Circuit Properties - AP Physics 2
Card 0 of 420
Consider the three circuits shown. In each circuit, the voltage source is the same, and all resistors have the same value. If each resistor represents a light bulb, which of the three circuits will produce the brightest light?
Consider the three circuits shown. In each circuit, the voltage source is the same, and all resistors have the same value. If each resistor represents a light bulb, which of the three circuits will produce the brightest light?
Tap to see back →
To answer this question, we first need to determine what we're looking for that will allow any given light bulb to produce bright light. The answer is power. The more energy that is delivered to the light bulb within a given amount of time, the brighter the light will be. Let's go ahead and look at the equation for power in a circuit.
And since we're told in the question stem that the voltage source is identical in each circuit, we're looking for the circuit that has the largest current.
In order to find which circuit has the largest current, we'll need to invoke Ohm's law.
What this shows is that a higher current will occur in the circuit that has the lowest total resistance. Thus, we'll need to determine what the total resistance is in each of the three circuits.
In circuit A, we can see that there is only a single resistor. Thus, we can give this circuit's total resistance a value of 
.
In circuit B, we have two resistors that are connected in series. Remember that when resistors are connected in this way, the overall resistance of the circuit increases. We find the total resistance by summing all the resistors connected in series.
Therefore, we can give circuit B a total resistance of 
.
Now, let's look at circuit C. We can see that there are two resistors connected in parallel, and each of these are connected in series with a third resistor. To solve for the total resistance of this circuit, we first need to determine the equivalent resistance of the two resistors connected in parallel. Once we find that value, then we can take into account the third resistor connected in series.
To solve for the resistance of the two resistors connected in parallel, we have to remember that they add inversely.
Now that we've found the equivalent resistance for the two resistors connected in parallel, we can consider the third resistor connected in series.
Thus, we can give circuit C a value of 
.
Now that we've found the total resistance for each circuit, we can obtain our answer. Since circuit A has the lowest total resistance, it will also have the greatest current. Consequently, it will also have the greatest power delivered to its resistor (the light bulb), thus causing the light coming from that bulb to be the brightest.
To answer this question, we first need to determine what we're looking for that will allow any given light bulb to produce bright light. The answer is power. The more energy that is delivered to the light bulb within a given amount of time, the brighter the light will be. Let's go ahead and look at the equation for power in a circuit.
And since we're told in the question stem that the voltage source is identical in each circuit, we're looking for the circuit that has the largest current.
In order to find which circuit has the largest current, we'll need to invoke Ohm's law.
What this shows is that a higher current will occur in the circuit that has the lowest total resistance. Thus, we'll need to determine what the total resistance is in each of the three circuits.
In circuit A, we can see that there is only a single resistor. Thus, we can give this circuit's total resistance a value of
In circuit B, we have two resistors that are connected in series. Remember that when resistors are connected in this way, the overall resistance of the circuit increases. We find the total resistance by summing all the resistors connected in series.
Therefore, we can give circuit B a total resistance of
Now, let's look at circuit C. We can see that there are two resistors connected in parallel, and each of these are connected in series with a third resistor. To solve for the total resistance of this circuit, we first need to determine the equivalent resistance of the two resistors connected in parallel. Once we find that value, then we can take into account the third resistor connected in series.
To solve for the resistance of the two resistors connected in parallel, we have to remember that they add inversely.
Now that we've found the equivalent resistance for the two resistors connected in parallel, we can consider the third resistor connected in series.
Thus, we can give circuit C a value of
Now that we've found the total resistance for each circuit, we can obtain our answer. Since circuit A has the lowest total resistance, it will also have the greatest current. Consequently, it will also have the greatest power delivered to its resistor (the light bulb), thus causing the light coming from that bulb to be the brightest.
Consider the circuit:
What is the rate of power consumption in the circuit if every resistor has a resistance of 
?
Consider the circuit:
What is the rate of power consumption in the circuit if every resistor has a resistance of
Tap to see back →
To calculate the power consumption of the circuit, we need to first reduce it to an equivalent circuit with a single resistor. Since each resistor has the same resistance, this solution will keep resistance calculations as multiples of 
until the circuit is fully reduced.
Start with the two branches in parallel. We can condense R3 and R4, then solve for the total resistance of R2, R3, and R4.
