, , 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.

, , 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:

, , 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

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.

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 power dissipated by the resistor will be

Using Ohm's Law:

And changing the equation to be exclusively in terms of voltage and resistance:

From this, it can be seen that doubling the voltage will quadruple the power.

The power dissipated by the resistor is

Using Ohm's Law

And changing the equation to be exclusively in terms of voltage and resistance

From this, it can be seen that doubling the resistance will halve the power dissipated if the voltage is kept constant.

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.

In series, resistance adds conventionally.

Using Ohm's law:

Using definition of electric power:

Plugging in values:

Use Ohm's law: 

Converting  to 

Using definition of electric power:

Plugging in values: