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 

To find the amperage, first find the combined resistances of the resistors in parallel:

After that, calculate the current using Ohm's Law:

To find the voltage, first find the combined resistances of the resistors in parallel:

Use Ohm's law to find the voltage.

Find the total resistance of the circuit, which can be determined using Ohm's law.

Now, the resistance of the second resistor can be found. Since the two resistors are in parallel, they're related to the total resistance as follows:

Rearrange and solve for 

Find the combined resistances for the resistors in parallel:

Combine these two combined series resistors to find the total resistance:

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

Therefore, the total resistance is:

Now, note that since  and  are in parallel, the voltage drop across them is the same. Use Ohm's law to relate current in terms of voltage and resistance.

Substitute into Ohm's law for the resistance across :