{"id":1695,"date":"2018-06-07T10:27:39","date_gmt":"2018-06-07T14:27:39","guid":{"rendered":"http:\/\/www.circuitcrush.com\/?p=1695"},"modified":"2021-06-30T17:29:40","modified_gmt":"2021-06-30T21:29:40","slug":"ninja-circuit-analysis-tricks","status":"publish","type":"post","link":"https:\/\/www.circuitcrush.com\/ninja-circuit-analysis-tricks\/","title":{"rendered":"3 Ninja Circuit Analysis Tricks"},"content":{"rendered":"<p>In the last post concerning circuit analysis (<a href=\"http:\/\/www.circuitcrush.com\/circuit-analysis\/\" target=\"_blank\" rel=\"noopener\">Simple Circuit Analysis Techniques You Should Know<\/a>) I went over some very basic but important techniques.<\/p>\n<p>First, we had a quick over view of resistors in series, parallel, and series-parallel circuits. Then, we talked about Ohms Law. Finally, Kirchhoff\u2019s Laws were introduced.<\/p>\n<p>This time, I\u2019m going to show you a few simple tricks for combining parallel resistors before I go into circuit analysis using some other very important (and slick) methods.<\/p>\n<p><!--more--><\/p>\n<h1><strong>Circuit Analysis 201: Slick Ninja Techniques<br \/>\n<\/strong><\/h1>\n<h2><strong>Supersleuth Shortcuts for Combining Parallel Resistors<br \/>\n<\/strong><\/h2>\n<p>Last time I promised to reveal a few tricks that work on certain parallel circuits. I\u2019ll start by saying that the shortcuts that follow not only apply to resistors, but also apply to other components.<\/p>\n<p>Recall from the first post in this series that parallel resistors add up with the formula below. We also know that inductors follow a similar pattern.<\/p>\n<p><img decoding=\"async\" class=\"size-full wp-image-1643 aligncenter\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Formula-Parallel-Resistors.jpg\" alt=\"Formula-Parallel-Resistors\" width=\"142\" height=\"61\" \/><\/p>\n<p>Of course, this can extend beyond 3 resistors to any number by adding more 1\/R terms on the bottom.<\/p>\n<p>The problem is that doing calculations like this is tedious and boring. Good thing there are common instances where it\u2019s totally unnecessary!<\/p>\n<p>Consider the example of two resistors in parallel (a common situation). I\u2019ll spare the derivation (though it\u2019s not hard), but equation 1 is a quick, easy to remember formula for this situation.<\/p>\n<figure id=\"attachment_1701\" aria-describedby=\"caption-attachment-1701\" style=\"width: 113px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" class=\"wp-image-1701 size-full\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/2-resistors-in-parallel-formula.jpg\" alt=\"2-resistors-in-parallel-formula\" width=\"113\" height=\"42\" \/><figcaption id=\"caption-attachment-1701\" class=\"wp-caption-text\">(eq. 1)<\/figcaption><\/figure>\n<p>Things get even better if there are multiple resistors of the same value in parallel.<\/p>\n<p>For example, most of you can probably see that two 100 \u03a9 resistors in parallel have a total resistance of 50 \u03a9. This suggest that if all the resistors in a parallel network are the same value, the total resistance is the value of one resistor divided by the number of resistors.<\/p>\n<p>So, if we have ten 100 \u03a9 resistors in parallel, the total resistance is 10 \u03a9.<\/p>\n<p>Feast your eyes upon figure 1 below.<\/p>\n<p><img fetchpriority=\"high\" decoding=\"async\" class=\"alignnone wp-image-1696\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Resistors-in-Parallel-Same-Value.png\" alt=\"Parallel Resistors Same Value\" width=\"354\" height=\"277\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Resistors-in-Parallel-Same-Value.png 512w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Resistors-in-Parallel-Same-Value-150x117.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Resistors-in-Parallel-Same-Value-300x235.png 300w\" sizes=\"(max-width: 354px) 100vw, 354px\" \/><\/p>\n<p><strong><em>Figure 1: 3 parallel resistors of the same value.