
#1
Dec1411, 09:39 PM

P: 37

**PICTURE ATTACHED AT BOTTOM**
1. The problem statement, all variables and given/known data The current in the 13.8Ω resistor in the attached picture is .795 A. Find the current in the other resistors in the circuit. 2. Relevant equations V=RI In a series circuit: V of Circuit = V1 + V2 + V3 +...Vn I of Circuit = I1 = I2 = I3 = ...In R of Circuit = R1 + R2 + R3 +... Rn In a parallel circuit: V of Circuit = V1 = V2 = V3 = ... Vn I of Circuit = I1 + I2 + I3 +... In 1/R of Circuit = 1/(R1) + 1/(R2) + 1/(R3) +... 1/(Rn) 3. The attempt at a solution For the 13.8Ω resistor, I found that Voltage = 10.971 by using V=IR  Because V of Circuit = (V1 = V2 = V3 = ... Vn), the Voltage for the 17.2Ω resistor is also 10.971 Based on this, the Current for the 17.2Ω resistor is .638: V=IR I=V/R I=(10.971)/(17.2) I=.638 After this, I proceeded to find the equivalent resistance for the entire circuit. To find this, I separated the circuits into groups. Group 1= the 15Ω and 12.5Ω resistors (they are in series). Group 2= the 13.8Ω and 17.2Ω resistors (they are parallel). Group 3= the 8.45Ω and 4.11Ω resistors (they are in series). When solved for, the equivalent resistance for each group is as follows: Group 1= 27.5Ω Group 2= 7.66Ω Group 3= 12.56Ω Note that Groups 2 and 3 are now parallel. We will combine them into Group 4. Group 4 Resistance = 4.75Ω Now we only have Groups 1 and 4 left in series. When combined, we will get the net resistance for the entire Circuit. This works out to be 32.56Ω .....So that's everything I could find, which is decent, but doesn't come anywhere near answering the question. I know I must be missing something somewhere. How do I figure out the individual currents for each resistor with the information I currently have? 



#2
Dec1411, 10:14 PM

Mentor
P: 11,404

What then is the total current for the subcircuit? 



#3
Dec1411, 11:11 PM

P: 37

Ah, of course!
Okay, so based on that... Current of the 8.45 Ω resistor should be (10.971)/(8.45) = 1.3 A and the 4.11 Ω will be (10.971)/(4.11) = 2.67 A So the total for the subcircuit will be .795 A +.638 A +1.3 A +2.67 A = 5.403 A .... Would that be correct? 



#4
Dec1511, 07:22 AM

Mentor
P: 11,404

Need help with a circuit problem? (Picture attached) 



#5
Dec1511, 07:46 AM

P: 37

(10.971)/(8.45+4.11) = .873 A And then the subcircuit would work out as .873 A + .795 A + .638 A = 2.3 A 



#6
Dec1511, 07:55 AM

Mentor
P: 11,404





#7
Dec1511, 02:54 PM

P: 37

Alrighty, here are my final results. Everything seems to check out:
13 Ω = .795 A 17.2 Ω = .638 A 8.41 Ω = .87 A 4.11 Ω = .87 A 15 Ω = 2.3 A 12.56 Ω = 2.3 A 



#8
Dec1511, 03:01 PM

Mentor
P: 11,404





#9
Dec1511, 03:22 PM

P: 37

thanks for all the help.



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