To solve this question we need to understand the principles of circuits. Circuit elements (such as resistors, capacitors, and inductors) connected in parallel have the same voltage whereas circuit elements connected in series have the same current flowing through them. In parallel circuits, the current hits nodes (regions where two circuit elements branch out and become parallel) and gets split into two different currents, each supplying the individual circuit elements; therefore, the current flowing through parallel connected circuit elements is always less than the total flow of the current. On the other hand, circuit elements in series have the same current flowing through them; therefore, the total current of the circuit flows through each element connected in series. This means that the current flowing through series circuit is higher.

Voltage drop across the parallel circuit elements is equal; therefore, the voltage equals the voltage supplied by the power source. In circuit elements in series, however, the voltage drop is smaller across each element; therefore, resistors in series have smaller voltage drop and larger current than their parallel circuit counterparts.

The amount of brightness that each light bulb will produce is proportional to the amount of current passing though it. If the lightbulbs were to be placed in series, the total amount of current passing though each bulb would be equal to the voltage of the battery divided by the sum of the resistances of both lightbulbs, given by:

If the lightbulbs were instead to be placed in parallel, the total amount of current passing though would be equal to the voltage across the lightbulb, equal to the voltage of the battery, divided by the internal resistance of the lightbulb:

From those two equations, it is clear to see that 

, and that there would be more light if the lightbulbs are placed parallel to one another. 

All of the current flowing through the circuit will go through R5, since it is not in parallel with anything. To find the current flowing through the circuit, we will need to first find the total equivalen resistance of the circuit.

To do this, we first we need to condense R3 and R4. They are in series, so we can simply add them to get:

Now we can condense R2 and R34. They are in parallel, so we will use the following equation:

The equivalent circuit now looks like:

Since everything is in parallel, we can simply add everything up:

Now that we have the total resistance of the circuit, we can use Ohm's law to find the total current:

Rearranging for current, we get:

Kirchoff's current law tells us that the same amount of current entering the junction after R1 must also leave the junction. We also know that the voltage drop across each path of the split is the same.

Consider the following circuit to help visualize things:

Using Ohm's law to expand the voltages, we get:

We now have two equations that we can solve simultaneously:

Since we are solve for I3, let's rearrange the first equation for I2:

Substituting this into the second equation, we get:

Rearranging to solve for I3:

We have all the values we need, so simply plug in the given values and solve:

Kirchoff's current law tells us that the same amount of current entering the junction after R1 must also leave the junction. We also know that the voltage drop across each path of the split is the same.

Consider the following circuit to help visualize things:

Using Ohm's law to expand the voltages, we get:

We now have two equations that we can solve simultaneously:

To find V2, we need to calculate I2, so let's rearrange the first equation for I3

Substituting this into the second equation, we get:

Rearranging for I2:

We have all the values we need, so simply plug in the given values and solve:

Now that we know I2, we can use Ohm's law to find the voltage drop across that resistor:

In order to calculate the current flowing through the circuit in each scenario, we need to calculate the equivalent resistance in each scenario.

SCENARIO 1: Switch open

When the switch is open, everything is series, so we can simply add all of the resistances up to find the total equivalent resistance. Note that R2 will be excluded.

Using Ohm's Law, we can now calculate the current flowing through the circuit:

Rearranging for current:

SCENARIO 2: Switch closed

The circuit now looks like this:

Since we have a few resistors in parallel, this calcluation will involve a few more steps.

First we need to condense R3 and R4. They are in series, so we can simply add them to get:

Now we can condense R2 and R34. They are in parallel, so we will use the following equation:

The equivalent circuit now looks like:

Since everything is in series, we can simply add everything up:

Now using Ohm's law to calculate the current flowing through the circuit:

Now, we simply take the different of the currents in the two scenarios:

We are asked to compare two different scenarios, each involving the calculation of equivalent resistance, which will use the following formula:

Scenario 1: With R3 Present

Now using Ohm's law:

Scenario 2: Without R3

Now using Ohm's law:

Calculate the change in current:

Although it is possible to solve this problem by calculating an equivalent resistance, calculating a total current through the circuit, and then using Kirchoff's junction rule to find the current through R1, it is much easier to simply use Kirchoff's Loop rule. What this rule says is that through any closed loop in a circuit, all voltages must add up to zero. Written as an equation:

If we consider the closed loop path consisting of only the power supply and R1, we can use Ohm's law to calculate the current:

The total current through the circuit is unnecessary to solve this problem. We only need to know the current through the resistor, as well as Kirchoff's loop rule. The loop rule states that for any closed loop through a circuit, the voltages add up to zero.

For this problem, we will consider the loop that consists solely of the voltage source and R2. From the rule, we know that 12V is lost through the resistor. Using Ohm's law, we can write:

Rearranging for R2, we get:

The voltage drop is related to the current and resistance via Ohm's law: