First, we need to calculate the equivalent resistance of the three resistors in parallel. To do this, we will use the following equation:

Now, to get the total equivalent resistance, we can simply add the two remaining values, since they are in series:

First, we need to calculate the current flow through R2 without the extra voltage attached. We will need to calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add them.

Then, we can use Ohm's law to calculate the current through the circuit:

Now that we have the current, we can calculate the additional current that the new voltage contributes:

There is only one resistor (R2) in the path of the new voltage, so we can calculate what that voltage needs to be to deliver the new current:

First, let's remind ourselves that the effective resistance of resistors in a series is  and the effective resistance of resistors in parallel is .

Start this problem by determining the effective resistance of resistors 2, 3, and 4:

 (This is because these three resistors are in series.)

Now, the circuit can be simplified to the following:

Next, we will need to determine the effective resistance of resistors  and 6:

Again, the circuit can be simplified:

From here, the effective resistance of the DC circuit can be determined by calculating the effective resistance of resistors , 1, and 5:

The first step to figuring out this problem is to figure out how resistances of light bulb correlate to the power rating. For a resistor, the power dissipated is:

Thus, there is an inverse relationship between the resistance of the lightbulb and the power rating. 

The second step is to take a look at circuit elements in series and in parallel. In series, they share the same current; in parallel they share the same voltage. Thus, for the two lightbulbs in series, the one with the higher resistance (lower wattage) will be brighter, and for a parallel configuration the one with the lower resistance (higher wattage) will be brighter.

The total resistance of resistors in a series is the sum of their individual resistances. In this case,

Resistors in series:

Resistors in parallel:

The equation for equivalent resistance for multiple resistors in parallel is:

Plug in known values and solve. 

Notice that for resistors in parallel, the total resistance is never greater than the resistance of the smallest element.

Since the resistors  and  form a parallel network, removing  from the circuit increases the resistance of that part of the circuit. Because the new circuit is the series combination of  and , the increased resistance leads to lower current in each of these resistors.

To determine total charge stored, we need to add up the capacitance of each capacitor(because they are capacitors in parallel) and multiply by the voltage difference. Recall that for capacitors,

For parallel plate capacitors:

Here, , which is the permittivity of empty space,  is the dielectric constant, which is  since there is only vacuum present, , which is the area of the parallel plates, and , which is the distant between the plates. 

Plug in known values to solve for capacitance.

Each of the two capacitors has capacitance 

Since the capacitors are in parallel, the total capacitance is the sum of each individual capacitance. The total capacitance in the circuit  is given by:

Plug this value into our first equation and solve for the total charge stored.

, where  is the total charge stored in the capacitor. Since 

First we need to determine the overall resistance of the circuit before we know how much current is flowing through. Since the resistors are in parallel, their resistances will add reciprocally:

where  is the total resistance of the circuit. 

Now that we've solved for , we know that the current  flowing through the circuit can be found using Ohm's law: