Kirchoff's Laws Problem - Power

In summary, the conversation discusses the rate of conversion of electrical energy to internal energy in two resistors, 4.56Ω and 5.91Ω, using Kirchoff's Laws. The attempt at a solution involves setting up current directions and utilizing the junction rule and loop rule, but the calculations do not match the expected results. The suggestion is made to use a nodal analysis approach instead.
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Homework Statement


At what rate is electrical energy converted to internal energy in the 4.56Ω and the 5.91Ω resistors in the figure?

Homework Equations


Kirchoff's Laws:
ƩI(in) = ƩI(out) Junction Rule

ƩΔV = 0 Loop Rule


The Attempt at a Solution


I have set the top branch current to go from right to left = I(3)
I have set the middle branch current to go from right to left = I(2)
I have set the bottom branch current to go from left to right = I(1)

The left middle junction results in the following I(3) + I(2) = I(1)...Junction#1
Loop #1 = 0 = 6.96V-5.91I(2)-4.56I(3)
Loop #2 = 0 = -9.56I(1)-5.91I(2)+1.86V

I get I1=2.7502
I2=-4.1340
I3=6.8842

The answers I get when using the P=I^(2)*R...do not yield results consistent with the online homework

I have tried this several different ways and the answer according to my homework do not match. Please provide some insight here.

Thank You.
 

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  • #2
I think you're not following your own stipulations for current directions. Once you choose them, if you start writing equations you need to follow through.

When you "pass though" a resistor in the direction of the assumed current flow, you expect a voltage drop. If you pass through against the flow then you get a voltage rise.

Just a thought: you might find a nodal analysis approach to be easy to apply here.
 

What are Kirchoff's Laws?

Kirchoff's Laws are two fundamental principles in electrical engineering that describe the behavior of electrical circuits. The first law, known as Kirchoff's Current Law, states that the total current entering a node in a circuit is equal to the total current leaving that node. The second law, known as Kirchoff's Voltage Law, states that the sum of all voltage drops in a closed loop circuit is equal to the sum of all voltage rises in that loop.

How do Kirchoff's Laws apply to power problems?

Kirchoff's Laws can be applied to power problems by using them to determine the values of different electrical quantities, such as voltage, current, and resistance, in a circuit. These values can then be used to calculate the power dissipated in different components of the circuit.

What is the difference between Kirchoff's Current Law and Kirchoff's Voltage Law?

Kirchoff's Current Law deals with the conservation of charge in a closed circuit, while Kirchoff's Voltage Law deals with the conservation of energy in a closed circuit. In other words, Kirchoff's Current Law is concerned with the flow of current, while Kirchoff's Voltage Law is concerned with the flow of energy.

How can I apply Kirchoff's Laws to solve a power problem?

To apply Kirchoff's Laws to solve a power problem, you first need to identify all the components in the circuit and their corresponding values, such as voltage and resistance. Then, you can use Kirchoff's Laws to create a system of equations, which can be solved to determine the unknown values. Finally, you can use these values to calculate the power dissipated in different components of the circuit.

What are some common mistakes when applying Kirchoff's Laws to power problems?

Some common mistakes when applying Kirchoff's Laws to power problems include forgetting to label the direction of current and voltage drops, not considering the direction of current in relation to the direction of voltage drops, and ignoring the signs of values in the equations. It is also important to correctly apply the laws to different parts of the circuit, such as series and parallel branches.

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