<\/em><\/strong><\/p>\n<p>You may recognize this picture from the first post in this series (with slight modifications). Assuming all three resistors are the same value, we can write:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1702 aligncenter\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/R-3.jpg\" alt=\"R\/3\" width=\"75\" height=\"42\" \/><\/p>\n<p>In general, we can write:<\/p>\n<figure id=\"attachment_1703\" aria-describedby=\"caption-attachment-1703\" style=\"width: 81px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1703 size-full\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Parallel-Resistance.jpg\" alt=\"Parallel-Resistance resistors of all same value\" width=\"81\" height=\"42\" \/><figcaption id=\"caption-attachment-1703\" class=\"wp-caption-text\">(eq. 2)<\/figcaption><\/figure>\n<p>Where <em>n<\/em> is the total number of resistors in the network.<\/p>\n<p><strong>Remember, equation 2 only works when all the resistors are the same value and equation 1 only works for <em>two<\/em> parallel resistors (even if they\u2019re not equal).<\/strong><\/p>\n<p>Enough about combining parallel resistors &#8212; Let get to the good stuff.<\/p>\n<h2><strong>Voltage Division: Ninja Circuit Analysis Trick 1<br \/>\n<\/strong><\/h2>\n<p>The need to find the voltage drop across a resistor is a need you\u2019ll often encounter in the world of electronics.<\/p>\n<p>Consider the circuit in figure 2.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1704\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Connected-Resistors-1.png\" alt=\"Series Resistors\" width=\"308\" height=\"164\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Connected-Resistors-1.png 308w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Connected-Resistors-1-150x80.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Connected-Resistors-1-300x160.png 300w\" sizes=\"(max-width: 308px) 100vw, 308px\" \/><\/p>\n<p><strong><em>Figure 2: Look familiar? This is the series circuit from the last post in this series.<\/em><\/strong><\/p>\n<p>If we want to calculate the voltage drop across each resistor, we need to perform several steps:<\/p>\n<ul>\n<li>Calculate total resistance by adding up all the resistors.<\/li>\n<li>Once we know the total resistance and the source voltage, we can calculate the current.<\/li>\n<li>Using the current, we can then calculate the voltage drop across each resistor using ohm\u2019s Law.<\/li>\n<\/ul>\n<p>Now this is all fine and good, but there is an easier way with less steps to do this.<\/p>\n<p>Voltage division allows you to calculate the voltage drop across any resistor without having to first calculate the current. This allows you to bypass the last two steps given above. That makes your life as an electronics enthusiast easier.<\/p>\n<h3 style=\"text-align: center;\">Become the Maker you were born to be. Try <a href=\"https:\/\/learnarduinonow.com\">Arduino Academy<\/a> for FREE!<\/h3>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter  wp-image-4238\" src=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2.png\" alt=\"\" width=\"386\" height=\"147\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2.png 828w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2-300x114.png 300w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2-150x57.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2-768x292.png 768w\" sizes=\"(max-width: 386px) 100vw, 386px\" \/><\/p>\n<p>Deriving the formula for voltage division from Ohm\u2019s Law isn\u2019t that difficult, but I\u2019ll leave it up to you to do so if you choose. Equation 3 gives the formula for voltage division:<\/p>\n<figure id=\"attachment_1705\" aria-describedby=\"caption-attachment-1705\" style=\"width: 103px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1705 size-full\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division.jpg\" alt=\"Voltage Division Formula\" width=\"103\" height=\"46\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division.jpg 103w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division-100x46.jpg 100w\" sizes=\"(max-width: 103px) 100vw, 103px\" \/><figcaption id=\"caption-attachment-1705\" class=\"wp-caption-text\">(eq. 3)<\/figcaption><\/figure>\n<p>Where V<sub>R<\/sub> is the voltage drop across the resistor of interest<\/p>\n<p>V<sub>S<\/sub> is the source voltage<\/p>\n<p>R is the value of the resistor of interest<\/p>\n<p>And R<sub>total<\/sub> is the total series resistance<\/p>\n<p>Although voltage division can give you the voltage drop across a particular resistor in a series circuit with any number of resistances greater than one, you\u2019ll often apply it to a circuit (or part of a circuit) with two resistors. In this case equation 3 becomes:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1706 