{"id":1640,"date":"2018-05-04T15:49:18","date_gmt":"2018-05-04T19:49:18","guid":{"rendered":"http:\/\/www.circuitcrush.com\/?p=1640"},"modified":"2021-06-30T17:33:07","modified_gmt":"2021-06-30T21:33:07","slug":"circuit-analysis","status":"publish","type":"post","link":"https:\/\/www.circuitcrush.com\/circuit-analysis\/","title":{"rendered":"Simple Circuit Analysis Techniques You Should Know"},"content":{"rendered":"<p>The simple circuit analysis techniques we\u2019ll cover in this post are basic, yet important.<\/p>\n<p>Anyone can copy a circuit or project from the Internet or a book. But if you want to be able to design your own circuits, understand how things work, and be able to competently troubleshoot when things go wrong (and they will!) an understanding of the basics of electronics and electricity is crucial.<\/p>\n<p><!--more--><\/p>\n<p>For example, consider the mega-popular Raspberry Pi. By itself, it can do little besides maybe teach you something about Linux and programming. But if you really want to get the most out of it (or any other micro or platform) you\u2019ll need to connect other things to it.<\/p>\n<p>Not only can learning the basics shed some light on how your platform or board of choice works (and maybe keep you from frying it), it can shed light on how the circuits you connect to it work.<\/p>\n<p>And there are many instances where you may be able to skip the microcontroller all together, using discrete components or ICs to accomplish your task. This can reduce the complexity, cost, and time required to bring your creation to life. Save your $25 Arduino for a more complex project, such as home automation. Of course, one needs to understand at least the basics of electronics to do this.<\/p>\n<h1><strong>Circuit Analysis 101<br \/>\n<\/strong><\/h1>\n<p>This article covers basic circuit analysis, so in that spirit we\u2019ll only be talking about DC circuits. Perhaps we\u2019ll cover AC circuits in another post.<\/p>\n<p>Also, we\u2019ll forget about things like capacitors and inductors for now and focus on circuits containing only resistors in an effort to keep things simple. Again, these types of circuits will probably make an appearance in another post.<\/p>\n<p>Note, however, that most of the techniques I present here will work on circuits containing capacitors and inductors as well as resistors. The only exception would be the way capacitors add up in series or parallel.<\/p>\n<p>If you\u2019ve been dabbling in electronics for more than 5 minutes, you likely know how to combine resistors. However, some of you may be brand-spanking new, so we\u2019ll quickly review this. I will assume, however, that you know the difference between series and parallel circuits and how current and voltage behave in each.<\/p>\n<p>Resistors in series simply add up. For example, the total resistance in the circuit from figure 1 is simply:<\/p>\n<p>R<sub>total <\/sub>= R1 + R2 + R3 + R4<\/p>\n<p> <\/p>\n<p><strong><em><img fetchpriority=\"high\" decoding=\"async\" class=\"alignnone size-full wp-image-1641\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Connected-Resistors.png\" alt=\"series connected resistors\" width=\"308\" height=\"164\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Connected-Resistors.png 308w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Connected-Resistors-150x80.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Connected-Resistors-300x160.png 300w\" sizes=\"(max-width: 308px) 100vw, 308px\" \/><\/em><\/strong><\/p>\n<p><strong><em>Figur<\/em><\/strong><strong><em>e 1: series connected resistors.<br \/>\n<\/em><\/strong><\/p>\n<p>The total resistance in a parallel circuit is always lower than the lowest-value resistor. So, if R1 were the smallest resistor in the circuit in figure 2, the total resistance would be less than R1.<\/p>\n<p>To get the total resistance in a parallel circuit, we add up the reciprocal of all the resistances and then take the reciprocal of this sum.<\/p>\n<p><img decoding=\"async\" class=\"alignnone wp-image-1642\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Resistors-in-Parallel.png\" alt=\"Parallel resistors\" width=\"244\" height=\"191\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Resistors-in-Parallel.png 512w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Resistors-in-Parallel-150x117.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Resistors-in-Parallel-300x235.png 300w\" sizes=\"(max-width: 244px) 100vw, 244px\" \/><\/p>\n<p><strong><em>Figure 2: resistors in parallel.