This section follows from the last one, so unless you've worked through that, or you've already got some experience with simple circuits, you might want to go back.

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.

For our first circuit, we'll place two resistors (labeled 1 & 2) in series in parallel with two more resistors (3 & 4) in series, like this:

The approach here is to first solve for the total resistance of each parallel branch, adding resistors 1 & 2, and 3 & 4: to get 200 Ω and 300 Ω, respectively. From that we can reduce the complexity of the circuit (we always come back to the original version at the end):

Now we can find the total resistance of that parallel element:

$$ \begin{align} R_{tot} &= \left( \frac{1}{200} + \frac{1}{300} \right)^{-1} = \left( \frac{5}{600} \right)^{-1} \\[5pt] &= 120 \, \Omega \end{align}$$

So the total resistance is 120 Ω, and we have our final simplified equivalent circuit,

... 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 **I _{200}** and the lower

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

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.

The next circuit has a pair of parallel resistors (50 Ω, 100 Ω) in series with a 20 Ω resistor. Our goal is to find all voltage drops and currents throughout the circuit. Here's an outline of the approach we'll take:

- Reduce the parallel pair to an equivalent single resistor,
- Add that to the 20 Ω resistance to get our single-resistor equivalent circuit,
- Calculate the overall current in the circuit (the current through the battery and the 20 Ω resistor),
- Calculate the voltage drops across the parallel element (one drop for the whole thing) and the 20 Ω resistor,
- Calculate the currents in each arm of the parallel element.

First we'll reduce the parallel resistors to a single resistance:

$$R_{par} = \left( \frac{1}{50} + \frac{1}{100} \right)^{-1} = \left( \frac{3}{100} \right)^{-1} = 33 \, \Omega$$

That gives us the equivalent series circuit:

which we can reduce, by adding 33Ω + 20Ω = 53Ω, to get the equivalent simple circuit (in the previous section, we learned that this is called **Thevenin's equivalent**),

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

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

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.

This circuit is composed of two parallel elements wired in series with a battery.

To calculate all currents and voltage drops for this circuit, we'll follow this road map:

- Calculate the equivalent resistances of each parallel element, resistors 1 & 2, and resistors 3 & 4,
- Form an equivalent series circuit and add the resistances to form the simplest equivalent circuit (one resistor and the 20 V battery),
- Calculate the overall current in the circuit,
- Calculate the voltage drops across each parallel element, and
- Use those drops to calculate the current in each resistor.

First we'll convert each of the parallel elements to single equivalent resistances. I'll call them R_{1-2} and R_{3-4} after the labeled resistors.

$$ \begin{align} R_{1-2} &= \left( \frac{1}{25} + \frac{1}{50} \right)^{-1} = \left( \frac{3}{50} \right)^{-1} = 16.7 \, \Omega \\[5pt] R_{3-4} &= \left( \frac{1}{50} + \frac{1}{75} \right)^{-1} = \left( \frac{125}{50\cdot 75} \right)^{-1} = 30 \, \Omega \end{align}$$

Now we have this equivalent circuit:

... which we can use to calculate the overall resistance of the circuit,

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

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:

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

Solve for all voltage drops and currents everywhere in these DC circuits:

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