Calculating Heat Loss in a Series Circuit

[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).

Homework Equations

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.
Thanks in advance for your guidance/help.

 

[Chestermiller]

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

 

[gneill]

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.

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.

 

[Brian blake science]

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!

 

[CWatters]

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

 

[phinds]

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