Using the relationship:

Here, is the resistivity,  is the resistance,  is cross sectional area, and  is the length.

The area of a circle is:

Substitute:

Plug in given values and solve.

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:

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).

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.

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.

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.

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.

 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.

 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:

 and are in series with each other, therefore, they will have the same current values. They also have the same resistance values. Thus, they will have the same power dissipated, as .