<\/em><\/strong><\/p>\n<p>Consider the circuit in figure 2. The total resistance would be:<\/p>\n<p><img decoding=\"async\" class=\"alignnone size-full wp-image-1643\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Formula-Parallel-Resistors.jpg\" alt=\"Formula-Parallel-Resistors\" width=\"142\" height=\"61\" \/><\/p>\n<p>As we\u2019ll find out in the next post, there are shortcuts that work on <em>some<\/em> parallel circuits.<\/p>\n<p>Many circuits you\u2019ll encounter in real life are series-parallel circuits which are a combination of the two types we just spoke of. An example of a series-parallel circuit is given in figure 3.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1645\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Parallel-Circuit.png\" alt=\"series-parallel circuit\" width=\"308\" height=\"163\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Parallel-Circuit.png 308w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Parallel-Circuit-150x79.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Series-Parallel-Circuit-300x159.png 300w\" sizes=\"(max-width: 308px) 100vw, 308px\" \/><\/p>\n<p><strong><em>Figure 3: a series-parallel circuit.<\/em><\/strong><\/p>\n<p>In figure 3, we see that R1 and R2 are in parallel while R3 and R4 are in series. The series combination of R3 and R4 is in parallel with R1 and R2.<\/p>\n<p>To combine resistors in this case, we would work right to left and simply determine which resistor combinations are in series and which are in parallel and use the appropriate formulas to get an equivalent total resistance.<\/p>\n<h2><strong>Ohm\u2019s Law<br \/>\n<\/strong><\/h2>\n<p>Ohm\u2019s Law states that <em>the voltage across a resistor is directly proportional to the current flowing through that resistor<\/em>. Another way to word it is <em>current flow in a circuit is directly proportional to source voltage and inversely proportional to the resistance of the circuit<\/em>. It shows a linear relationship between voltage and current.<\/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=\"389\" height=\"148\" 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: 389px) 100vw, 389px\" \/><\/p>\n<p>This is one of the easiest to remember and most basic rules in electronics. There are two nifty charts associated with Ohm\u2019s Law \u2013 remembering them will make your life easier.<\/p>\n<p> <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1646\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law.png\" alt=\"Ohm's Law\" width=\"262\" height=\"200\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law.png 604w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law-600x457.png 600w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law-150x114.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law-300x228.png 300w\" sizes=\"(max-width: 262px) 100vw, 262px\" \/><\/p>\n<p><strong><em>Figure 4: Ohm\u2019s Law.<\/em><\/strong><\/p>\n<p>The first one is given in figure 4 where V is for voltage, I is for current, and R is resistance. Note that sometimes you may see voltage represented as E rather than V.<\/p>\n<p>This chart is easy to remember and tells us how to solve for any of the three values as long as two are known.<\/p>\n<p>Want to know the voltage? Multiply I x R. Need the current? Divide V by R: V\/R.<\/p>\n<p>Here\u2019s a handy trick, take what ever parameter you need to solve for (i.e. voltage, current, resistance) and cover that with your thumb. That will give you the formula you need.<\/p>\n<p>For example, if I want to know the voltage, I\u2019d imagine myself covering the V with my thumb leaving me to multiply I x R. Similarly, if I wanted current, I\u2019d cover the \u201cI\u201d and divide V by R.<\/p>\n<p>Another easy to remember version of Ohm\u2019s law concerns power in a circuit and is given below in figure 5.<\/p>\n<p> <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1647\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law-Power.png\" alt=\"Ohm's law w power\" width=\"260\" height=\"206\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law-Power.png 513w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law-Power-150x119.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Ohms-Law-Power-300x239.png 300w\" sizes=\"(max-width: 260px) 100vw, 260px\" \/><\/p>\n<p><strong><em>Figure 5: Ohm\u2019s Law with power.<\/em><\/strong><\/p>\n<p>Notice figure 5 uses E for voltage instead of V. You can use either as your mnemonic, but I find PIE easier to remember than PIV.<\/p>\n<p>We use this triangle the same way we use the other one. For example, if I need to know current and know the power and voltage, I simply divide the power by the voltage: I = P\/E.