aligncenter\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division_2.jpg\" alt=\"Voltage Division\" width=\"122\" height=\"48\" \/><\/p>\n<p>Similarly:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1707 aligncenter\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division_3.jpg\" alt=\"\" width=\"121\" height=\"45\" \/><\/p>\n<p>Note that we can also use voltage division to find the drop across two or more series resistors. For example, if we want to know the voltage drop across the series combination of R<sub>2<\/sub> and R<sub>3<\/sub> from figure 2, we can use voltage division to find that simply by treating R<sub>2<\/sub> and R<sub>3<\/sub> as one, larger resistor.<\/p>\n<p>Let\u2019s do a couple examples using voltage division.<\/p>\n<p><strong><u>Ex. 1<\/u><\/strong><\/p>\n<p>Find the voltage drop across R<sub>3<\/sub> in the figure below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1709\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division.png\" alt=\"Voltage-Division Series Circuit\" width=\"333\" height=\"187\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division.png 333w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division-150x84.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Division-300x168.png 300w\" sizes=\"(max-width: 333px) 100vw, 333px\" \/><\/p>\n<p><strong><u>Answer:<\/u><\/strong><\/p>\n<p>All we need to do is apply equation 3:<\/p>\n<p>V<sub>R3<\/sub> = 12[(7.5k)\/(5k + 10k + 7.5k)]<\/p>\n<p>Doing the arithmetic, we get:<\/p>\n<p>V<sub>R3 <\/sub>= 4 V<\/p>\n<p>The second example is a circuit (or more likely part of a circuit) that shows up a lot in the world of electronics.<\/p>\n<p><strong><u>Ex. 2:<\/u><\/strong><\/p>\n<p>The circuit in the figure below is a voltage divider. A voltage divider is a simple circuit which turns a large voltage into a smaller one. They are very common and are one of the most fundamental circuits in electronics.<\/p>\n<p>If V<sub>in<\/sub> is 5 V, what is V<sub>out<\/sub>?<\/p>\n<p><strong><u><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1710\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Divider.png\" alt=\"Voltage Divider\" width=\"171\" height=\"326\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Divider.png 200w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Divider-79x150.png 79w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Divider-157x300.png 157w\" sizes=\"(max-width: 171px) 100vw, 171px\" \/><\/u><\/strong><\/p>\n<p><strong><u>Answer:<\/u><\/strong><\/p>\n<p>This problem is a piece of cake, yet a very important and practical depiction of reality.<\/p>\n<p>V<sub>out<\/sub> = V<sub>in<\/sub>(R<sub>2<\/sub>\/(R<sub>1<\/sub> + R<sub>2<\/sub>)) = 5(10k\/(10k + 20k)) = 5(10k\/30k)<\/p>\n<p>Therefore:<\/p>\n<p>V<sub>out<\/sub> = 1.67 V<\/p>\n<p>Voltage dividers and voltage division in general show us a basic yet important concept in electronics. They show us that the source voltage in a series circuit is divided between the resistors in direct proportion to their resistance. The bigger the resistor compared to the others, the larger the voltage drop.<\/p>\n<p>This, of course, is in accordance with Ohm\u2019s Law which tells us that voltage equals current multiplied by resistance, or V = I x R.<\/p>\n<p>Note that potentiometers (and other things like light dependent resistors) often find use in voltage dividers as they\u2019re being adjustable makes the divider more versatile and easy to change if need be.<\/p>\n<p>Some that are new to electronics may be wondering why we should even bother with voltage dividers.<\/p>\n<p>Why not just start with the voltage that we need in the first place?<\/p>\n<p>The problem with this is that many circuits need more than one voltage and building a power supply to deliver each one is impractical and expensive.<\/p>\n<h3><strong>The Ninja Gets Loaded: A Voltage Divider Caveat<br \/>\n<\/strong><\/h3>\n<p>Consider the voltage divider from example 2 above. By itself, it can do little more than act as an aid for the study of electronics.<\/p>\n<p>In other words, the reason we build voltage dividers in the first place is so that we can attach something to them, and this is the norm.<\/p>\n<p>So, when we attach a load to a voltage divider, the circuit in example 2 ends up looking like the circuit in figure 3.