac{10}{21\Omega}$

$Z_1=2.1\Omega$

Since and are in series, their combined resistance is:

$R_{tot}=Z_1+Z_2=2.1\Omega+6\Omega=8.1\Omega$

Use Ohm's law to find the current.

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

$I=\frac{12V}{8.1\Omega}$

If the current through a $10\Omega$ resistor is , what is the voltage across the resistor?

$\frac{5}{3}V$

Explanation

UseOhm's law.

Plug in known values and solve.

$V=6A\cdot 10\Omega$

Consider the circuit:

Circuit_4

If each resistor has a value of $2\Omega$ , how much current is flowing through the circuit?

Explanation

First we need to calculate the equivalent resistance of the circuit using the following expression for condensing parallel resistors:

$\frac{1}{R_{eq}}=\sum\frac{1}{R} = \frac{1}{2\Omega}+\frac{1}{2\Omega}+\frac{1}{2\Omega}+\frac{1}{2\Omega}$

$R_{eq}=0.5\Omega$

Now we can use Ohm's law to calculate the current flowing through the circuit:

$I = \frac{V}{R}=\frac{12V}{0.5\Omega} = 24A$

A light bulb requires 60 W to function properly. If it is connected to a powersupply of 120 A and functions properly, then what is the resitance of the light bulb?

$4m\Omega$

$0.5\Omega$

$7m\Omega$

$1m\Omega$

$1\Omega$

Explanation

First, identify the given information:

Two equations are required for this problem:

1.) Ohm's law,

2.) Electrical power

Using the equation for electrical power, we can rearrange to solve for :

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

At this point, we can substitute in the known values and determine the voltage:

$V=\frac{60W}{120A}=0.5V$

Ohm's law can then be rearranged to solve for the resistence of the light bulb:

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

The known voltage value then can be substituted into Ohm's law to determine the resistance of the light bulb:

$R=\frac{0.5V}{120A}=0.00416\Omega\approx 4m\Omega$

A circuit has a power source and a $25\Omega$ resistance. What is the current in the circuit?

Explanation

We use Ohm's law, , to find the current in the circuit. In Ohm's law is the voltage in the circuit, is the current in the circuit and is the circuit's resistance.

Solving the equation for , we have

A student assembles a circuit with a resistor and a voltage source. He realizes that he needs to increase the amount of electrons flowing through the circuit to reach his goal. What can the student do to achieve this?

I. Change the voltage source to an alternating source

II. Use a new voltage source with higher voltage

III. Use another resistor with lower resistance

II and III

I and II

I and III

I, II and III

Explanation

Recall that the definition of current is the amount of electrons flowing through a circuit in a given amount of time. We can increase the amount of electrons flowing through a circuit (for a given time) if we increase the current. Using Ohm’s law, we can determine conditions that will increase the current.

Solving for current we get

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

This means we can get a high current if we increase voltage or decrease resistance. Note that that changing the voltage source from a direct source to an alternating source will generate an alternating current with similar amplitude; it won’t increase the total current of the system and, subsequently, the amount of electrons flowing through the circuit per given time.

If a closed circuit connected to a battery has a resistance of $2: \Omega$ , what is the current flowing through this circuit?

Explanation

This question can be solved by making use of Ohm's law, which states that the voltage difference across a circuit is proportional to the current flowing through the circuit, as well as to the resistance of the circuit. Written in equation form, we have:

Solving for current, we can rearrange to obtain:
$i=\frac{V}{R}=\frac{12: V}{2: \Omega }=6:Amperes$

A battery has an internal resistance of $2.5\Omega$ . If the current within the battery were to be measured using a multimeter, what magnitude would the meter record?

Explanation

We use Ohm's law, , to find the current in the circuit. In Ohm's law is the voltage in the circuit, is the current in the circuit and is the circuit's resistance.

Solving the equation for , we have

Consider the circuit diagram shown. In this circuit, the values of $R_{1}$ , $R_{2}$ , and $R_{3}$ are known, but the value of $R_{x}$ is variable.

Physics voltmeter problem

Which of the following expressions would give a situation in which the voltmeter in the diagram would read zero?

$R_x=\frac{R_{2}R_{3}}{R_{1}}$

$R_x=\frac{R_{1}R_{3}}{R_{2}}$

$R_x=\frac{R_{1}R_{2}}{R_{3}}$

$R_x=R_{1}R_{2}R_{3}$

$R_x=\frac{1}{R_{1}R_{2}R_{3}}$

Explanation

To answer this question, we'll need to find an expression for the value of the variable resistor that would make the voltmeter read zero.

First, it's important to realize what situation would result in a reading of zero from the voltmeter. For there to be no reading, that means that there cannot be any voltage difference between the top row of resistors and the bottom row. For this to happen, the voltage drop for the resistors on the left of the voltmeter must be equal, and the same is true for the two resistors to the right of the voltmeter. In other words, both rows of resistors will experience the same voltage decrease as current flows through, thus the difference of voltage drop in the top and bottom row will be identical.

So let's consider the top and bottom resistors on the left side first. In the top left corner, the voltage of the first resistor will be $I_{top}R_{1}$ from Ohm's law. Moreover, the voltage drop of the bottom left resistor will be $I_{bottom}R_{2}$ . These two voltages will need to be equal to one another in order to have the voltmeter read zero.

$I_{top}R_{1}=I_{bottom}R_{2}$

Now let's take a look at the other resistors on the right. The voltage of the third resistor will be $I_{top}R_{3}$ and the voltage of the variable resistor will be $I_{bottom}R_{x}$ . Just as before, these two resistors will also need to be equal in voltage.

$I_{bottom}R_{x}=I_{top}R_{3}$

Now that we have the two expressions shown above, we can isolate the term for the variable resistance in terms of the other three resistors to find our answer.

$I_{bottom}=\frac{I_{top}R_{1}}{R_{2}}$