<\/p>\n<p>Using these two figures, one can come up with many different formulas for Ohm\u2019s Law depending on what parameters we know or don\u2019t know about a circuit. The complete version is shown in figure 6.<\/p>\n<p> <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1648\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/ohms-law-complete.gif\" alt=\"Ohm's lAw all formulas\" width=\"405\" height=\"369\" \/><\/p>\n<p><strong><em>Figure 6: all the formulas that one can derive from Ohm\u2019s Law.<\/em><\/strong><\/p>\n<p>You could memorize figure 6 (or post it at your bench), but any of the formulas above can be easily derived from figures 4 and 5, which are much easier to remember.<\/p>\n<p>Let\u2019s take P = I<sup>2<\/sup> x R as an example.<\/p>\n<p>From figure 5 we can easily see that P = I x E (or P = I x V; take your pick).<\/p>\n<p>But, figure 4 tells us that E = I x R.<\/p>\n<p>Substituting that into the first equation we get: P = I x (I x R) which gives P = I<sup>2<\/sup> x R. Easy peasy.<\/p>\n<p>According to Ohm\u2019s Law, power, voltage, current, and resistance each have three ways one can solve for them.<\/p>\n<p>The important thing to remember is that regardless of which of the three formulas you choose to use for solving for the parameter you need (i.e. voltage) is that all three formulas should produce the same answer. So, it does not matter which one you pick as long as you have the other two values you need to use for whatever of the three formulas you choose.<\/p>\n<p>So, if I need to find the power and I know the current and resistance, I obviously need to use P = I<sup>2<\/sup> x R since I do not know the voltage.<\/p>\n<p>You can use Ohm\u2019s Law to analyze a variety of simple circuits. Let\u2019s do a few examples.<\/p>\n<p><u>Ex 1:<\/u> How much current flows through the resistor in figure 7?<\/p>\n<p> <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1649\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Simple-Circuit.png\" alt=\"Ohm's Law circuit analysis\" width=\"308\" height=\"160\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Simple-Circuit.png 308w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Simple-Circuit-150x78.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Simple-Circuit-300x156.png 300w\" sizes=\"(max-width: 308px) 100vw, 308px\" \/><\/p>\n<p><strong><em>Figure 7<\/em><\/strong><\/p>\n<p>To get the answer, we remember (using our handy mnemonic from figure 4) that the total current in a circuit like this equals the voltage divided by the resistance:<\/p>\n<p>I = V\/R = 9\/3,000 = 3 mA.<\/p>\n<p>This seems like a silly example, but real batteries exhibit an internal resistance (though it\u2019s usually less than 3 k\u03a9!) and this sometimes must be factored into your design.<\/p>\n<p>Let\u2019s do another, more practical example you\u2019re almost sure to see.<\/p>\n<p><u>Ex 2:<\/u> LEDs don\u2019t like too much current and should always have a resistor in series with them. How much current flows though the LED in figure 8? Assume a 5 V source, a 0.7 V drop across the LED, and that R = 120 \u03a9.<\/p>\n<p> <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1651\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/LED-and-Resistor.jpg\" alt=\"Resistor and LED\" width=\"320\" height=\"225\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/LED-and-Resistor.jpg 961w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/LED-and-Resistor-600x421.jpg 600w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/LED-and-Resistor-150x105.jpg 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/LED-and-Resistor-300x211.jpg 300w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/LED-and-Resistor-768x539.jpg 768w\" sizes=\"(max-width: 320px) 100vw, 320px\" \/><\/p>\n<p><strong><em>Figure 8<\/em><\/strong><\/p>\n<p>Since this is a series circuit the same amount of current will flow through the resistor and the LED. Because 0.7 V drops across the LED, we know that 4.3 V drops across the resistor. Using this information, we get the current through the LED:<\/p>\n<p>I = V\/R = 4.3\/120 = 35.83 mA.<\/p>\n<p>Note that \u2013 depending on the LED you pick \u2013 you may need to use a larger resistor to keep the current within the LED\u2019s limitations. Ohm\u2019s Law to the rescue again\u2026<\/p>\n<p><u>Bonus question (feel free to answer in comments):<\/u> how much power does the resistor dissipate? Hint: eat your PIE\u2026<\/p>\n<h1><strong>Circuit Analysis and Kirchhoff\u2019s Laws<br \/>\n<\/strong><\/h1>\n<p>Mr. Georg Simon Ohm is not the only one who came up with his own circuit analysis rules. Another scientist \u2013 Gustav Kirchhoff \u2013 also spent some time observing electricity and came up with his own rules.