Ohm's Law - AP Physics 1

Card 0 of 20

Question

Consider the circuit:

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

Answer

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:

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

Compare your answer with the correct one above

Question

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?

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

Use Ohm's law.

Plug in known values and solve for resistance.

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

Compare your answer with the correct one above

Question

What is the voltage across a resistor with a resistance of $10\Omega$ that has a current of going through it?

Answer

Use Ohm's Law.

$V=20A\cdot10\Omega =200V$

Compare your answer with the correct one above

Question

What is the current through a resistor if the resistor has a resistance of $40\Omega$ and the voltage across the resistor is ?

Answer

Use Ohm's law.

$I=\frac{60V}{40\Omega}=1.5A$

Compare your answer with the correct one above

Question

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

Answer

UseOhm's law.

Plug in known values and solve.

$V=6A\cdot 10\Omega$

Compare your answer with the correct one above

Question

$V_{in}=12V$

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?

Answer

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.

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

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

Compare your answer with the correct one above

Question

$V_{in}=12V$

, and the voltage measured from a point between and to the ground is

In the circuit above, what is the resistance of ?

Answer

Begin by finding the total resistance in the circuit.

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

Now note that the voltage identified in the problem is the same as the voltage drop across the second resistor:

$Z_2=\frac{V_D}{I}=\frac{8V}{2A}$

$Z_2=4\Omega$

Compare your answer with the correct one above

Question

$V_{in}=12V$

The voltage measured from a point between and to the ground is

What is the resistance of ?

Answer

Begin by finding the total resistance in the circuit.

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

Now note that the voltage identified in the problem is the same as the voltage drop across the second resistor:

$Z_2=\frac{V_D}{I}=\frac{8V}{2A}$

$Z_2=4\Omega$

Now, since and combine to form the total resistance:

$Z_1=Z-Z_2=6\Omega -4\Omega=2\Omega$

Compare your answer with the correct one above

Question

A resistor with a resistance of $4 \Omega$ has a current flowing through it of 5A. What is the potential drop across the resistor?

Answer

Ohm's law states that the potential drop across a resistor is equal to the product of the current flowing through the resistor and the resistance of the resistor:

We were given the current, I, and the resistance, R, so we simply multiply the two together to get our final answer.

Compare your answer with the correct one above

Question

Determine the voltage drop across wire that is connected to two resistors in series with resistances $5\Omega$ and $10\Omega$ , with a current flowing through the circuit of ?

Answer

By Ohm's law:

$\Delta V=IR_{total}$ , where $\Delta V$ is the voltage drop across the wire. is the current flowing through the wire, and $R_{total}$ is the total resistance within the circuit.

Since resistors are in series:

$R_{total}=R_1+R_2$ , where and are the resistances of the two resistors.

In our case:

$R_{total}=5\Omega+10\Omega=15 \Omega$

Therefore:

$\Delta V=(2.5A)(15\Omega)=37.5V$

Compare your answer with the correct one above

Question

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

Answer

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:
$i=\frac{V}{R}=\frac{12: V}{2: \Omega }=6:Amperes$

Compare your answer with the correct one above

Question

A current of passes through a circuit. A single resistor in this circuit has a resistance of $2.5 \Omega$ . What is the voltage drop across this resistor?

Answer

We need to use Ohm's law here which is given by:

Where is the voltage in Volts, is the current in Amperes, and is the resistance in Ohms. We know the current in the circuit as well as the resistance of the resistor. We substitute our known values and solve for Voltage which will give us the voltage drop across the resistor.

$V = (10 A)(2.5\Omega )= 25V$

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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?

Answer

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

.

Compare your answer with the correct one above

Question

What voltage is required to produce a current in a circuit with a $5\Omega$ resistance?

Answer

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.

In our problem,

$V=IR=(30Amps)(5\Omega )=150V$

Compare your answer with the correct one above

Question

If a circuit has a voltage of and a current of , what is the resistance of the circuit?

Answer

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 Ohm's law for resistance gives us

$R=V/I=60/15=4\Omega$ .

Compare your answer with the correct one above

Question

$V=2.5V, \ R_1=70\Omega , \ I=24mA$

If the voltage drop across $R_{1}$ is , what is the resistance or $R_{2}$ ?

Answer

A few rules of circuits that will help here for series circuits are:

$R_{eq}=R_{1}+R_{2}+ ... +R_{n}$

$\varepsilon=\Delta V_{1}+\Delta V_{2}+...+\Delta V_{n}$

$I_{battery}=I_{1}=I_{2}=...=I_{n}$

Method 1:

The current through $R_{1}$ is given and that will be the same current going through $R_{2}$ since there are no current junctions. Since the sum of all voltage drops must equal the emf of the battery, the voltage drop across $R_{2}$ can be found:

$\Delta V_{2}=\varepsilon-\Delta V_{1}=0.75V$

We can use Ohm's law to find the resistance:

$\Delta V_{2}=I_{2} R_{2} \rightarrow R_{2}=30\Omega$

Method 2:

Ohm's law:

$\varepsilon=I_{battery} R_{eq}$

The resistance values are added to get $R_{eq}$ , so

$\varepsilon=I_{battery} \left ( R_{1}+R_{2} \right )$

Solve for $R_{2}$ :

$R_{2}=\frac{\varepsilon }{I_{battery}}-R_{1}=30\Omega$

Compare your answer with the correct one above

Question

A circuit consists of a single voltage source and a single resistor. When is fed through the circuit, a current of is measured through the resistor. What is the measured current if a voltage of is fed through the circuit?

Answer

Using ohm's law, the resistance is determined to be $\frac{V}{I}$ which is calculated to be $2\Omega$ . Ohm's law is used again to find the current at with the same resistance.

Compare your answer with the correct one above

Question

Which of the following statements is true?

Answer

To answer this question we need to use the definition of Ohm’s law.

where is voltage, is current, and is resistance. As the equation suggests, to determine the effect of voltage on current we need information regarding the resistance. For example, increasing the voltage will increase the current if resistance decreases or if resistance stays the same. On the other hand, increasing the voltage will decrease the current if the resistance increases drastically; therefore, we cannot determine the effect of voltage on current without knowing anything about the resistance (we need to know if resistance increases, decreases, or stays the same). Similarly, we cannot determine the effect of resistance on the current without knowing about the voltage.

Compare your answer with the correct one above

Tap the card to reveal the answer