Mesh analysis and power dissipation

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In mesh analysis, the power dissipation in resistors must equal the power supplied by voltage sources. To solve for power dissipation, first determine the current in each branch and calculate the power for each resistor using the formula PR = I²R. Sum these power values to find the total power dissipated in the network. Additionally, calculate the power contribution from voltage sources with PS = Vs⋅I, ensuring to consider the orientation of the voltage source and the direction of current. Finally, confirm that the total power dissipated equals the total power supplied, demonstrating that ∑PR = ∑PS holds true.
Andriy B
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Hi I'm taking an intro course to electrical and computer engineering and I'm having trouble figuring out part c of number one. I have attached a copy of my solutions to parts a and b which I believe are right (correct me if I'm wrong please) however when I calculate the power dissipation of each component and add them up I get a large number. Thank you in advance.
 

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" Determine the power dissipation of each component and show that the sum of the power dissipated equals zero "

Somebody was drunk while writting that assignement ? Dissipative components in that mesh are resistors. Sum of powers dissipated in them must be equal to sum of powers delivered by voltage sources.
 
So what's the correct way to solve part c?
 
After you find current of each branch, calculate power dissipated by every resistor (PR=IR2R).Sum these powers up to find total power dissipated in network. Power contribution of voltage source calculate as PS=Vs⋅I, where I is current through the source (pay attention to voltage source orientation and current flowing through it, depending on network parameters it may be negative). Then show that ∑PR= ∑PS holds
 

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