Power dissipation in resistive circuit

In summary, the conversation is about a circuit with a single resistor connected to a battery and another identical resistor added in series. The goal is to find the power dissipated by the second circuit in terms of the power dissipated by the first circuit. The answer should not depend on the variables I, R, U, or t. The best approach is to use the equations V = I*R and P = V^2/R for both circuits and find the ratio of the powers. It is important to not give full solutions to problems in the homework help section.
  • #1
csimon1
17
0

Homework Statement



A single resistor is wired to a battery as shown in the diagram below. (Figure 1) Define the total power dissipated by this circuit as P(subscript 0). Now, a second identical resistor is wired in series with the first resistor as shown in the second diagram to the left (figure 2). What is the power, in terms of P(subscript 0) dissipated by this circuit?

Homework Equations



I am not sure! This problem seems so easy -- I don't think I'm really understanding what they want from me? This is the feedback I got.
The correct answer does not depend on the variables: I, R
So I tried it again...
The correct answer does not depend on the variables: U, t


The Attempt at a Solution



I uploaded the pictures from the question. Can anyone help?
 

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  • #2
[EDIT] Deleted what I had previously said: "Will your program allow you to input the answer in terms of the variables V and R?"

I think I understand now. The problem statement is asking you for a power ratio. A power divided by a power (not an actual power itself). So once you find the two powers, divide one by the other.
 
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  • #3
Huh?? Well, my first guess was IsquaredR and then it was U/t...because both those formulas give power. What do you mean a power ratio?
 
  • #4
The answer not a formula. Don't understand...
 
  • #5
csimon1 said:
Huh?? Well, my first guess was IsquaredR and then it was U/t...because both those formulas give power. What do you mean a power ratio?

That's fine to start with equations such as

[tex] P _0 = (I _0)^2 R [/tex].

For the first circuit. But you'll find it easier in this case, if you stick with the given variables (Both circuits are given only in terms of V and R, so you might try to stick with those). Since you already know

[tex] V = I _0 R [/tex]

you'll find it easier to use

[tex]
P _0 = \frac{V^2}{R}
[/tex]

Now do the same thing for the second circuit, except call it [tex] P _1 [/tex].

It think your physics program is looking for [tex] P_1 / P_0 [/tex]. Give this a try and I think you'll see what I mean.
 
  • #6
You should also try to put 0.5P0 since by adding one resistor of the same magnitude in series the power should go down by half, since P = V^2/(Sum of R)
 
  • #7
camm1090 said:
You should also try to put 0.5P0 since by adding one resistor of the same magnitude in series the power should go down by half, since P = V^2/(Sum of R)

Don't give people full solutions to problems in the homework help section, no matter how simple the problem may be.

EDIT: It's against forum rules, and it is also counterproductive for the original poster, who learns nothing and gains nothing from having other people do his/her homework for him/her.
 
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1. What is power dissipation in a resistive circuit?

Power dissipation in a resistive circuit refers to the amount of energy that is converted to heat as a result of the flow of electric current through a resistor. This heat is caused by the resistance in the circuit, which converts electrical energy into thermal energy.

2. How is power dissipation calculated in a resistive circuit?

Power dissipation in a resistive circuit can be calculated using Ohm's Law, which states that power (P) is equal to the product of current (I) and voltage (V), or P = I x V. In a resistive circuit, the voltage can be calculated using the formula V = IR, where R is the resistance in the circuit.

3. What factors affect power dissipation in a resistive circuit?

The main factor that affects power dissipation in a resistive circuit is the resistance of the circuit. The higher the resistance, the more power will be dissipated as heat. Other factors that can affect power dissipation include the current flowing through the circuit and the voltage applied to the circuit.

4. How does power dissipation impact the performance of a resistive circuit?

Power dissipation can have a significant impact on the performance of a resistive circuit. If too much power is dissipated as heat, it can cause the circuit to overheat and potentially fail. It can also affect the accuracy of the circuit's output, as the heat generated can alter the resistance and change the flow of current.

5. How can power dissipation be reduced in a resistive circuit?

There are several ways to reduce power dissipation in a resistive circuit. One method is to decrease the resistance in the circuit, which will result in less power being converted to heat. Another option is to use a material with a lower resistivity, such as copper, to decrease the overall resistance. Additionally, using higher voltage and lower current can also reduce power dissipation. Finally, proper circuit design and cooling techniques can also help to reduce power dissipation.

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