DC Circuits 2

### Combining series and parallel resistors in the same circuit

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.

### Example 1

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 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.

### Example 2

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 Vpar, the voltage drop across our 33 Ω parallel element, and Vser 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.

### Example 3

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 R1-2 and R3-4 after the labeled resistors.

\begin{align} R_{1-2} &= \left( \frac{1}{25} + \frac{1}{50} \right)^{-1} \\[5pt] &= \left( \frac{3}{50} \right)^{-1} = 16.7 \, \Omega \\[8pt] R_{3-4} &= \left( \frac{1}{50} + \frac{1}{75} \right)^{-1} \\[5pt] &= \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}

### Practice circuits

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

 1 Solution First we'll reduce the 6Ω and 12Ω resistors to a single 18Ω resistor: Now reduce the parallel resistors like this: $$R_{||} = \left( \frac{1}{18} + \frac{1}{18} \right)^{-1} = \left( \frac{2}{18} \right)^{-1} = 9 \, \Omega$$ That gives us the following intermediate circuit: the total resistance of the circuit is then $R_{tot} = 24 + 9 = 33 \, \Omega.$ So our simplest equivalent circuit is Now the total current is $I = \frac{V}{R} = \frac{6 \, V}{33 \, \Omega} = 0.1818 \, A.$ Now we can back up to our intermediate circuit, The voltage drops across these resistors are \begin{align} V_{||} &= (0.1818 \, A)(9 \, \Omega) = 1.636 \, V \\[5pt] V_{24} &= (0.1818 \, A)(24 \, \Omega) = 4.364 \, V, \end{align} Where V|| is the voltage drop across the two parallel resistors in the first reduced circuit. Now we can calculate the currents through each resistor using Ohm's law. They are: \begin{align} I_{18} &= \frac{V}{R} = \frac{1.636 \, V}{18 \, \Omega} = 0.0909 \, A \\[5pt] I_{24} &= \frac{4.364 \, V}{24 \, \Omega} = 0.1818 \, A \end{align} Finally, we can go back to the original circuit and calculate the voltage drops across the 6Ω and 12Ω resistors: \begin{align} V_6 &= IR = (0.0909 \, A) (6 \, \Omega) = 0.5454 \, V \\[5pt] V_{12} &= (0.0909 \, A) (12 \, \Omega) = 1.0908 \, V \end{align} 2 Solution First take care of the two series resistances in that parallel set to arrive at this intermediate circuit: Now these parallel resistors can be combined: $$R_{||} = \left( \frac{1}{120} + \frac{1}{200} + \frac{1}{80} \right)^{-1} = 38.71 \, \Omega$$ The resulting simplest equivalent circuit is The total current is then $I = \frac{V}{R} = \frac{8 \, V}{38.71 \, A} = 0.2067 \, A.$ Now we can step back and calculate the currents through each arm of the parallel element. Recall that in the intermediate circuit, the voltage drops across our 120, 200 & 80 Ω resistors is the same: 8 Volts. \begin{align} I_{120} &= \frac{V}{R} = \frac{8}{120} = 0.0668 \, A \\[5pt] I_{200} &= \frac{8}{200} = 0.040 \, A \\[5pt] I_{80} &= \frac{8}{80} = 0.10 \, A \end{align} Finally, we can use the currents through each branch to calculate the voltage drops across each resistor. No calculation is needed, though. Note that in the top and bottom branch, the voltage drop of 8V is split between equivalent resistances, 4V each, so the voltage drops are: \begin{align} V_{60} &= 4 \, V \\[5pt] V_{200} &= 8 \, V \\[5pt] V_{40} &= 4 \, V \end{align} 3 Solution First we need to convert the parallel element (the two 80Ω resistors) to a single equivalent resistor: $$R_{||} = \left( \frac{2}{80} \right)^{-1} = 40 \, \Omega$$ The next equivalent circuit is then Now sum the resistors in series (40 + 40 = 80Ω, 20 + 40 = 60Ω) to further reduce the equivalent circuit: Now find the total resistance by adding these parallel resistances: $$R_{tot} = \left( \frac{1}{80} + \frac{1}{60} \right)^{-1} = 34.28 \, \Omega$$ The final equivalent circuit is Now the total current is $I = \frac{V}{R} = \frac{10 \, V}{34.28 \, \Omega}$ $= 0.2917 \, A.$ Backing up to the previous equivalent circuit, the currents through the 80 and 60 Ohm resistors are: \begin{align} I_{80} &= frac(V){R} = \frac{10}{80} = 0.125 \, A \\[5pt] I_{60} &= \frac{10}{60} = 0.1667 \, A \end{align} Now we can take another step back toward the original circuit and calculate the voltage drops across the 20 and 40Ω resistors (or equivalents). These are: \begin{align} V_{40, top} &= 5V \\[5pt] V_{40, bottom} &= IR = (0.1667 \, A)(40 \, \Omega) = 6.67 \, V \\[5pt] V_{20, bottom} &= (0.1667 \, A)(20 \, \Omega) = 3.33 \, V, \end{align} where V{40, top} is shared equally by two resistors, for 5V each. We didn't need the current through that branch to figure that out. Finally, the current through each branch of the parallel element (the 80Ω resistors) of our original circuit is just half of the current through that branch, or $0.125/2 = 0.0625 \, A.$ Here is a summary of our results:
 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}

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