i have I3=I1+I2 -5i1 +21v -.1i1 -20i2 -40i2 + .5i2 +47v pretty sure it\'a wrong.
ID: 1643105 • Letter: I
Question
i have I3=I1+I2
-5i1 +21v -.1i1 -20i2 -40i2 + .5i2 +47v
pretty sure it'a wrong. can you please give me a step by step on how to find i1. please write a little explanation with it that can explain sign changes and other important things
HomeStudent: apthomp@okstate.edu MyA Class Management Help Chapter 21-22A Homework Begin Date: 6/29/2017 12:00:00 AM - Due Date: 7/5/2017 11:5900 PM End Date: 8/3/2017 12:00:00 AM (11%) Problem 4: Consider the circuit in the diagram, with sources of emf listed below Randomized Variables 0.10 5.0 20 =5.5 V 8,-27V 0.50 40 0.20 78 i 0.05 h ©theexpertta.con 33% Part (a) Find 11 in amps. Grade Summary Deductions Potential 0% 100% ted tan() | | cosO |-asin() sinO Submissions Attempts remaining: (09o per attempt) cotan() cotan asinacos0 4 5 6Explanation / Answer
i3 = i1 + i2
Now kirchoff's voltage law in upper loop 'aedcba'
in the direction of current +ve sign, otherwise -ve sign
E2 - E1 = (r2 + R2)*i2 - (r1 + R1 + R5)*i1
47 - 21 = (0.5 + 40)*i2 - (0.1 + 5 + 20)*i1
26 = 40.5*i2 - 25.1*i1
Now kirchoff's voltage law in lower loop
E2 + E3 - E4 = (i2 + R2)*i2 + (r3 + r4 + R3)*i3
47 + 5.5 - 27 = (0.5 + 40)*i2 + (0.05 + 0.2 + 78)*i3
25.5 = 40.5*i2 + 78.25*i3
Solving these three equations:
26 = 40.5*i2 - 25.1*i1
25.5 = 40.5*i2 + 78.25*i3
i3 = i1 + i2
25.5 = 40.5*i2 + 78.25*(i1 + i2)
25.5 = 118.75*i2 + 78.25*i1
26 = 40.5*i2 - 25.1*i1
i1 = -0.333 Amp.
i2 = 0.435 Amp.
i3 = 0.102 Amp.
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