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

\(\displaystyle R_{eq} = R_{1}+R_{2}+...\)

when in series, they add via their reciprocal

\(\displaystyle R_{eq}^{-1} = R_{1}^{-1} + R_{2}^{-1}+...\)

Using these rules, we can first combine all the resistors in series (\(\displaystyle R_{1}\) and \(\displaystyle R_{2}\), \(\displaystyle R_{4}\) and \(\displaystyle R_{5}\right)\)), which can be diagrammed as such:

Using the parallel rule, find to total equivalent resistance.

\(\displaystyle R_{eq} = \left(R_{12}^{-1} + R_{3}^{-1} + R_{45}^{-1}\right)^{-1} = \left(\frac{1}{35\Omega}+\frac{1}{15\Omega}+\frac{1}{10\Omega}\right )^{-1} = 5.1\Omega\)

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

\(\displaystyle Q=CV\)

Since the voltage drop across both \(\displaystyle C_{1}\) and \(\displaystyle C_{2} + C_{3}\) 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

\(\displaystyle C_{23} = (C_{2}^{-1}+C_{3}^{-1})^{-1} = 3.3\mu F\)

Plugging into the first equation.

\(\displaystyle Q_{23} = C_{23}V_{23} = 30\mu C\)

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

\(\displaystyle Q_{3} = Q_{23} = 30\mu C\)

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 \(\displaystyle R_{3}\).

These initial conditions give us our first equation:

\(\displaystyle I_{1}+I_{2} = I_{3}\)

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:

\(\displaystyle -50I_{1} = 10V - 15V+35I_{2} = 0\)

The loop on the right gives us:

\(\displaystyle -35I_{2} + 15V-40I_{3} = 0\)

Since we're looking to find the current that flows through \(\displaystyle R_{3}\) we will need to solve for \(\displaystyle I_{3}\). The easiest way to do this would be to take the first equation and replace either \(\displaystyle I_{1}\) or \(\displaystyle I_{2}\) in one of the other two equations. The following solution will replace \(\displaystyle I_{1}\) in the first equation to accomplish this.

\(\displaystyle -50(I_{3}-I_{2}) + 35I_{2}-5V = 0\)

Rearrange.

\(\displaystyle 17I_{2} - 10I_{3} = 1V\)

Simplify the second equation.

\(\displaystyle 7I_{2}+8I_{3} = 3V\)

Lastly, solve for \(\displaystyle I_{2}\) and set the two equations equal to each other.

\(\displaystyle \frac{1}{17}+\frac{10}{17}I_{3} = \frac{3}{7}-\frac{8}{7}I_{3}\)

\(\displaystyle I_{3} = \frac{22}{103} = 0.21A\)

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

\(\displaystyle C_{eq}^{-1} = C_{1}^{-1} + C_{2}^{-1} + ...\)

Capacitors in parallel add as such:

\(\displaystyle C_{eq} = C_{1} + C_{2} + ...\)

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

\(\displaystyle C_{345} = (C_{3}^{-1} + C_{4}^{-1} + C_{5}^{-1})^{-1} = (\frac{1}{20\mu F}+\frac{1}{5\mu F}+\frac{1}{5\mu F})^{-1} = 2.2\mu F\)

The equivalent circuit is shown below:

Add the capacitors in parallel.

\(\displaystyle C_{eq} = C_{1} + C_{2} + C_{345} = 10\mu F + 30\mu F + 2.2\mu F = 42\mu F\)

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 \(\displaystyle I_{1}\), the current flowing through \(\displaystyle R_{2}\) and \(\displaystyle R_{3}\) as \(\displaystyle I_{2}\), and the current through \(\displaystyle R_{1}\) as \(\displaystyle I_{3}\). The first loop on the left, when written out using the loop rule, gives:

\(\displaystyle 9V-30\Omega I_{3} = 0\)

Solve for current.

