How Does Current Flow in a Multi-Loop Circuit with Multiple Voltage Sources?

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SUMMARY

The discussion focuses on analyzing current flow in a multi-loop circuit with multiple voltage sources, specifically addressing the circuit in Figure 27-69. The calculated currents are i1 = -2.44 A and i2 = -1.33 A, indicating their directions are downward. Energy transfer rates are determined to be 5.33 W for the 4.00 V battery (supplying energy) and 41.6 W for the 11.0 V battery (also supplying energy). The approach to solving the circuit involves using Kirchhoff's laws and equivalent resistance for parallel resistors.

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1. the problem
In Figure 27-69 (see attached), the ideal each resistance is 4.50 and V1 = 11.0 V.


(a) What is the size and direction of current i1? (Take upward to be positive.)
(No Response)[-2.44] A
(b) What is the size and direction of current i2? (Take upward to be positive.)
(No Response)[-1.33] A
(c) At what rate is energy being transferred at the 4.00 V battery and is the battery supplying or absorbing energy?
(No Response)[5.33] W (No Response)[ supplying ]
(d) At what rate is energy being transferred at the 11.0 V battery and is the battery supplying or absorbing energy?
(No Response)[41.6] W (No Response)[ supplying ]


Attempted Solution

Help needed with part b.

for part a: we ignored the right hand part of the cirucuit, and just did the left-most loop. Then we summed the voltages and set them to zero, and got the correct answer.

for part b: We looked at the two right loops and combined them by saying that there are two resistors in parallel which can be replaced by [tex]R_{eq}=\frac{1}{R}+\frac{1}{R})^{-1}[/tex]. Then we tried ignoring the left loop and summing the voltages in the middle loop (with the top right resistor replaced by R_eq) and setting everything equal to zero. That's the wrong answer.
Frankly, I don't understand why [tex]i_2\neq i_1[/tex].
And the bloody final is tomorrow:cry:
 

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Here's a better diagram, I think, which hopefully will be approved a little more quickly. :wink:

for part b: We looked at the two right loops and combined them by saying that there are two resistors in parallel which can be replaced by

In the new diagram, the two resistors being combined are R3 and R4.
 

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