Electric Circuit Analysis: Finding I1, I2, and E with Ammeter Reading of 4.11 A

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AI Thread Summary
The discussion revolves around solving a circuit analysis problem involving an ammeter reading of 4.11 A and resistances R1, R2, and R3 with an electromotive force E1 of 11 V. Participants clarify the application of Kirchhoff's laws to derive equations for the currents I1 and I2, emphasizing the need to account for the combined current through R3. The calculations reveal that I1 is found to be 4.72 A and I2 is 6.28 A after correcting earlier mistakes. The final value for E2 is determined to be 39.9 V based on the derived currents and resistances. The thread highlights the importance of careful interpretation of circuit components and relationships in solving for unknowns.
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Last Problem...PLEASE HELP Me!

Homework Statement



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The ammeter reads I=4.11 A. Find I1. (R1=1.4 Ω, R2=4.6 Ω, R3=1 Ω, ε1=11 V.)
Find I2
Find E

Homework Equations





The Attempt at a Solution


I need someone to walk me through this because conceptually I don't understand what to do here...
 
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For I1
I = E/R
But which R's do I include? r1 and r2 or all of them?
 
For I1 you would use r1 and r3. If you look at the arrows for the current on either side, you can see that the current comes from either E1 or E2, flows through r1 or r2 respectively, and both currents flow through r3. Does that help?
 
Thats what I did at first I added R1 and R3 to get 2.4 then to get I I did..
I = 11/2.4 = 4.58 A..but it says its wrong??
 
Okay, my apologies, I didn't read the problem correctly. You should just use the resistance of r1--it is just looking for the flow of current across that resistor.
 
I did that too...
I = 11/1.4 = 7.86
Says its wrong though...not sure what else to do.
Does the I in the middle have to be taken into account for anything?
 
Kirchoffs Laws. you have two loops e1 and e2 the voltage drop across each loop must sum to zero That means for left Loop E1 the voltage accros R1 and Voltage across R3 must sum to E1. Voltage across R1 is R1 x I1 Voltage across R3 is R3 x (I1 + I2) THAT is the TRICK the Current in R3 is the sum of two currents I1 and I2. doing same fro right loop we get E2 = I2xR2 + (I1+I2)x R3 and from left loop E1 = I1xR1 +(I1+I2)x R3 simplify and group terms..E1 = I1x(R1+R3) + I2xR3 and E2 = I2x(R2+R3) + I1xR3 use values of r1 r2 and r3 E1 = I1x2.4 + I2x 4.6 and E2 = 5.6 x I2 + 1 x I1 This is two equations in two unkowns BUT he already told us e1 = 11 and I = to I1 + I2 = 4.11 so E1 = I1x2.4 +I2x4.6 so I1 = 4.11 - I2 so e1 = 11 = 2.4x4.11-2.4x I2 + 4.6 x I2----- 11 = 9.86 + 2.2 x I2 ... I2 = 0.518 Amps
 
wow okay...
so I have 3 equations...
1.) I1 = I2 + I3
2.) E -I1r1 - I3R3
3.) E2 -I2r2 - I3R3

And how do I solve for I1 and I2?
 
It is solved already We just worked at getting I2 = .518 Now go back to second loop eequation and sustitute what we have so far E2 = .518x(R2) + (I1+I2)xR3 But he gave you I = I3 = I1 + I2 = 11 so E2 = .518x4.6 + 11 x 1 or E2 = 13.383
 
  • #10
It says wrong for E2 = 13.383
 
  • #11
What the problem calls I you are calling I3, so let's re-write E1 = I1R1 + I3R3 E2 = I2R2 +I3R3 He gave us I = 4.11 and E1 = 11V 11V = I1x1.4 + 4.11x 1 so I1 = (11-4.11)/1.4 so I1 = 4.72 A so I3 = 11 = I1 + I2 so 11 = 4.72 + I2 so I2 = 6.28 sorry I screwed up last try
 
  • #12
Dis regard, I1 = 4.72 I2 = 6.28 so E2 = i2xr2 +i3xr3 6.28x4.6 + 11 x 1 = 39.9
 
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