\(\displaystyle I_{3} = .3A\)

The second loop gives us:

\(\displaystyle -25\Omega I_{2}-15\Omega I_{2} + 30\Omega I_{3} = 0\)

Plug in the value for \(\displaystyle I_{3}\)

\(\displaystyle I_{2} = \frac{3}{4}I_{3} = 0.225A\)

Find the power dissipated through \(\displaystyle R_{2}\)

\(\displaystyle P=I^{2}R=I^{2}_{2}R_{2}=0.76W\)

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

\(\displaystyle \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{1\Omega }+\frac{1}{4\Omega }\)

\(\displaystyle \frac{1}{R_T}=\frac{5}{4\Omega }\)

\(\displaystyle R_T=\frac{4}{5}\Omega\)

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

\(\displaystyle I=\frac{V}{R}=\frac{5V}{\frac{4}{5}\Omega}\)

\(\displaystyle I=6.25A\)

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

\(\displaystyle \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2\Omega }+\frac{1}{3\Omega }\)

\(\displaystyle \frac{1}{R_T}=\frac{5}{6\Omega }\)

\(\displaystyle R_T=\frac{6}{5}\Omega\)

Use Ohm's law to find the voltage.

\(\displaystyle V=IR\)

\(\displaystyle V=30A\left(\frac{6}{5}\Omega\right)=36V\)

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

\(\displaystyle V=IR\)

\(\displaystyle R_T=\frac{V}{I}\)

\(\displaystyle R_T=\frac{5}{1.25}\Omega=4\Omega\)

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:

\(\displaystyle \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}\)

Rearrange and solve for \(\displaystyle R_2\)

\(\displaystyle \frac{1}{R_2}=\frac{1}{R_T}-\frac{1}{R_1}\)

\(\displaystyle \frac{1}{R_2}=\frac{1}{4\Omega }-\frac{1}{6\Omega }=\frac{6-4}{24\Omega }=\frac{1}{12\Omega }\)

\(\displaystyle R_2=12\Omega\)

Find the combined resistances for the resistors in parallel:

\(\displaystyle \frac{1}{R_A}=\frac{1}{R_1}+\frac{1}{R_2}\)

\(\displaystyle \frac{1}{R_A}=\frac{1}{3\Omega}+\frac{1}{6\Omega}=\frac{9}{18\Omega}\)

\(\displaystyle R_A=2\Omega\)

\(\displaystyle \frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}\)

\(\displaystyle \frac{1}{R_B}=\frac{1}{2\Omega}+\frac{1}{16\Omega}=\frac{18}{32\Omega}\)

\(\displaystyle R_B=\frac{16}{9}\Omega\)

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

\(\displaystyle R_T=R_A+R_B=2\Omega+\frac{16}{9}\Omega=\frac{34}{9}\Omega\)

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

\(\displaystyle \frac{1}{R_A}=\frac{1}{R_1}+\frac{1}{R_2}\)

\(\displaystyle \frac{1}{R_A}=\frac{1}{1\Omega}+\frac{1}{2\Omega}=\frac{3}{2\Omega}\)

\(\displaystyle R_A=\frac{2}{3}\Omega\)

\(\displaystyle \frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}\)

\(\displaystyle \frac{1}{R_B}=\frac{1}{3\Omega}+\frac{1}{4\Omega}=\frac{7}{12\Omega}\)

\(\displaystyle R_B=\frac{12}{7}\Omega\)

Therefore, the total resistance is:

\(\displaystyle R_T=R_A+R_B=\frac{2}{3}\Omega+\frac{12}{7}\Omega=\frac{50}{21}\Omega\)

Now, note that since \(\displaystyle R_3\) and \(\displaystyle R_4\) are in parallel, the voltage drop across them is the same. Use Ohm's law to relate current in terms of voltage and resistance.

\(\displaystyle I=\frac{V_T}{R_T}\)

Substitute into Ohm's law for the resistance across \(\displaystyle R_3\):

\(\displaystyle V_{3,4}=\frac{V}{R_T}R_B\)

\(\displaystyle V_{3,4}=10V\left(\frac{21}{50\Omega}\right)\left(\frac{12\Omega}{7}\right)\)

\(\displaystyle V_{3,4}=7.2V\)