The equivalent circuit now has three resistors in series (R1, Req, and R5), so we can simply add them all up:
Plug in the value for R:
Now we can use the equation for power:
Substituting in Ohm's law for current, we get:
To calculate the power consumption of the circuit, we need to first reduce it to an equivalent circuit with a single resistor. Since each resistor has the same resistance, this solution will keep resistance calculations as multiples of
Start with the two branches in parallel. We can condense R3 and R4, then solve for the total resistance of R2, R3, and R4.
The equivalent circuit now has three resistors in series (R1, Req, and R5), so we can simply add them all up:
Plug in the value for R:
Now we can use the equation for power:
Substituting in Ohm's law for current, we get:
Calculate the power consumed across resistor 
.
Calculate the power consumed across resistor
Tap to see back →
To calculate power, we need two of the following three quantities: voltage, current, and resistance.
In this case, since we are lacking the voltage, let's try to find the current.
We can use Kirchoff's junction law to calculate current 
.
The current coming into the junction = the current coming out of the junction.
Let's take a look at the central junction to the right of resistor 
.
Now that we know 
and 
, we can calculate power across the resistor.
To calculate power, we need two of the following three quantities: voltage, current, and resistance.
In this case, since we are lacking the voltage, let's try to find the current.
We can use Kirchoff's junction law to calculate current
The current coming into the junction = the current coming out of the junction.
Let's take a look at the central junction to the right of resistor
Now that we know
Elements A-D represent light bulbs.
Which of the following is true about these two circuits? Assume voltage sources have the same value and all the light bulbs are all identical.
Elements A-D represent light bulbs.
Which of the following is true about these two circuits? Assume voltage sources have the same value and all the light bulbs are all identical.
Tap to see back →
Since bulbs A and B are in parallel, they will have the same voltage, and since the bulbs are identical in resistance, they will have the same current running through them and will be just as bright.
Let's say the voltage source as a value of 
and each bulb has a resistance of 
.
The current going through bulbs A and B is 
.
However, the current going through bulbs C and D is 
.
The current going through bulbs C and D is half as much as the other two, so their brightness will be less.
So, bulbs A and B will be brigher than bulbs C and D.
Since bulbs A and B are in parallel, they will have the same voltage, and since the bulbs are identical in resistance, they will have the same current running through them and will be just as bright.
Let's say the voltage source as a value of
The current going through bulbs A and B is
However, the current going through bulbs C and D is
The current going through bulbs C and D is half as much as the other two, so their brightness will be less.
So, bulbs A and B will be brigher than bulbs C and D.
You have 4 resistors, 
, 
, 
, and 
, set up like this:
Their resistance are as follows:
If the battery has 8V, what is the total power dissipated through the resistors?
You have 4 resistors,
Their resistance are as follows:
If the battery has 8V, what is the total power dissipated through the resistors?
Tap to see back →
The equation for power is
In order to get the power, we need the current. To find the current, we need to get the total resistance, and use Ohm's Law (
).
To find the total resistance, remember the equations for adding resistors is this:
Resistors 
and 
are in series, resistors 
and 
are in parallel, and resistors 
and 
are in series.
Now, we can find the current.
Finally, we can find the power.
Therefore, the power is 16W (watts).
The equation for power is
In order to get the power, we need the current. To find the current, we need to get the total resistance, and use Ohm's Law (
To find the total resistance, remember the equations for adding resistors is this:
Resistors
Now, we can find the current.
Finally, we can find the power.
Therefore, the power is 16W (watts).
If the circuit above is connected to a 
battery, what is the total power dissipated by the circuit?
If the circuit above is connected to a
Tap to see back →
The equation for power dissipated in a circuit is
The three resistors are in parallel with each other, so the total resistance is
Use Ohm's law to find current.
Finally, solve for power.
The equation for power dissipated in a circuit is
The three resistors are in parallel with each other, so the total resistance is
Use Ohm's law to find current.
Finally, solve for power.
A single 
resistor is added in series to a circuit with a 
battery. Determine the power dissipated by the resistor.
A single
Tap to see back →
Use Ohm's law:
Converting 
to 
Using definition of electric power:
Plugging in values:
Use Ohm's law:
Converting
Using definition of electric power:
Plugging in values:
In the circuit above, find the power being dissipated by 
.
In the circuit above, find the power being dissipated by
Tap to see back →
First, find the total resistance of the circuit.
and 
are in parallel, so we find their equivalent resistance by using the following formula:
Next, add the series resistors together.
Use Ohm's law to find the current in the system.
The current through 
and 
needs to add up to the total current, since they are in parallel.
