Energy in Circuits: The power lost to joule heating.

In summary, the question asks for the power lost to Joule heating in a high-voltage transmission line with a 600A current and a length of 60km and diameter of 2.5cm. The equation to find this power is P=I^2R, where R can be found using the equation R=P(L/A). The potential difference can be found using V=1/(4piEo)(q/r), and the charge can be determined using Gauss's Law. The final answer for the power lost to Joule heating is 1.2x10^6W.
  • #1
BillJ3986
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Homework Statement


The aluminum cable of a high-voltage transmission line carries a current 600A. The cable is 60km long, and it has a diameter of 2.5 cm. What is the power lost to Joule heating in this cable?


Homework Equations


Should I use the equation V=1/(4piEo)integral(dq/r) to find the potential? And in saying that I also need to find the charge in order to find the potential difference? Do I have to use Gauss's Law to find the charge?


The Attempt at a Solution


I know that the equation to find the power lost to joule heating is P=I^2R
in order to find the resistance, I solved solved for the potential difference using V=1/(4piEo)(q/r), I assumed that since the cable is aluminum that the Q= 1.6x10^-19C. my answer for potential difference is 1.2x10^-7 which I know is wrong.


 
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  • #2
I already answered the question. I had to look up resistivity for aluminum since the cable is made up of aluminum, and use the equation R=P(L/A). After finding the resistance. I was able to find out the power lost to Joule heating by using P=I^2(R). P=600^2(3.42)= 1.2X10^6W.
 

1. What is joule heating and how does it relate to energy in circuits?

Joule heating is the process in which electrical energy is converted into heat energy as it passes through a conductor. This is due to the resistance of the conductor, which causes the flow of electrons to slow down and release energy in the form of heat. This heating effect is an important consideration in the analysis and design of circuits, as it can affect the efficiency and performance of the circuit.

2. How is power lost in a circuit due to joule heating?

The power lost in a circuit due to joule heating is calculated using the formula P = I^2R, where P is the power lost, I is the current flowing through the circuit, and R is the resistance of the circuit. As the current increases or the resistance increases, the power lost due to joule heating also increases.

3. What factors can affect the amount of power lost to joule heating in a circuit?

The amount of power lost to joule heating in a circuit is affected by several factors, including the current flowing through the circuit, the resistance of the circuit, and the length and thickness of the conductor. Other factors such as the temperature and material of the conductor can also have an impact on the power lost to joule heating.

4. How can the effects of joule heating be minimized in a circuit?

To minimize the effects of joule heating in a circuit, it is important to reduce the resistance of the circuit by using materials with lower resistivity and increasing the thickness of the conductor. Additionally, using thicker and shorter conductors can help reduce the amount of power lost due to joule heating.

5. What are some real-life applications of joule heating in circuits?

Joule heating has various applications in everyday life, such as in electric heaters, toasters, and hair dryers, where electrical energy is converted into heat energy to perform a specific function. It is also used in electronic devices, such as computers and smartphones, where the heat generated due to joule heating must be managed to prevent damage to the components.

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