Solving Current in a Circuit with Kirchhoff's Rules

AI Thread Summary
The discussion focuses on solving a circuit using Kirchhoff's rules, specifically to find the currents through a 14.4Ω resistor and a 23.2Ω resistor. The user applied Kirchhoff's voltage law to derive equations for the circuit, resulting in a current of 1.74A through the 14.4Ω resistor and -0.71A through the 23.2Ω resistor. Feedback from other users confirmed that the calculations were correct, although there was a note about the labeling of currents in the diagram. Overall, the user expressed uncertainty about their approach but received validation for their solutions. The discussion illustrates the application of Kirchhoff's rules in circuit analysis effectively.
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Homework Statement



The currents are flowing in the direction indicated by the arrows. A negative current denotes flow opposite to the direction of the arrow. Assume the batteries have zero internal resistance.

http://img123.imageshack.us/img123/2045/599c898aa10950d7027807dyc7.th.jpg

A) Find the current through the 14.4Ω resistor and the 8.6 V battery at the top of the circuit. Answer in units of A.
B) Find the current through the 23.2 Ω resistor in the center of the circuit. Answer in units of A.

Homework Equations


Kirchhoff's Junction and Loop Rules


The Attempt at a Solution




By applying the Kirchhoff's voltage law to the above loop I get
8.6V - (23.2Ω)I1 - (14.4Ω)I = 0

23.2I1 + 14.4I = 8.6 ........(1)

Applying the Kirchhoff's voltage law to the down loop I get

16.4V + (23.2Ω)I1 = 0

I1 = - 0.71A

Substitute this value in the eq(1)

8.6V + (23.2)(0.71) = 14.4I

I = 1.74A

Therefore the current passes through the battery 8.6V and the resistance 14.4Ω is 1.74A

For Part B, I thought that the current would be equal to I1 which I already solved for in Part A, so the current would be -0.71 A.

I've pretty much worked everything out but I was hoping if someone could tell me if I was doing it right or not. I'm not too confident when it comes to these problems. Any comments would be very helpful. Thanks!
 
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Everything looks good, except for the picture, where all the currents are labeled by the same label (I). The answers are both correct.
 
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