If you're ready, we'll do some examples of circuits that combine series and parallel elements in the same loop. They can take a few more steps – and those have to be done in the right order, but the logic is really just the same.

... from which we can calculate the current through the battery (I say "through the battery" because it will be different in the branches of the parallel paths once we work our way back to there).

$$I = \frac{V}{R} \: \color{#E90F89}{\rightarrow} \: I = \frac{10 \, V}{120 \, \Omega} = 0.083 \, A$$

Now we can step back and calculate the current in each arm of the parallel paths. I'll label the upper arm I₂₀₀ and the lower I₃₀₀, using the total series resistance of each as a label to keep things straight (you could also use "upper" & "lower" ... or whatever you want).

$$ \begin{align} I_{200} = \frac{V}{R_{200}} = \frac{10 \, V}{200 \, \Omega} &= 0.050 \, A \\[5pt] I_{300} = \frac{V}{R_{300}} = \frac{10 \, V}{300 \, \Omega} &= 0.033 \, A \\[5pt] \hline \color{#E90F89}{\text{total current:}} &\phantom{00} 0.083 \, A \end{align}$$

The total currents, calculated differently, agree. Ahhh .... science!

Finally, we calculate the voltage drops in the resistors of each arm using the appropriate current for each branch, 0.05A for the top, 0.033A for the bottom. They are

$$ \begin{align} V_1 = (0.050 \, A)(100 \, \Omega) &= 5 \, V \\[5pt] V_2 = (0.050 \, A)(100 \, \Omega) &= 5 \, V \\[5pt] V_3 = (0.033 \, A)(100 \, \Omega) &= 3.3 \, V \\[5pt] V_4 = (0.033 \, A)(200 \, \Omega) &= 6.6 \, V \\[5pt] \end{align}$$

Notice that $V_1 + V_2 = 10 \, V$ and $V_3 + V_4 = 10 \, V.$ And that's the whole circuit – everything (well, almost) we'd want to know about it.

Now the total current is

$$I = \frac{V}{R} \: \color{#E90F89}{\rightarrow} \: I = \frac{5 \, V}{53 \, \Omega} = 0.094 \, A$$

Stepping back to the first equivalent circuit (the one with two resistors in series), we can calculate the voltage drop across the resistors. I'll call them V_par, the voltage drop across our 33 Ω parallel element, and V_ser the drops across the 20 Ω resistor:

$$ \begin{align} V_{par} = (0.094 \, A)(33 \, \Omega) &= 3.11 \, V \\[5pt] V_{ser} = (0.094 \, A)(20 \, \Omega) &= 1.89 \, V \\[5pt] \hline \color{#E90F89}{\text{total current:}} &\phantom{00} 5 \, V \end{align}$$

Notice that the sum of the voltage drops is equal to the battery voltage. Now that we have the drop across the parallel resistors (remember that parallel resistors share a common voltage drop), all that's left is to calculate the current through them:

$$ \begin{align} I_{50} = \frac{V}{R_{200}} = \frac{3.11 \, V}{50 \, \Omega} &= 0.062 \, A \\[5pt] I_{100} = \frac{V}{R_{300}} = \frac{3.11 \, V}{100 \, \Omega} &= 0.031 \, A \\[5pt] \hline \color{#E90F89}{\text{total current:}} &\phantom{00} 0.093 \, A \end{align}$$

There's a little rounding error there, but the sum of currents calculated this way agrees with the total current we calculated above. As a general rule, you should keep as many digits as possible (e.g. by not clearing your calculator) during the calculation and round at the end. Only worry about significance of digits when you report the result.

$$R_{series} = 16.7 \, \Omega + 30 \, \Omega = 46.7 \, \Omega$$

and the Thevenin's equivalent circuit looks like this:

Then, as usual, we calculate the overall current flowing through the circuit using I = V/R:

$$I = \frac{V}{R} \: \color{#E90F89}{\rightarrow} \: I = \frac{20 \, V}{46.7 \, \Omega} = 0.428 \, A$$

Stepping back to the previous series equivalent circuit, we can use Ohm's law calculate the voltage drops across each of our parallel elements:

$$ \begin{align} V_{1-2} = (0.428 \, A)(16.7 \, \Omega) &= 7.15 \, V \\[5pt] V_{3-4} = (0.428 \, A)(30 \, \Omega) &= 12.85 \, V \\[5pt] \hline \color{#E90F89}{\text{total current:}} &\phantom{00} 20 \, V \end{align}$$