Also, the voltage drop across them need to be equal, since they are in parallel.
Set up a system of equations.
Solve.
The equation for power is as follows:
First, find the total resistance of the circuit.
Next, add the series resistors together.
Use Ohm's law to find the current in the system.
The current through
Also, the voltage drop across them need to be equal, since they are in parallel.
Set up a system of equations.
Solve.
The equation for power is as follows:
In the circuit above, find the power being dissipated by 
.
In the circuit above, find the power being dissipated by
Tap to see back →
First, find the total resistance of the circuit.
and are in parallel, so we find their equivalent resistance by using the following formula:
Next, add the series resistors together.
Use Ohm's law to find the current in the system.
The current through and needs to add up to the total current, since they are in parallel.
Also, the voltage drop across them need to be equal, since they are in parallel.
Set up a system of equations.
Solve.
The equation for power is as follows:
First, find the total resistance of the circuit.
Next, add the series resistors together.
Use Ohm's law to find the current in the system.
The current through
Also, the voltage drop across them need to be equal, since they are in parallel.
Set up a system of equations.
Solve.
The equation for power is as follows:
In the circuit above, find the power being dissipaited by 
.
In the circuit above, find the power being dissipaited by
Tap to see back →
First, find the total resistance of the circuit.
and are in parallel, so we find their equivalent resistance by using the following formula:
Next, add the series resistors together.
Use Ohm's law to find the current in the system.
Because it is not in parallel, the total current in the circuit is equal to the current in 
.
The equation for power is as follows:
First, find the total resistance of the circuit.
Next, add the series resistors together.
Use Ohm's law to find the current in the system.
Because it is not in parallel, the total current in the circuit is equal to the current in
The equation for power is as follows:
Calculate the power being dissipated by 
Calculate the power being dissipated by
Tap to see back →
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combining 
with 
, 
with 
, 
with 
.
Then, combining 
with 
and 
:
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
The voltage drop across parallel resistors must be the same, so:
Using ohms law:
It is also true that:
Using Subsitution
Solving for 
:
Using the definition of electrical power, where 
is current and 
is the resistance of the component in question:
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combining
Then, combining
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
The voltage drop across parallel resistors must be the same, so:
Using ohms law:
It is also true that:
Using Subsitution
Solving for
Using the definition of electrical power, where
Calculate the power being dissipated by 
Calculate the power being dissipated by
Tap to see back →
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine with , with , with .
Then, combining with and :
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
The voltage drop across parallel resistors must be the same, so:
Using ohms law:
It is also true that:
Using Subsitution
Solving for 
:
Plugging in values
Using the definition of electrical power, where is current and is the resistance of the component in question:
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine
Then, combining
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
The voltage drop across parallel resistors must be the same, so:
Using ohms law:
It is also true that:
Using Subsitution
Solving for
Plugging in values
Using the definition of electrical power, where
Calculate the power being dissipated by 
Calculate the power being dissipated by
Tap to see back →
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine with , with , with .
Then, combining with and :
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
We know that the voltage drop across parallel resistors must be the same, so:
Using ohms law:
It is also true that:
Using Subsitution:
Solving for 
:
Plugging in values:
Using the definition of electric power, where 
is current and 
is the resistance of the component in question.
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine
Then, combining
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
We know that the voltage drop across parallel resistors must be the same, so:
Using ohms law:
It is also true that:
Using Subsitution:
Solving for
Plugging in values:
Using the definition of electric power, where
Calculate the power being dissipated by 
Calculate the power being dissipated by
Tap to see back →
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine with , with , with .
Then, combining with and :
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
It is true that the voltage drop across parallel resistors must be the same, so:
Using ohms law:
It is also true that:
Using Subsitution:
Solving for 
:
Pluggin in values:
Using the definition of electric power, where 
is current and 
is the resistance of the component in question.
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine
Then, combining
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
It is true that the voltage drop across parallel resistors must be the same, so:
Using ohms law:
It is also true that:
Using Subsitution:
Solving for
Pluggin in values:
Using the definition of electric power, where
Calculate the power being dissipated by 
Calculate the power being dissipated by
Tap to see back →
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine with , with , with .
Then, combining with and :
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
It is true that the voltage drop across parallel resistors must be the same, so:
Using ohms law
It is also true that:
Using Subsitution:
Solving for 
:
Plugging in values:
Using the definition of electrical power, where 
is current and 
is the resistance of the component in question.
