Question

Kirchoff's Laws 4. Now click on this link, which is another circuit I modified and created for this lab: http://tinyurl.com/y8toda84 A. Do not Click run/Stop. This is a circuit that has elements that are neither series nor parallel, as well as multiple voltage sources. The best way for us to analyze this circuit is to use Kirchoff's laws. From the text here are Kirchoff's laws: 1) The sum of the currents entering any junction must equal the sum of the currents leaving that junction. (This rule is often referred to as the junction rule.) 2) The sum of the potential differences across all the elements around any closed circuit loop must be zero. (This rule is usually called the loop rule.) In terms of how we use these rules, there are a number of steps we must take. 1) Assign unknown current values and arbitrary directions to these currents in every separate branch. We see three branches here, so we have to assign three unknown currents. Here I have placed 11, 12, and 13 and associated directions with them. The directions are arbitrary and do not have to be the correct directions. The steps of Kirchoff's Laws will provide us with solutions that are correct regardless of what directions we choose. The answers that we find will be correct in value, but may be negative. If they are negative, that tells us our arbitrary direction was incorrect, but the value you find will still be correct As you practice more and more with circuits, you will get better at choosing directions that are more likely to be correct. 2) The next step is to label the resistors as voltage drops using your current direction as reference. The side of the resistor that the current flows into (as determined by your current directions) we label +, and the other side we label as. We can also label Voltage sources the same way, but the long line side is + and the short line side is - 3) Finally, we create loops. There are three possible loops that can be made in this circuit. The entire perimeter, and the two loops I show in this figure. I also added direction on these loops, (left clockwise, right counter clockwise) also added points A and B in the corners, which indicates my starting and ending place when I create the loop equations. 200 0-10: 200 1 100 -

I need aome help with this part of my lab. thank you!

Answer 1

+5 -200 I, ܀ loop B 3 loop A 400 I3 +5-200 I3-600I, = 0 0 75-800 Iz 300 I2 -400 I3 +5V-200. Iz I, + Iz = I3 subsititute ③ in C e e 5-2001,-400(I, + Iz) +5-2005, +73) – 5001, co € 5-800 I2 - 300 Iz-400fIf+I3) +5-200 (I, + Iz) -- put like terms together i 10 – (200+400+ 200+500) I, -foo+200) Iz 1300 II 600 Iz A 10- (800 +300+400+200) I2-(400+200) I, co 10-1700 Iz-600 I, =0 multiply by 1700 & © by 600 2210000 I, 1020000 Iz 6000 360000 I 1020000 Iz= =0 1o zo 3 2 17000 zo + t 11000 - 1850000 I, EO

3.79x107- = I = 11000 = 5.9*X10-3 = 5.94 mA 18500vo -) 12 = 10-1300 (5.94 x10-3) I3 = I, + Iz (5-94 +3.79)x10-3- 9.73 mA Goo = 3-79m A 200rtop) 800 Rooor (bottom) 5oor Current 5-94mA 3-79 mA 9-73mA 3-79uA 9-73 mA 5.94mA 2.971 1.188 V 1-137 V 3-8921 3.0324 1.946 V Voltage -30or qoor ل

This is part B.

The lab is on Kirchoffs law, part B is solved as a part of this lab.