Presentation on theme: "Kirchhoff's Rules Continued"— Presentation transcript:
1 Kirchhoff's Rules Continued Gustav KirchhoffGerman Physicist
2 Circuit DefinitionsNode – any point where 2 or more circuit elements are connected togetherWires usually have negligible resistanceEach node has one voltage (w.r.t. ground)Branch – a circuit element between two nodesLoop – a collection of branches that form a closed path returning to the same node without going through any other nodes or branches twice
3 Problem-Solving Strategy Applying Kirchhoff’s Rules to a Circuit: 1. Assign labels and symbols to all the known and unknown quantities.2. Assign directions to the currents in each part of the circuit. Although the assignment of current directions is arbitrary, you must stick with your original choices throughout the problem as you apply Kirchhoff ’s rules.3. Apply the junction rule to any junction in the circuit. The rule may be applied as many times as a new current (one not used in a previously found equation) appears in the resulting equation.4. Apply Kirchhoff’s loop rule to as many loops in the circuit as are needed to solve for the unknowns. In order to apply this rule, you must correctly identify the change in electric potential as you cross each element in traversing the closed loop. Watch out for signs!5. Solve the equations simultaneously for the unknown quantities, using substitution or any other method familiar to the student.6. Check your answers by substituting them into the original equations.
5 Example Circuit(1): Solve for the currents through each resistor And the voltages across each resistor usingSeries and parallel simplification.
6 Example Circuit(1): The 6 and 4 ohm resistors are in series, so are combined into 6+4 = 10Ω
7 Example Circuit(1): The 8 and 10 ohm resistors are in parallel, so are combined into 8∙10/(8+10) =14.4 Ω
8 Example Circuit The 10 and 4.4 ohm resistors are in series, so are combined into 10+4 = 14.4Ω
9 Example Circuit(1): Writing KVL, I1∙14.4Ω – 50 v = 0 +I1∙14.4Ω-Writing KVL, I1∙14.4Ω – 50 v = 0Or I1 = 50 v / 14.4Ω = 3.46 A
10 Example Circuit(1): If I1 = 3.46 A, then I1∙10 Ω = 34.6 v +15.4 v-If I1 = 3.46 A, then I1∙10 Ω = 34.6 vSo the voltage across the 8 Ω = 15.4 v
11 Example Circuit(1): If I2∙8 Ω = 15.4 v, then I2 = 15.4/8 = 1.93 A +15.4 v-If I2∙8 Ω = 15.4 v, then I2 = 15.4/8 = 1.93 ABy KCL, I1-I2-I3=0, so I3 = I1–I2 = 1.53 A
12 Example (2)The circuit here has three resistors, R1, R2, and R3 and two sources of emf, Vemf,1 and Vemf,2This circuit cannot be resolved into simple series or parallel structuresTo analyze this circuit, we need to assign currents flowing through the resistors.We can choose the directions of these currents arbitrarily.
13 Example (2):At junction b the incoming current must equal the outgoing currentAt junction a we again equate the incoming current and the outgoing currentBut this equation gives us the same information as the previous equation!We need more information to determine the three currents – 2 more independent equations
14 Example (2)To get the other equations we must apply Kirchhoff’s Loop Rule.This circuit has three loops.LeftR1, R2, Vemf,1RightR2, R3, Vemf,2OuterR1, R3, Vemf,1, Vemf,2
15 Example - Kirchhoff’s Laws (4) Going around the left loop counterclockwise starting at point b we getGoing around the right loop clockwise starting at point b we getGoing around the outer loop clockwise starting at point b we getBut this equation gives us no new information!
16 Example - Kirchhoff’s Laws (5) We now have three equationsAnd we have three unknowns i1, i2, and i3We can solve these three equations in a variety of ways
17 Apply the junction rule to point c. I 1 is directed into Example (3):Find the currents in the circuit shown in Figure by using Kirchhoff’s rules.Solution:Apply the junction rule to point c. I 1 is directed intothe junction, I 2 and I 3 are directed out of the junction.Select the bottom loop, and traverse it clockwise startingat point a, generating an equation with the loop rule:Select the top loop, and traverse it clockwise from point c. Notice the gain across the 9.0- resistor, because it is traversed against the direction of the current
18 Rewrite the three equations, rearranging terms and dropping units for the moment, for convenience:Solve eq. 3 for I2 ,substitute into eq. 1:Substitute I3 back into eq. 3 to get I2From eq.1Substitute the latter expression Substitute I3 into eq. 2 to get I1into eq. 2 and solve for I3
19 Homework :Suppose the 6.0V battery is replaced by a battery of unknown emf, and an ammeter measures I1 =1.5 A. Find the other two currents and the emf of the battery.Answers I2 =0.96 A, I3 =0.54 A, e=11 VI1 =1.5 AeV
21 2.5. What is the current through the 4- resistor in this circuit? a) Ab) Ac) 1.0 Ad) 1.3 Ae) 1.5 A
22 2.6. What is the current through the 1- resistor in this circuit? a) 2.8 Ab) 3.0 Ac) 3.4 Ad) 4.0 Ae) 4.8 A
23 e) 6 V (8 ) I1 (6 ) I4 6 V (2 ) I5 (4 ) I3 = 0 2.7. Which one of the following equations is not correct relative to the other four equations determined by applying Kirchoff’s Rules to the circuit shown?a) I2 = I1 + I4b) I2 = I3 + I5c) 6 V (8 ) I1 (5 ) I2 (4 ) I3 = 0d) 6 V (6 ) I4 (5 ) I2 (2 ) I5 = 0e) 6 V (8 ) I1 (6 ) I4 6 V (2 ) I5 (4 ) I3 = 0Not a loop!
24 2. 8. Given is the multi-loop circuit on the right 2.8.Given is the multi-loop circuit on the right. Which of the following statements cannot be true:A)B)C)D)Junction ruleNot a loop!Upper right loopLeft loop
25 2.7. Consider the three resistors and the battery in the circuit shown. Which resistors, if any, are connected in parallel?a) R1 and R2b) R1 and R3c) R2 and R3d) R1 and R2 and R3e) No resistors are connected in parallel.