The equivalent resistance of resistors in series is calculated by , while for parallel resistors, . This clearly shows that resistors in a series configuration will generate a much higher, or maximum, equivalent resistance value.

In this question, we're presented with a circuit diagram. We're told the values of the voltage source as well as each of the three resistors. We're asked to find the current that is flowing through resistor B.

To begin this problem, we'll need to figure out how much current is flowing through the entire circuit. But before we can do this, we'll need to figure out what the equivalent resistance of the circuit is. To do this, we'll need to look at how the circuit is configured so that we can see which resistors are in series, and which are in parallel.

We can see that resistors A and B are in parallel with each other. Furthermore, both A and B are in series with resistor C. So to begin, we'll need to find the equivalent resistance for resistors A and B. Using the equation for parallel resistors, we can calculate the equivalent resistance for resistors A and B.

Now that we've found the equivalent resistance for A and B, we can use this value in combination with resistor C, which is connected in series.

Because we've considered all the resistors in the circuit, this value is the total resistance of the circuit. Now, we can use Ohm's law in order to find the current running through the circuit.

Now that we have the total amount of current flowing through the circuit, we can look at each individual resistor in order to determine the amount of current flowing through that resistor. We can see that after traveling past the voltage source, there is a node that branches off, leading to resistor A and B. Because each of these resistors has the same resistance, the current through each must also be the same. Therefore, the current will split in half at the node, causing half the current to go to resistor A, and the other half to go to resistor B. Thus, the current flowing through resistor B is .

In this question, we're presented with a circuit that has a number of resistors connected in series and in parallel, and we're asked to find the total resistance of the circuit.

In looking at this diagram, the first thing that we need to do is find the equivalent resistance for each group of resistors that are connected in series, which is the top two rows. And remember, resistors add directly when arranged in series.

Then, we can do the same thing for the middle row.

Now that we have a value of resistance for each row, we can now consider all three rows together as being connected in parallel. Remember, resistors arranged in parallel will add inversely.

In order to find the equivalent resistance, you must take small steps and slowly work towards finding the total equivalent resistance.You can combine resistors if they are in parallel. The expression to find parallel equivalent resistance is:

You can also combine resistors in series, or one after the other. The expression to find resistors in series is:

To start off, you can combine R2 and R3, since they are both in parallel:

Next, you can combine some of the series resistors together. 

After this, you can combine the following resistors in parallel:

Then, combine  and  in series:

Now get rid of the last parallel by combining  and  in parallel:

Now finally, add the remaining resistors in series to find the equivalent resistance:

The goal of this question is to realize that when two resistors are connected in parallel, the equivalent resistance is lower than either of the two original resistors. But when two resistors are connected in series, the equivalent resistance is the sum of the two original resistors. Therefore to minimize our equivalent resistance, we want all the resistance to be in the parallel resistors, leaving  for the resistor in series.

When resistors are in series, they add normally, such as

when in series, they add via their reciprocal

Using these rules, we can first combine all the resistors in series ( and ,  and ), which can be diagrammed as such:

Using the parallel rule, find to total equivalent resistance.

The relationship between a capacitor's charge and the voltage drop across it is:

Since the voltage drop across both  and  are the same, we just have to worry about the right part of the circuit. Capacitors are the opposite of resistors when it comes to finding equivalent capacitance, so for capacitors in series the two capacitors on the right will add as such

Plugging into the first equation.

Since the two capacitors are in series they must share the same charge as the equivalent capacitor.

The first step to solving this problem is to utilize Kirchoff's rules to write a set of equations to find the unknowns. Below two loops are diagrammed, and we assign 3 currents to the circuit. The first and second pass up through batteries 1 and 2 respectively, and the third current passes down through .

These initial conditions give us our first equation:

Now using the loop rule and setting all voltage changes across a loop equal to zero, we get these two equations for the two loops. The loop on the left gives us:

The loop on the right gives us:

Since we're looking to find the current that flows through  we will need to solve for . The easiest way to do this would be to take the first equation and replace either  or  in one of the other two equations. The following solution will replace  in the first equation to accomplish this.

Rearrange.

Simplify the second equation.

Lastly, solve for  and set the two equations equal to each other.

Capacitors add opposite of the way resistors add in a circuit. That is, for capacitors in series they add as such:

Capacitors in parallel add as such:

Use this information to add all the capacitors in series together. The only branch this applies to is the right hand branch.

The equivalent circuit is shown below:

Add the capacitors in parallel.

The first step to solving a circuit problem would be to identify and calculate for all the currents that are unknown using Kirchoff's Laws. By drawing loops in the circuit and setting the voltage drop across a loop equal to zero we can calculate for the unknown currents and solve for the power. The two loops are indicated here in this diagram.

Designate the current flowing through the battery as , the current flowing through  and  as , and the current through  as . The first loop on the left, when written out using the loop rule, gives:

Solve for current.

The second loop gives us:

Plug in the value for 

Find the power dissipated through