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1712\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Divider-Loaded.jpg\" alt=\"Voltage Divider Loaded\" width=\"234\" height=\"306\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Divider-Loaded.jpg 286w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Divider-Loaded-115x150.jpg 115w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Voltage-Divider-Loaded-230x300.jpg 230w\" sizes=\"(max-width: 234px) 100vw, 234px\" \/><\/p>\n<p><strong><em>Figure 3: voltage divider with load attached.<\/em><\/strong><\/p>\n<p>The load in figure 3 could be anything, it doesn\u2019t matter \u2013 loads are modeled as a resistance attached to the circuit. This is because every circuit or piece of equipment offers a certain amount of resistance.<\/p>\n<p>But what happens to V<sub>out<\/sub> when we attach this load?<\/p>\n<p>The short answer is that V<sub>out<\/sub> changes because we\u2019re \u201cloading\u201d the circuit.<\/p>\n<p>How much it changes depends on the load. Let\u2019s take a closer look at this.<\/p>\n<p><strong><u>Ex. 3:<\/u><\/strong><\/p>\n<p>Let\u2019s assume that R<sub>Load<\/sub> in figure 3 is 30 k and that V<sub>in<\/sub> is still 5 V. When the circuit was unloaded, V<sub>out<\/sub> was 1.67 V, what is it now?<\/p>\n<p><strong><u>Answer:<\/u><\/strong><\/p>\n<p>The series circuit we once had has now become a series-parallel circuit with R<sub>2 <\/sub>|| R<sub>Load<\/sub> .<\/p>\n<p>By applying the supersleuth shortcut for two parallel resistors, we see that their combination offers 7,500 \u03a9. Also, since they\u2019re in parallel, the same voltage appears across both.<\/p>\n<p>Now, we use voltage division again to see what this new voltage is:<\/p>\n<p>5((7.5k)\/(7.5k + 20k)) = 1.36 V.<\/p>\n<p>We \u201clost\u201d some of the output voltage.<\/p>\n<p>Let\u2019s do another example, only this time we\u2019ll make R<sub>Load<\/sub> much bigger than 10 k.<\/p>\n<p><strong><u>Ex. 4:<\/u><\/strong><\/p>\n<p>Assume V<sub>in<\/sub> is still 5 V and that R<sub>Load<\/sub> is now 1 M\u03a9. Find V<sub>out<\/sub>.<\/p>\n<p>The parallel combo of 10 k and 1 M is 9900.99 \u03a9 (using our supersleuth formula, of course).<\/p>\n<p>Again, using voltage division, we get a V<sub>out<\/sub> of about 1.66 V.<\/p>\n<p>This value is a lot closer to the value we originally wanted, which was 1.67 V.<\/p>\n<p>This suggests that when using a simple voltage divider such as this, you\u2019ll want to use a load with a high input resistance if possible to avoid loading the circuit.<\/p>\n<p>If not possible, you\u2019ll need to calculate how much current the load will pull and then alter the voltage divider resistor values to offset the loading effect.<\/p>\n<p>Just for kicks, let\u2019s see what happens to the output voltage if the load resistance is 500 \u03a9.<\/p>\n<p><strong><u>Ex 5:<\/u><\/strong><\/p>\n<p>The parallel combination of 500 and 10 k yields about 476 \u03a9.<\/p>\n<p>Using voltage division, we get an output voltage of only about 0.15 V when we thought we\u2019d have 1.67 V!<\/p>\n<p>Now we know why having a load with a high input impedance (or resistance) is often desirable.<\/p>\n<h3><strong>One More Thing About Voltage Dividers<br \/>\n<\/strong><\/h3>\n<p>Don\u2019t use a voltage divider as a voltage supply for any load that requires even a modest amount of power. Not only would this be extremely inefficient, you\u2019re likely to burn up the resistors and destroy it.<\/p>\n<h2><strong>Current Division: Ninja Circuit Analysis Trick 2<br \/>\n<\/strong><\/h2>\n<p>Though not as often used as voltage division, current division is another trick which can come in handy. Unlike voltage division, which works on series resistors, current division works on parallel combinations of resistors.<\/p>\n<p>Current division tells us that the total current in a parallel network is shared by the resistors in inverse proportion to their resistance. You\u2019ll see what I mean in a minute.<\/p>\n<p>It comes in handy when you know the input current but not the input voltage.<\/p>\n<p>Consider the circuit in figure 4.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1713\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Current-Division.png\" alt=\"Currrent Division\" width=\"381\" height=\"192\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Current-Division.png 381w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Current-Division-150x76.