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine
Then, combining
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
It is true that the voltage drop across parallel resistors must be the same, so:
Using ohms law
It is also true that:
Using Subsitution:
Solving for
Plugging in values:
Using the definition of electrical power, where
Calculate the power being dissipated by 
Calculate the power being dissipated by
Tap to see back →
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine with , with , with .
Then, combining with and :
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
It is true that the voltage drop across parallel resistors must be the same, so:
Using ohms law
It is also true that:
Using Subsitution:
Solving for 
:
Pluggin in values:
Using the definition of electrical power, where is current and is the resistance of the component in question:
The first step is to find the total resistance of the circuit.
In order to find the total resistance of the circuit, it is required to combine all of the parallel resistors first, then add them together as resistors in series.
Combine
Then, combining
Ohms is used law to determine the total current of the circuit
Combing all voltage sources for the total voltage.
Plugging in given values,
It is true that the voltage drop across parallel resistors must be the same, so:
Using ohms law
It is also true that:
Using Subsitution:
Solving for
Pluggin in values:
Using the definition of electrical power, where
What is the power being dissapaited by 
?
What is the power being dissapaited by
Tap to see back →
, 
, and 
are in parallel, so they are added by using:
Plugging in given values:
, 
, and 
are in series. So they are added conventionally:
Plugging in values:
First, it is necessary to find the total current of the circuit. Using Ohm's law:
Solving for 
:
Because 
, 
and 
are in parallel,
Also, the voltage drop must be the same across all three since they are in parallel.
Using Ohm's law again and substituting:
Using algebraic subsitution:
Solving for 
Plugging in values
Using the definition of electrical power, where is current and is the resistance of the component in question:
Plugging in values
Plugging in given values:
Plugging in values:
First, it is necessary to find the total current of the circuit. Using Ohm's law:
Solving for
Because
Also, the voltage drop must be the same across all three since they are in parallel.
Using Ohm's law again and substituting:
Using algebraic subsitution:
Solving for
Plugging in values
Using the definition of electrical power, where
Plugging in values
What is the power being dissapaited by 
?
What is the power being dissapaited by
Tap to see back →
, , and are in parallel, so they are added by using:
Plugging in given values:
, , and are in series. So they are added conventionally:
Plugging in values:
First, it is necessary to find the total current of the circuit. Using Ohm's law:
Solving for :
Because , and are in parallel,
Also, the voltage drop must be the same across all three since they are in parallel.
Using Ohm's law again and substituting:
Using algebraic subsitution:
Solving for 
Plugging in values:
Using the definition of electrical power, where 
is the current and 
is the resistance of the component in question:
Plugging in given values:
Plugging in values:
First, it is necessary to find the total current of the circuit. Using Ohm's law:
Solving for
Because
Also, the voltage drop must be the same across all three since they are in parallel.
Using Ohm's law again and substituting:
Using algebraic subsitution:
Solving for
Plugging in values:
Using the definition of electrical power, where
What is the power being dissapaited by 
?
What is the power being dissapaited by
Tap to see back →
, , and are in parallel, so they are added by using:
Plugging in given values:
, , and are in series. So they are added conventionally:
Plugging in values:
First, it is necessary to find the total current of the circuit. Using Ohm's law:
Solving for :
Because , and are in parallel,
Also, the voltage drop must be the same across all three since they are in parallel.
Using Ohm's law again and substituting:
Using algebraic subsitution:
Solving for 
:
Plugging in values:
Using the definition of electric power, where 
is current and 
is reistance.
Plugging in given values:
Plugging in values:
First, it is necessary to find the total current of the circuit. Using Ohm's law:
Solving for
Because
Also, the voltage drop must be the same across all three since they are in parallel.
Using Ohm's law again and substituting:
Using algebraic subsitution:
Solving for
Plugging in values:
Using the definition of electric power, where
What is the power being dissapaited by 
?
What is the power being dissapaited by
Tap to see back →
, , and are in parallel, so they are added by using:
Plugging in given values:
, , and are in series. So they are added conventionally:
Plugging in values:
First, it is necessary to find the total current of the circuit. Using Ohm's law:
Solving for 
:
The total current of the circuit is also the current through 
Using the definition of electric power, where 
is current and 
is the resistance of the component in question:
Plugging in given values:
Plugging in values:
First, it is necessary to find the total current of the circuit. Using Ohm's law:
Solving for
The total current of the circuit is also the current through
Using the definition of electric power, where