To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top  and the bottom  is the same and equal to the voltage across the circuit, .

The voltage drop across  can be found as:

To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top  and the bottom  is the same and equal to the voltage across the circuit, .

Furthermore, the current that passes through  must be the same as the current that passes through .

Therefore, the current that passes through them can be found by rearranging Ohm's law, solving for current.

The quickest way to approach this problem is to realize that the voltage drop across  is the same as the voltage drop across the combined resistances of  and . Since this parallel combination is the only presence of resistance in the circuit, this voltage drop must be the total voltage of the circuit, .

Therefore, the current across  is:

Realize that the voltage drop across the combined resistances of  and  must be equal to the voltage of the circuit, since the parallel combination is the only presence of resistance in the circuit. This voltage drop must be the total voltage of the circuit, .

The current across  and  is the same, and is given as:

Begin by finding the resistance of the parallel connection:

The total current is then found using Ohm's law:

The symbol for a capacitor is written as a break in the circuit separated by two parallel lines of equal length as shown below. This loosely resembles the most common type of capacitor, a parallel plate capacitor.

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.