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Current-Division-300x151.png 300w\" sizes=\"(max-width: 381px) 100vw, 381px\" \/><\/p>\n<p><strong><em>Figure 4: current division.<\/em><\/strong><\/p>\n<p>It\u2019s easy to see that the two parallel resistors divide the current provided by the source amongst themselves, but how much current goes through each resistor?<\/p>\n<p>The answer is easy if we use the current division formula:<\/p>\n<figure id=\"attachment_1714\" aria-describedby=\"caption-attachment-1714\" style=\"width: 91px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1714 size-full\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Current-Division_1.jpg\" alt=\"Current-Division Formula\" width=\"91\" height=\"43\" \/><figcaption id=\"caption-attachment-1714\" class=\"wp-caption-text\">(eq. 4a)<\/figcaption><\/figure>\n<p>Similarly:<\/p>\n<figure id=\"attachment_1715\" aria-describedby=\"caption-attachment-1715\" style=\"width: 88px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1715 size-full\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Current-Division_2.jpg\" alt=\"Current Division Formula\" width=\"88\" height=\"42\" \/><figcaption id=\"caption-attachment-1715\" class=\"wp-caption-text\">(eq. 4b)<\/figcaption><\/figure>\n<p>Notice the striking similarity to voltage division. But watch out \u2013 the numerator in the fraction is the opposite of that for voltage division!<\/p>\n<p>As is similar to voltage division, current division can extend to a parallel network with any number of resistors great than one.<\/p>\n<p>Do we really need an example problem?<\/p>\n<p>I don\u2019t think so.<\/p>\n<h2><strong>Superposition: Ninja Circuit Analysis Trick 3<br \/>\n<\/strong><\/h2>\n<p>Sorry &#8212; we\u2019re not talking about anyone\u2019s sexual escapades here. What we <em>are <\/em>talking about is perhaps even better than that.<\/p>\n<p>The superposition theorem not only finds use in electronics, but also in physics, economics, and more.<\/p>\n<p>This principle helps us analyze <em>linear<\/em> circuits with more than one independent source by calculating the contribution of each independent source separately. It also provides some insight into in determining the contribution of each source to the parameter under investigation.<\/p>\n<p>What\u2019s an independent source?<\/p>\n<p>Glad you asked.<\/p>\n<p>Unlike a <em>dependent<\/em> source, an <em>independent<\/em> voltage or current source\u2019s output does not depend on some other parameter or source in the circuit.<\/p>\n<p>For example, a battery is an independent voltage source, while something like an amplifier (among other things) is usually considered a dependent source. Right now, we\u2019ll stick with independent sources in our study of circuit analysis.<\/p>\n<p>The procedure for using superposition in circuit analysis is as follows:<\/p>\n<ul>\n<li>Remove all independent voltage sources and replace with a short.<\/li>\n<li>Remove all independent current sources and replace with an open. In other words, just remove the current sources.<\/li>\n<li>You should now have only one remaining source. Calculate the contribution that this source makes to the parameter of interest.<\/li>\n<li>When all independent voltage and current sources have been considered, add the results obtained for each.<\/li>\n<li>If you do run into a dependent source in your travels, these stay active. Do not short or remove them. Grab a bottle of aspirin, you may need it.<\/li>\n<\/ul>\n<p>Before we do a few examples, realize that superposition may result in more work if the circuit you\u2019re analyzing has more than two or three sources. However, it is useful for reducing the complexity of certain circuits.<\/p>\n<p>Another thing to bear in mind is that to use superposition, the circuit or parameter you\u2019re looking for must be linear. So, you couldn\u2019t use superposition to find the individual powers components consume and add them up. Power (P = I<sup>2<\/sup>R) is not linear due to the I<sup>2<\/sup> term. You could, however, use superposition to calculate the voltage or current and then use Ohm\u2019s Law to get the power.<\/p>\n<p>Here\u2019s a few examples to clear things up.<\/p>\n<p><strong><u>Ex. 6:<\/u><\/strong><\/p>\n<p>Let\u2019s revisit that fictitious circuit from the last circuit analysis post. This time, we\u2019ll find the current <em>I <\/em>using superposition.