Engineering Help with circuits (Current, Voltage Drop, and Power Dissipation)

AI Thread Summary
The discussion focuses on calculating power dissipation and voltage drop for three resistors (120, 80, and 50 ohms) connected to a 12V battery. The initial calculations for voltage drops and power dissipation were incorrect due to premature rounding. The correct total resistance was found to be 250 ohms, leading to a current of approximately 0.048A. Accurate voltage drops were recalculated as 6V, 4V, and 2.5V, with power dissipation values adjusted accordingly. The importance of avoiding early rounding in calculations was emphasized for accurate results.
sloan13
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Homework Statement



Basically, I need to know the power dissipation (W) for three resistors using aw 12V battery: 120, 80, and 50 ohms.

https://webwork2.uncc.edu/webwork2_files/tmp/Fall2014-Engr1201-Common/img/2ff2f0ce-cb10-3786-8109-c42346130316___1a2e5eeb-8657-3983-941f-b7d8a9123afa.gif

Homework Equations


I think the formula for voltage drop is Voltage Drop = Resistance * Current, and I thought the formula for power dissipation was V^2/R.

The Attempt at a Solution



It also asks for the current, which I think is 0.05A and voltage drop across all three resistors.

I think the answers for voltage drop are 6, 4, and 2.5 V, and the current was 0.05A.

I thought the power dissipation for the three resistors were 1.2, 1.8, and 2.9 W.

The website says I got "at least one wrong." Thanks, WeBWork.
 
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I just got the power dissipation problems. All I need is the voltage drop. Apparently, V=IR doesn't apply here?
 
sloan13 said:

Homework Statement



Basically, I need to know the power dissipation (W) for three resistors using aw 12V battery: 120, 80, and 50 ohms.

https://webwork2.uncc.edu/webwork2_files/tmp/Fall2014-Engr1201-Common/img/2ff2f0ce-cb10-3786-8109-c42346130316___1a2e5eeb-8657-3983-941f-b7d8a9123afa.gif

Homework Equations


I think the formula for voltage drop is Voltage Drop = Resistance * Current, and I thought the formula for power dissipation was V^2/R.


The Attempt at a Solution



It also asks for the current, which I think is 0.05A and voltage drop across all three resistors.

I think the answers for voltage drop are 6, 4, and 2.5 V, and the current was 0.05A.

I thought the power dissipation for the three resistors were 1.2, 1.8, and 2.9 W.

The website says I got "at least one wrong." Thanks, WeBWork.

Adding your voltage drops gives 12.5V, which is more than your voltage source.

I also got a different current when I divided 12V/Rtotal. Can you show your work in those two steps?
 
Last edited by a moderator:
berkeman said:
Adding your voltage drops gives 12.5V, which is more than your voltage source.

I also got a different current when I divided 12V/Rtotal. Can you show your work in those two steps?

12V/250(Rtotal) = 0.048A which I rounded to 0.05A.

Vdrop1 = 120 * .05 = 6
Vdrop2 = 80 * .05 = 4
Vdrop3 = 50* .05 = 2.5

Do you think I just rounded too early?
 
Yea, I reentered them without rounding, and they were correct. Thank you, berkeman.
 
sloan13 said:
Yea, I reentered them without rounding, and they were correct. Thank you, berkeman.

Glad to help. Yeah, definitely don't round anything until the end. :smile:
 
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