Kirchhoff's Rules Worksheet

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Northeastern University *

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109

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Electrical Engineering

Date

Feb 20, 2024

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pdf

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Worksheet 8 - Kirchhoff's Circuit Rules Note: As we have been suggesting throughout this unit, it will help you to do as much of the worksheet as possible at the boards. ● Remember, Kirchhoff’s rules are used to solve the voltages and currents of a circuit. Each application of Kirchhoff’s Loop Rule and Kirchhoff’s Junction Rule gives you a new equation that can be used to solve a system of equations. ● Kirchhoff’s Loop Rule states that for any closed loop of a circuit, the total voltage change must equal zero. In other words, if you designate one point of the circuit to be at V=0, then if you make a closed loop around the circuit and total up all of those components, returning to the same spot as before will give you V=0. Think about this like going on a hike through mountainous terrain. If you start at your car and go on a hike, you will go through many elevation gains and elevation drops. However, if you return back to your car at the end, the total net elevation difference will be zero. ● Kirchhoff’s Junction Rule states that the amount of current that goes into a junction must be equal to the amount of current that leaves that junction. This is an application of conservation of charge: all electrons that come into a junction must also leave that junction. ● To solve a system of equations, you must have as many equations as you do unknown variables. For example, in a circuit in which you do not know three currents, you must construct three unique equations in order to solve the system of equations. ● When looking at a circuit, you may not know which direction the current is going before solving the circuit. This means that you do not know whether or not a voltage drop or a voltage gain will occur across a resistor. This is ok! At the end of the problem if you have labeled a current going the wrong way, the answer you will get will be negative; this is a sign that you should indicate your current going the other way. Page 1

Part 1: Series-Parallel combination circuits Consider the following circuit as our starting circuit. The purpose of this exercise is to help you understand how to break up a series-parallel combination circuit to find the current and potential difference across each resistor. To do this, we will first combine different resistors and then work our way back up. Think about how these four resistors are connected to each other. Which ones are in parallel and which ones are in series? Instructions - follow steps 1-3 to contract the circuit to one equivalent resistor. Then go back up (steps 4-7) to label every resistor in every diagram with its current and voltage. Step 1: Combine two resistors together first. Which two should you choose? Redraw the circuit with three resistors and label the resistors with their resistance. (To redraw, use the Google drawing below. Simply erase one resistor by putting a white box over it to have only three resistors.) Don’t worry about the current and voltage labels until step 6. Resistance calculation: Page 2

Step 2: Now, combine two more resistors together, and redraw the circuit with only two resistors. Again, use white boxes to cover up resistors you have merged and don’t need. You may need to then add a line to represent a wire. Use the voltage/current labels in step 5. Resistance calculation: Page 3

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Step 3 : Finally, combine the last two and redraw the circuit. You should only have one equivalent resistor left. Step 4 : Now we are going to come back up. First, determine the total current in the circuit. Hint: the total potential difference is the voltage of the battery, and the total resistance should be the equivalent resistance you found in Step 3. I total = _____ A General rules : (1) When you expand a resistor back to two resistors in series, the currents are all the same and the two voltages add to the voltage across the one equivalent resistor. Find the voltages using Ohm’s law. (2) When you expand a resistor back to two resistors in parallel, the voltages are all the same and the two currents add to the current through the one equivalent resistor. Find the currents using Ohm’s law. Step 5 : Now that we have calculated the total current of the circuit, go back to the two resistors in Step 2 and find the current through each. Think about how they are connected. What is the same for both and what is different? Find the current and potential difference through each resistor. Page 4

Step 6 : We will now further expand the circuit to three resistors found in Step 1. Find the current and potential difference through each resistor. Think about how the new resistor is connected to the two resistors in the last step. What is the same for the new resistor and what is different? Step 7 (LAST STEP!) : Finally, go back to our starting circuit. Find the current through and potential difference across each resistor. Question 1.1 : Let’s say you add a new resistor in parallel with the 20 ohm resistor. The total… A. Current would increase B. Battery voltage would increase C. Current would decrease D. Battery voltage would decrease Page 5

Part 2: Another approach to solving this circuit 1. Because the 40 Ω resistor is connected to all the other resistors in parallel, we will analyze it first. What is the potential difference across and the current through the 40 Ω resistor? 2. Now let’s consider the two resistors in parallel (20Ω and 60Ω), what is the ratio of the current through the two resistors? I 20 : I 60 = ?:? 3. Because the 20 Ω and 60 Ω are connected to the 5Ω resistor in series, the two components (20Ω+60Ω vs. 5Ω) will have the same current. We could solve for current using the Loop Rule. We could also find the combined resistance of the resistors in parallel, then find the equivalent resistance of all three resistors on the right side of the circuit. And lastly, we could find the combined current of the three resistors and use the ratio we found in item 2 to find the current through Page 6

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each resistor. Give it a try and compare your answer to the other approach. You should have the same values. The main takeaway of this exercise is that there are different ways you could break down a complicated series-parallel combination circuit into smaller components. Always start by thinking about how the resistors are connected to each other, then apply rules and equations for either in series or in parallel to those smaller components to solve the question. Part 3: Solving a more complex circuit In the above circuit schematic, R 1 = 4 Ω , R 2 = 4 Ω , R 3 = 8 Ω , R 4 = 2 Ω , R 5 = 4 Ω , R 6 = 2 Ω , R 7 = 1 Ω , and the battery supplies 12V. Question 3.1 : Calculate the equivalent resistance of the circuit and the total current supplied by the battery. Page 7

Question 3.2 : First, label all of the currents in the below diagram. Indicate which of the currents you labeled are equivalent to each other due to Kirchhoff’s Junction Rule. Then, write two unique Junction Rules for this diagram. Next, write Kirchhoff’s Loop Rule for the two indicated Loops. The larger loop is Loop 1 and the smaller loop is Loop 2. Question 3.3 : The removal of which of the following resistors from the circuit would result in an increase in the current supplied by the battery? There may be multiple correct answers. [ ] R 1 [ ] R 2 [ ] R 3 [ ] R 4 [ ] R 5 [ ] R 6 [ ] R 7 Page 8