Calculating Heat Loss in a Series Circuit

• Renue
In summary: I used the difference in P = V^2/R, the power dissipated in the circuit, to calculate the heat loss.Heat energy loss/t = (12^2/6) - (6^2/4) = 15W.Apparently, I am incorrect; but I don't know the right direction to solve this problem.Thanks in advance for your guidance/help.
Renue

Homework Statement

Determine the heat loss in the circuit that is connected in series with two batteries (12 V and 6V) and two resistors (6ohms and 4 ohms).

The Attempt at a Solution

I used the difference in P = V^2/R, the power dissipated in the circuit, to calculate the heat loss.
Heat energy loss/t = (12^2/6) - (6^2/4) = 15W.
Apparently, I am incorrect; but I don't know the right direction to solve this problem.

What is the total voltage change across the batteries? What is the total resistance of the resistors? What is the current?

Renue said:
I used the difference in P = V^2/R, the power dissipated in the circuit, to calculate the heat loss.
Heat energy loss/t = (12^2/6) - (6^2/4) = 15W.
Apparently, I am incorrect; but I don't know the right direction to solve this problem.

How did you decide that there was 12 V across the 6 Ω resistor and 6 V across the 4 Ω resistor? What is the current flowing in the series circuit? Can it be different for different components? Try writing KVL for the loop.

CWatters
See you used the formula , P=V^2/R but as you mentioned the resistors are connected in series. The formula you used was when the resistors are connected in parallel. Thus wrong formula= wrong answer

Now don't just abuse . Say the right formula. I think most of the people would have this ridiculous thought in their mind . Try using the formula -
P=I^2Rt
Hope it helped!

The equation P=V^2/R works just fine for ALL circuits (series or parallel) if you use the correct values for V and R.

See post #3

cnh1995
draw a circuit diagram and show your calculations for the loop current and how you arrive at it and derive the power from it

1. What is heat loss in a circuit?

Heat loss in a circuit is the process of energy being converted into heat and dissipating from the circuit into the surrounding environment.

2. What causes heat loss in a circuit?

Heat loss in a circuit is primarily caused by resistance, which is the opposition to the flow of electricity in a material. As electricity flows through a circuit, the resistance in the wires and components causes energy to be converted into heat.

3. How does heat loss affect the performance of a circuit?

Heat loss can negatively impact the performance of a circuit by reducing its efficiency and causing components to overheat. This can lead to malfunctions, damage, and even safety hazards.

4. How can heat loss be minimized in a circuit?

Heat loss can be minimized by using materials with lower resistance, ensuring proper insulation and ventilation in the circuit, and using efficient designs and components that generate less heat.

5. Can heat loss be completely eliminated in a circuit?

No, heat loss cannot be completely eliminated in a circuit. However, it can be managed and reduced through proper design, materials, and maintenance of the circuit.

• Introductory Physics Homework Help
Replies
3
Views
655
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
886
• Introductory Physics Homework Help
Replies
2
Views
375
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
22
Views
2K
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
325