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

In summary, this person tried to solve a homework problem and got some wrong answers. They re-did the problem and got the correct answers.
  • #1
sloan13
73
0

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.
 
Last edited by a moderator:
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  • #2
I just got the power dissipation problems. All I need is the voltage drop. Apparently, V=IR doesn't apply here?
 
  • #3
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:
  • #4
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?
 
  • #5
Yea, I reentered them without rounding, and they were correct. Thank you, berkeman.
 
  • #6
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:
 

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

1. What is the difference between current and voltage?

Current refers to the flow of electric charge, while voltage is the force that drives this flow. In other words, current is the movement of electrons, while voltage is the pressure or potential difference that causes electrons to move.

2. How do I calculate voltage drop in a circuit?

Voltage drop can be calculated using Ohm's law, which states that voltage drop (V) is equal to current (I) multiplied by resistance (R). This can also be represented as V=IR.

3. What factors affect power dissipation in a circuit?

Power dissipation is affected by the amount of current flowing through the circuit and the resistance of the components in the circuit. The higher the current and resistance, the higher the power dissipation will be.

4. How do I measure current in a circuit?

Current can be measured using an ammeter, which is a device that is placed in series with the circuit and measures the flow of electrons. It is important to note that the ammeter should have a very low resistance to avoid affecting the current in the circuit.

5. What is the relationship between current, voltage, and power?

The relationship between current, voltage, and power can be expressed as P=IV, where P is power, I is current, and V is voltage. This means that the power dissipated in a circuit is directly proportional to both the current and voltage in the circuit.

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