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1655\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/KVL-Practice-Problem.jpg\" alt=\"Circuit analysis with Kirchhoff\u2019s Voltage law\" width=\"298\" height=\"230\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/KVL-Practice-Problem.jpg 672w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/KVL-Practice-Problem-600x463.jpg 600w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/KVL-Practice-Problem-150x116.jpg 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/KVL-Practice-Problem-300x232.jpg 300w\" sizes=\"(max-width: 298px) 100vw, 298px\" \/><\/p>\n<p>We pick one of the two voltage sources, short it out, then use KVL.<\/p>\n<p>Let\u2019s short the 20 V source. The circuit now looks like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1718\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition.jpg\" alt=\"Superposition 1\" width=\"301\" height=\"232\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition.jpg 672w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-600x463.jpg 600w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-150x116.jpg 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-300x232.jpg 300w\" sizes=\"(max-width: 301px) 100vw, 301px\" \/><\/p>\n<p>Using KVL:<\/p>\n<p>50 = -2I \u2013 I\u00a0 <strong>&#8212;-&gt;<\/strong> 50 = -3I<\/p>\n<p>I = -16.67 A<\/p>\n<p>Now, we\u2019ll put the 20 V source back and short the 50 V source.<\/p>\n<p>Using KVL again:<\/p>\n<p>-20 = -2I &#8211; I\u00a0\u00a0 <strong>&#8212;-&gt;<\/strong> -20 = -3I<\/p>\n<p>I = 6.67 A<\/p>\n<p>Now we\u2019ll add the two results together:<\/p>\n<p>6.67 A \u2013 16.37 A = -10 A which is the same answer we got last time.<\/p>\n<p>Example 7 is another interesting circuit analysis problem using superposition.<\/p>\n<p><strong><u>Ex. 7:<\/u><\/strong><\/p>\n<p>Find the current <em>I<\/em> in the figure below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1721\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_2.jpg\" alt=\"Superposition-Circuit-Analysis\" width=\"360\" height=\"267\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_2.jpg 646w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_2-600x445.jpg 600w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_2-150x111.jpg 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_2-300x222.jpg 300w\" sizes=\"(max-width: 360px) 100vw, 360px\" \/><\/p>\n<p>Here we have a voltage source and a current source. Recall that to use superposition, we short voltage sources and remove current sources. We\u2019ll start by considering the voltage source by itself. Removing the current source, we have the circuit below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1722\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_3.jpg\" alt=\"Superposition-Circuit-Analysis 2\" width=\"364\" height=\"270\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_3.jpg 646w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_3-600x445.jpg 600w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_3-150x111.jpg 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_3-300x222.jpg 300w\" sizes=\"(max-width: 364px) 100vw, 364px\" \/><\/p>\n<p>The first thing to note about the figure above is that R3 is just hanging out there in space. Since no current flows through R3, no voltage drops across it and we can safely ignore it. This gives us a series circuit composed of the voltage source, R1, and R2.<\/p>\n<p>Using Ohm\u2019s Law, we can easily find the current the voltage source contributes to the circuit:<\/p>\n<p>I = V\/R = 10\/3k = 3.33 mA<\/p>\n<p>Now, we\u2019ll reinsert the current source and short the voltage source which gives us the circuit below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1724\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_4.jpg\" alt=\"Superposition-Circuit-Analysis 3\" width=\"379\" height=\"281\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_4.jpg 646w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_4-600x445.jpg 600w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_4-150x111.jpg 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Superposition-Circuit-Analysis_4-300x222.jpg 300w\" sizes=\"(max-width: 379px) 100vw, 379px\" \/><\/p>\n<p>This nut\u2019s a bit tougher to crack.