PHYS 202 1st Edition Lecture 11Outline of Last Lecture I. Kirchhoff’s Node and Loop RulesOutline of Current Lecture II. How to Solve Problems with Kirchhoff’s RulesIII. Example ProblemsCurrent Lecture Solving problems Using Kirchhoff’s Rules:Node Rule: When you begin solving a problem using Kirchoff’s rules, the first thing you have to do is arbitrarily assign directions to the currents in the system.Example, if you have the following circuit layout: You can assign it currents that run like this; I3 I1 I2 I1 I2These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Or that run like this; I3 I1 I2 I1 I2Because the current will ALWAYS flow from an area of lower potential to an area of higher potential, once we solve for our currents, we should get a positive value for all I’s. If the value we get for I is positive, then we picked the correct direction in which the current is flowing. If we get a negative value for I, then the current is flowing in a different direction than we originally thought. To begin solving for I, remember Kirchoff’s node rule. The current flowing into a node is always equal to the current flowing out of the node. This means that you can always set two of your currents equal to a third.Example:Assume you have assigned your circuit the following currents and node labels; A B C I3 I1 I2 D I1 E I2 FNotice that only node B and node E are connected to all three currents. Therefore, you will select one of these nodes and make your first current solving equation based on how the currents are flowing in and out of the node. If we select node B, we notice that I2 is flowing into the node, and I1 and I3 are flowing out of the node. If we select node E, then we notice that I1 and I3 are flowing into the node and I2 is flowing out of the node. Therefore, it doesn’t matter which node we start with, we begin with the equation I2=I1+I3.Now that we have our starting point, we can start applying Kirchhoff’s loop rule to solve for our currents as they move around the circuit. To do this, we’ll divide the circuit into two loops. In the case of the circuit above, these will be loop ADEBA and loop BEFCB. To get the equation for each loop, we follow the current around the loop and notice what it encounters between nodes. When the current moves through a battery, we add +E or –E to our equation. If the current moves from the positive plate to the negative plate of the battery (the long line to the short line) then we use –E. If the current moves from the negative plate to the positive plate of the battery, then weuse +E. If the current moves through a resistor, then we add -IR or +IR to the equation. If we are following the current in the opposite direction of our arbitrarily assigned direction, then IR is positive. Otherwise, IR is negative. If the current passes through neither a resistor, nor a battery, then we don’t add anything to our equation (or we add or subtract 0 for book keeping purposes)So, for the circuit above;Loop ADEBA = AD+ DE + EB + BAAD: We move through a resistor in the same direction as we assigned to the current, so we get –I1RDE: The current doesn’t move through anything, so we get -0EB: The current moves through a resistor in the opposite direction of the one assigned, so we get +I3RB A: The current moves through a battery from the small line (- ) to the long line (+), so we get +ETherefore; Loop ADEBA = AD+ DE + EB + BA = -I1R-0+I3R+EWe’ll then plug in the corresponding values for the resistance and the voltage of the battery and set the equation equal to 0. Now we can solve until we get some kind of equivalence equation for I1 and I3. This will be our second equation. We will repeat this process for loop BEFCB, now coming up with an equivalence equation for I2 and I3. Now, we can solve algebraically with our three equations to find the value of our currents. Example Problem:Consider The following complex circuit: A E1=7.3V B E2=1V C I3 R1=10Ω R3=5Ω R2=10Ω I1 I2 F D EFind the potential difference (V)between node C and node DVCD=I2R2=-E2-i3R3Node Rule: Incoming current at d is I1+I3=I2 Equation 1Loop BAFDB:+E1-I1R1+0+I3R3=07.3-10I1+5I3 => I3=1/5(10I1-7.3)  Equation 2Loop DECBD:0-I2R2-E2-I3R3=0-10I2-1-5I3=0 => -5I2-0.5=I3 Equation 3Put equation 1 and 3 together to get:I3=(-2-2I1)/3Using equation 2 and 31/5(10I1-7.3)=-(.2-2I1)/3 I1=.522AIf I know I1, I can plug this value into equation number 2 to find I3. Once I know both I1 and I3 I can find