Find currents and voltages - 2 batteries & 3 resistors

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SUMMARY

The discussion focuses on solving a circuit problem involving two batteries and three resistors, specifically finding the currents I1, I2, I3, and the voltages across each resistor. The resistors are defined as R1 = 50 ohms, R2 = 220 ohms, and R3 = 150 ohms. Using Kirchhoff's loop and junction laws, the currents were calculated as I1 = 0.093 A, I2 = 0.024 A, and I4 = 0.069 A, with the voltages across the resistors determined to be 4.65 V for R1, 5.28 V for R2, and 10.35 V for R3. The confusion regarding the introduction of I4 was clarified, confirming that in a series circuit, the same current flows through all components.

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Homework Statement



upload_2016-4-20_22-13-18.png


Find I1, I2, I3 as well as the voltages across each resistor.

R1= 50 ohms
R2= 220 ohms
R3= 150 ohms

Homework Equations


Kirchhoff's loop law
Kirchhoff's junction law

The Attempt at a Solution


Loop 1: Starting from the negative terminal of the 10V battery (ccw)

10V - 50 (I1) - 220 (I2) = 0

Loop 2: Starting from junction a (ccw). Note I added an I4 through R3 to try and make things easier for me to visualize .

- 150 (I4) + 5 V + 0 (I3) + 220 (I2) = 0

Junction law
I1 = I2 + I4Loop 1
10V - 50 I2 - 50 I4 - 220 I2 =0
10V - 270 I2 - 50 I4 = 0

Loop 2
I2= (150 I4 - 5 V )/220 = 0.682 I4 - 0.0227 A

Loop 1
10V - 270 (0.682 I4 - 0.0227 A) - 50 I4 = 0
- 234.14 I4 = -16. 129
I4 = 0.069 A
I2= 0.024 A
I1= 0.093 A
I3 = ??

Voltages across :
R1 ==> 4.65 V
R2 ==> 5.28 V
R3 ==> 10.35 VI can't find I3. Is it possible that R2 "receives" a portion of I3 downstream (...I5. ...) in addition to having I2 go through R2 it upstream? ? Or maybe I3 is infinitely big (ideal wire model, goes from the positive terminal of the 5V battery and then directly to the negative terminal of the 10V battery)?
 
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I don't know why you've introduced I4. The branch has two components in series, so they MUST carry the same current, and that's already labelled I3.

upload_2016-4-20_22-23-20.png
 
gneill said:
I don't know why you've introduced I4. The branch has two components in series, so they MUST carry the same current, and that's already labelled I3.Oooh. I see it now. Thanks!

View attachment 99436
 
Resolved.
 
Last edited:

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