<\/p>\n<p>Let\u2019s use KCL and add another current (call it I2) flowing into node 0 (I would\u2019ve used node 1 but I\u2019s and 1\u2019s look a lot alike, ya know). I didn\u2019t add I2 in the picture, so you\u2019ll just have to imagine it flowing into node 0.<\/p>\n<p>Using KCL, we get:<\/p>\n<p>1 mA = (V<sub>0<\/sub> \/ 1k) + (V<sub>0<\/sub> \/ 3k) where V<sub>0 <\/sub>is the voltage at node 0.<\/p>\n<p>If you\u2019re good at algebra, you can solve this by hand. I\u2019m lazy, so I used an online equation solver to get V<sub>0<\/sub>:<\/p>\n<p>V<sub>0<\/sub> = 0.67 V<\/p>\n<p>Now that we know V<sub>0<\/sub>, let\u2019s find I2:<\/p>\n<p>I2 = 0.67\/1k = 0.67 mA<\/p>\n<p>Since the total current in this circuit is 1 mA, we just need to subtract 0.67 mA from 1 mA to get 0.33 mA.<\/p>\n<p>However, notice that <em>I<\/em> is flowing the opposite direction it should be as the 1 mA from the current source should split at the bottom then flow up towards R1 and R2.<\/p>\n<p>No big deal &#8212; we just need to slap a negative sign in front of our answer: I = -0.33 mA.<\/p>\n<p>This is important, because now we need to add up the individual contributions to<em> I<\/em> from each source and if the signs are incorrect, we\u2019ll get a wrong answer.<\/p>\n<p>I<sub>total<\/sub> = 3.33 mA \u2013 0.33 mA = 3mA.<\/p>\n<p>The current from one source slightly opposes the current from the other for a total net of 3 mA.<\/p>\n<h2><strong>Ninja Circuit Analysis Concluded<br \/>\n<\/strong><\/h2>\n<p>This is probably one of the longest posts I\u2019ve written so far, but I wanted to not only be sure my circuit analysis skills were up to snuff, I wanted to select the right examples.<\/p>\n<p>The next post on circuit analysis will either cover Thevenin and Norton or it\u2019ll go over nodal and mesh analysis. I\u2019m not sure yet but maybe that\u2019ll keep you on your toes until next time ?<\/p>\n<p>While you\u2019re waiting leave a comment and tell us:<\/p>\n<p>Which circuit analysis technique is your favorite?<\/p>\n<p>Which don\u2019t you like? Why?<\/p>\n<h2 style=\"text-align: center;\">Become the Maker you were born to be. Try <a href=\"https:\/\/learnarduinonow.com\">Arduino Academy<\/a> for FREE!<\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-4238\" src=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2.png\" alt=\"\" width=\"828\" height=\"315\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2.png 828w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2-300x114.png 300w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2-150x57.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/FB_Cover2-768x292.png 768w\" sizes=\"(max-width: 828px) 100vw, 828px\" \/><\/p>\n<a target=\"_blank\" href=\"https:\/\/www.drpeterscode.com\/index.php\"><img src=\"https:\/\/www.circuitcrush.com\/wp-content\/plugins\/dpabottomofpostpage\/apixel1x1.jpg\" ><\/a><table><\/table>","protected":false},"excerpt":{"rendered":"<p>In the last post concerning circuit analysis (Simple Circuit Analysis Techniques You Should Know) I went over some very basic but important techniques. First, we had a quick over view of resistors in series, parallel, and series-parallel circuits. Then, we talked about Ohms Law. Finally, Kirchhoff\u2019s Laws were introduced. This time, I\u2019m going to show [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":1728,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[107,9],"tags":[108],"class_list":{"0":"post-1695","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-circuit-analysis","8":"category-electronics","9":"tag-circuit-analysis","10":"entry"},"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ninja-Circuit-Analysis-Tricks.jpg","_links":{"self":[{"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/posts\/1695","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/comments?post=1695"}],"version-history":[{"count":2,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/posts\/1695\/revisions"}],"predecessor-version":[{"id":4300,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/posts\/1695\/revisions\/4300"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/media\/1728"}],"wp:attachment":[{"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/media?parent=1695"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/categories?post=1695"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/tags?post=1695"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}