Now it's just a matter of using the appropriate voltage drops to calculate the currents through each resistor:

$$ \begin{align} I_{1} = \frac{V}{R_{1}} = \frac{7.15 \, V}{25 \, \Omega} &= 0.286 \, A \\[5pt] I_{2} = \frac{V}{R_{2}} = \frac{7.15 \, V}{50 \, \Omega} &= 0.143 \, A \\[5pt] \hline \color{#E90F89}{\text{total current:}} &\phantom{00} 0.429 \, A \end{align}$$

The total current here has a bit of roundoff error, but that's typical. The currents in resistors 3 & 4 are:

$$ \begin{align} I_{3} = \frac{V}{R_{3}} = \frac{12.85 \, V}{50 \, \Omega} &= 0.257 \, A \\[5pt] I_{4} = \frac{V}{R_{4}} = \frac{12.85 \, V}{75 \, \Omega} &= 0.171 \, A \\[5pt] \hline \color{#E90F89}{\text{total current:}} &\phantom{00} 0.428 \, A \end{align}$$

4.	Solution First we'll reduce the parallel (120 and 100Ω) resistors to a single resistance: $$R_{||} = \left( \frac{1}{120} + \frac{1}{100} \right)^{-1} = 54.54 \, \Omega$$ That gives us this equivalent circuit: The total resistance is just the resulting series resistance, 84.54Ω, giving the simplest equivalent circuit: The total current is then $I = \frac{V}{R} = \frac{5 \, V}{84.54 \, \Omega} = 0.0591\, \Omega.$ Now we can calculate the voltage drops across our equivalent 54.54Ω and 30Ω resistors: $$ \begin{align} V_{54.54} &= IR = (0.0591 \, A)(54.54 \, \Omega) \\[5pt] &= 3.22 \, V \\[5pt] V_{30} &= (0.0591)(30) = 1.78 \, V. \end{align}$$ Finally, we can return to the original circuit and calculate the current through the parallel resistors: $$ \begin{align} I_{120} &= \frac{V}{R} = \frac{3.22 \, V}{120 \, \Omega} = 0.0269 \, A \\[5pt] I_{100} &= \frac{3.22 \, V}{100 \, \Omega} = 0.0322 \, A \end{align}$$
5.	Solution This circuit is drawn in a funky way, but one that is actually fairly common in circuit diagrams. It can be redrawn like this: Summing the series resistors gives a simpler circuit: Summing the parallel resistors, $R_{||} = \left(\frac{2}{20} \right)^{-1}$ $= 10\, \Omega,$ which gives the simplest equivalent: Now the total current is $I = \frac{V}{R} = \frac{5}{10} = 0.5 \, A.$ Backing up one equivalent circuit, it's easy to see that half of this current (0.25 A) flows through each 20Ω branch of the parallel portion. Finally, the voltage drop across each 10Ω resistor are: $$V_{10} = IR = (0.25 \, A)(10 \, \Omega) = 2.5 \, V$$ No calculation was actually required for that, if you think about it.
6.	Solution It might help to first redraw this circuit so that things are a little clearer: Now we can add any resistors in series to arrive at this equivalent circuit: Summing the parallel resistors gives $$R_{||} = \left( \frac{2}{20} + \frac{1}{10} \right)^{-1} = 5 \, \Omega$$ That gives us another simplification in the circuit: Now summing the final two resistances gives us the equivalent circuit Now the total current is $I = \frac{V}{R} = \frac{5 \, V}{55 \, \Omega} = 0.0909 \, A.$ From here we can back up and calculate the voltage drops across the 50Ω resistor and the group of parallel resistors: $$ \begin{align} V_{50} &= IR = (0.0909\, A)(50 \, \Omega) = 4.545 \, V \\[5pt] V_{||} &= (0.0909\, A)(5 \, \Omega) = 0.4545 \, V \end{align}$$ Now we can use the voltage drop across the parallel element to calculate the currents through each branch. The top and bottom branches are the same: $$ \begin{align} I_{top,bottom} &= \frac{V}{R} = \frac{0.4545}{20} = 0.0227 \, A \\[5pt] I_{middle} &= \frac{0.4545}{10} = 0.04545 \, A \end{align}$$ Now the voltage drops. First across the 50Ω resistor: $V = IR = 0.0909(50) = 4.545 \, V.$ Now the voltage drop across each branch of the parallel element is just 5V minus that, or 0.455 V. In the upper and lower branches of the original circuit, that voltage is split across two equivalent resistors. Overall, we have $$ \begin{align} V_{10, top/bottom} &= 0.227 \, V \\[5pt] V_{10, middle} &= 0.4545 \, V \end{align}$$