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Can you explain please? Activity: Kirchhoff's Rules Score: Name (print): Date Loop rule: the sum of the voltage drops around any loop is zero. E1

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Activity: Kirchhoff's Rules Score: Name (print): Date Loop rule: the sum of the voltage drops around any loop is zero. E1 = 18 V Junction rule: the sum of the currents at any junction (node) is zero. c d Loop Rule Signs: Pick a loop, a direction, and a starting point. 0.5 0 Resistors: going through a resistor with the current gives a voltage drop or R2 - sign; against the current gives a + sign. 2.5 0 Batteries: Going forward (- to +) through a battery gives a + sign, 12 backwards a - sign. a M There are three loops: the top loop (abcdea), the bottom loop 6.0 0 (aefgha), and the outside loop (abcdefgha). Only two are needed. 13 R3 1.5 0 Top Loop: Write down the Kirchhoff's loop rule equation for the top loop going clockwise from point a. Use symbols not numbers. M 0.5 Q E2 = 45 V Outside Loop: Now go counterclockwise around the outside loop starting again a point a. Symbols not numbers. There are two junctions, also called nodes: a and b. You only need one. Junction a: Write down Kirchhoff's junction rule equation for the junction at point a. Solution: You now have three equations in the three unknowns /1, 12, and 13. The solution is simple algebra. You should do it on the back and put the answers here. Some of the currents may be negative. This merely indicates that the current flows opposite to the (arbitrary) choice made on the diagram. DO NOT change it. 11 = 13 = This Activity has forced the choice of loops, directions of traverse, junctions, and currents. Normally you will have to make these choices for yourself. In doing the loop rule for resistors, compare your direction of travel around the loop to the direction of the current in order to determine the sign. You don't know the actual directions of the currents until after you solve the circuit, so make a reasonable but arbitrary choice, mark it on your diagram, and stick with it. You could choose to do any two of the three possible loops. The third is just a linear combination of the other two. You need only one of the two possible junctions

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