<\/p>\n<p>Ohm\u2019s Law is great but quickly becomes unwieldy when analyzing all but the simplest circuits. That\u2019s where Kirchhoff\u2019s Laws come in. The two together form a powerful suite of tools for analyzing more complex circuits.<\/p>\n<p>Mr. Kirchhoff is famous for two laws: Kirchhoff\u2019s Current Law (KCL) and Kirchhoff\u2019s Voltage Law (KVL).<\/p>\n<h2><strong>Kirchhoff\u2019s Current Law or KCL <\/strong><\/h2>\n<p>Kirchhoff\u2019s Current Law states that <em>the algebraic sum of currents entering a node is 0<\/em>.<\/p>\n<p>To put it another, perhaps easier to understand way, it says that <em>the sum of all currents entering a node is equal to the sum of all currents leaving a node<\/em>. After all, electrons don\u2019t simply evaporate and disappear!<\/p>\n<p>Figure 9 illustrates this.<\/p>\n<p> <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1652\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Kirchoffs-Current-Law-KCL.png\" alt=\"Kirchhoff\u2019s Current Law\" width=\"358\" height=\"195\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Kirchoffs-Current-Law-KCL.png 358w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Kirchoffs-Current-Law-KCL-150x82.png 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Kirchoffs-Current-Law-KCL-300x163.png 300w\" sizes=\"(max-width: 358px) 100vw, 358px\" \/><\/p>\n<p><strong><em>Figure 9: Kirchhoff\u2019s Current Law.<\/em><\/strong><\/p>\n<p>Here we see I1 and I2 entering node 3, while I3 is leaving the node.<\/p>\n<p>It\u2019s easy to see that I1 + I2 = I3 or we can write it as I1 +I2 \u2013 I3 = 0, which shows us that the algebraic sum does indeed equal 0.<\/p>\n<p>Let\u2019s do an example problem using KCL.<\/p>\n<p><u>Ex 3:<\/u> Find the battery current I<sub>B<\/sub> in figure 10.<\/p>\n<p> <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1653\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/KCL-Practice-Problem.jpg\" alt=\"Circuit Analysis with Kirchhoff\u2019s Current Law\" width=\"373\" height=\"201\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/KCL-Practice-Problem.jpg 579w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/KCL-Practice-Problem-150x81.jpg 150w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/KCL-Practice-Problem-300x162.jpg 300w\" sizes=\"(max-width: 373px) 100vw, 373px\" \/><\/p>\n<p><strong><em>Figure 10<\/em><\/strong><\/p>\n<p>Using Kirchhoff\u2019s Current Law at node 1 gives us:<\/p>\n<p>I<sub>B<\/sub> = I<sub>1<\/sub> + I<sub>2<\/sub><\/p>\n<p>Ohm\u2019s Law tells us that:<\/p>\n<p>I<sub>1<\/sub> = V\/R = 20\/5 = 4 A<\/p>\n<p>And, since the resistors are in parallel, 20 V shows at both:<\/p>\n<p>I<sub>2<\/sub> = 20\/15 = 1.33 A<\/p>\n<p>Therefore:<\/p>\n<p>I<sub>B<\/sub> = 4 A + 1.33 A = 5.33 A<\/p>\n<h2><strong>Kirchhoff\u2019s Voltage Law or KVL<br \/>\n<\/strong><\/h2>\n<p>Kirchhoff\u2019s Voltage Law states that <em>the algebraic sum of all voltages around a closed path or loop is 0<\/em>.<\/p>\n<p>To put it another way, <em>the sum of all voltage rises around a closed loop equals the sum of the voltage drops around the loop<\/em>.<\/p>\n<p>We can also phrase it as <em>the sum of the voltage drops is a series circuit is equal to the total voltage applied<\/em>.<\/p>\n<p>Figure 11 illustrates this.<\/p>\n<p> <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1654\" src=\"http:\/\/www.circuitcrush.com\/wp-content\/uploads\/Kirchoffs-Voltage-Law-KVL.jpg\" alt=\"Kirchhoff\u2019s Voltage law\" width=\"311\" height=\"230\" srcset=\"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Kirchoffs-Voltage-Law-KVL.jpg 253w, https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Kirchoffs-Voltage-Law-KVL-150x111.jpg 150w\" sizes=\"(max-width: 311px) 100vw, 311px\" \/><\/p>\n<p><strong><em>Figure 11: Kirchhoff\u2019s Voltage law.<\/em><\/strong><\/p>\n<p>Because figure 11 is a series circuit it is a closed loop and the same current I flows through all the elements.<\/p>\n<p>Using KVL, we can write: V<sub>s<\/sub> \u2013 VR<sub>1<\/sub> \u2013 VR<sub>2<\/sub> = 0 or we can rearrange and write: V<sub>s<\/sub> = VR<sub>1<\/sub> + VR<sub>2<\/sub><\/p>\n<p>Both say the same thing.<\/p>\n<p>Let\u2019s do a quick practice problem.<\/p>\n<p><u>Ex 4:<\/u> How much current flows through the circuit in figure 12?