Find currents and voltages - 2 batteries & 3 resistors

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Homework Help Overview

The problem involves analyzing a circuit with two batteries and three resistors, specifically finding the currents (I1, I2, I3) and the voltages across each resistor. The circuit is governed by Kirchhoff's loop and junction laws.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply Kirchhoff's laws to set up equations for the circuit. They introduce an additional current (I4) to aid in visualization but question its necessity. Other participants point out the redundancy of I4, suggesting that I3 should suffice for the analysis.

Discussion Status

The discussion has progressed with participants clarifying the relationships between the currents in the circuit. There is acknowledgment of the misunderstanding regarding the introduction of I4, leading to a resolution of that specific point.

Contextual Notes

Participants are navigating the complexities of current flow in series and parallel components, questioning assumptions about current distribution and the implications of ideal circuit models.

kamhogo
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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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