<\/p>\n<p> <\/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=\"315\" height=\"243\" 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: 315px) 100vw, 315px\" \/><\/p>\n<p><strong><em>Figure 12<\/em><\/strong><\/p>\n<p>Those who have taken engineering courses may recognize this type of circuit as being a fictitious, engineering-esqe circuit which exists only on paper to mess with your mind. You can always tell these because they usually have at least two off-the-wall sources and are either impossible to build as shown or never encountered in real life.<\/p>\n<p>Anyway, the direction of current flow is already given as clockwise \u2013 this matters. The polarity of the voltages also matter. The polarity of the resistors in this fictitious circuit are given to you. Normally, we assume voltage to drop across a resistor. For figure 12 we\u2019ll treat voltage rises as positive and voltage drops as negative as we follow the path of the current.<\/p>\n<p>If the current\u2019s path is not given in a situation like this (which is likely), I like to assume it travels clockwise, but you can pick counter-clockwise, too. You <em>must<\/em> be <em>consistent<\/em> with your method of choice or you won\u2019t get the right answer. For more about this, see <a href=\"http:\/\/www.circuitcrush.com\/conventional-current-vs-electron-flow\/\" target=\"_blank\" rel=\"noopener\">Conventional Current vs. Electron Flow: Which is Correct?<\/a><\/p>\n<p>Using KVL, we can write:<\/p>\n<p>20 \u2013 V<sub>1<\/sub> \u2013 50 \u2013 V<sub>2<\/sub> = 0<\/p>\n<p>Ohm\u2019s Law tells us that:<\/p>\n<p>V<sub>1<\/sub> = I x R = I x 1 = I<\/p>\n<p>And<\/p>\n<p>V<sub>2<\/sub> = I x 2<\/p>\n<p>So, 20 \u2013 I \u2013 50 \u2013 2I = 0<\/p>\n<p>Rearranging algebraically, we get:<\/p>\n<p>-I \u2013 2I = 50 \u2013 20 which after simplifying is -3I = 30<\/p>\n<p>Solving for I, we get -10 A.<\/p>\n<p>The negative sign tells us that the current is actually flowing the opposite direction we assumed it was, since it is impossible to have a negative number of electrons. If we were to draw the arrow representing current the opposite way and redo our analysis correctly, our answer would be 10 A. Notice that the absolute value (or the amount) of the current that flows is the same either way.<\/p>\n<p><u>Bonus question:<\/u> given resistors with the right power rating and voltage sources are available, is it possible to build this circuit with the exact parameters shown in the figure? Why or why not?<\/p>\n<h1><strong>More Circuit Analysis Tricks to Come<br \/>\n<\/strong><\/h1>\n<p>As usual, this post has grown into a 2,000+ word monster, and there\u2019s a lot of circuit analysis tricks we haven\u2019t covered.<\/p>\n<p>In part 2 of this series we\u2019ll look at some other tricks and applications to extend the things we learned here.<\/p>\n<p>We\u2019ll look at handy tricks like voltage division, current division, and superposition. Then (perhaps in part 3) we\u2019ll get into some more advanced methods like nodal and mesh analysis and also Thevenin and Norton.<\/p>\n<p>Truthfully, one can easily devote a whole post to each of these advanced techniques. We\u2019ll see about that when the time to discuss each comes.<\/p>\n<p>Meanwhile, comment and tell us about your favorite circuit analysis trick or technique.<\/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>The simple circuit analysis techniques we\u2019ll cover in this post are basic, yet important. Anyone can copy a circuit or project from the Internet or a book. But if you want to be able to design your own circuits, understand how things work, and be able to competently troubleshoot when things go wrong (and they […]<\/p>\n","protected":false},"author":1,"featured_media":1657,"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],"tags":[108,110,111,109],"class_list":{"0":"post-1640","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-circuit-analysis","8":"tag-circuit-analysis","9":"tag-kcl","10":"tag-kvl","11":"tag-ohms-law","12":"entry"},"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/www.circuitcrush.com\/wp-content\/uploads\/Circuit-Analysis-2.jpg","_links":{"self":[{"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/posts\/1640","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=1640"}],"version-history":[{"count":1,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/posts\/1640\/revisions"}],"predecessor-version":[{"id":4302,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/posts\/1640\/revisions\/4302"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/media\/1657"}],"wp:attachment":[{"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/media?parent=1640"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/categories?post=1640"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.circuitcrush.com\/wp-json\/wp\/v2\/tags?post=1640"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}