Solving a Circuit Problem with Ideal Batteries: Finding Energy Dissipation Rate

  • Thread starter Thread starter Silimay
  • Start date Start date
  • Tags Tags
    Circuit
Click For Summary
SUMMARY

The discussion centers on solving a circuit problem involving two ideal batteries with EMF values of 3 V and 1 V, and resistances of 5 Ω, 2 Ω, and 4 Ω. The user, Silimay, initially misapplied the loop rule and made algebraic errors, leading to an incorrect calculation of current (i_1 = 0.95 A) and power dissipation in resistor R_1. The correct approach requires consistent sign usage in equations and proper application of the power formula P = i²R to find the energy dissipation rate in R_1, which should yield 0.346 W.

PREREQUISITES
  • Understanding of Kirchhoff's loop rule
  • Familiarity with Ohm's law and power calculations (P = i²R)
  • Basic algebra skills for solving systems of equations
  • Knowledge of circuit components, specifically resistors and ideal batteries
NEXT STEPS
  • Review Kirchhoff's laws for circuit analysis
  • Practice solving systems of equations in electrical circuits
  • Learn about power dissipation in resistive circuits
  • Explore circuit simulation tools like LTspice for visualizing circuit behavior
USEFUL FOR

Electrical engineering students, circuit designers, and hobbyists looking to improve their problem-solving skills in circuit analysis and power calculations.

Silimay
I have a circuit problem I'm having trouble with. I've attached an image of it.

[tex]Emf_1 = 3 V; Emf_2 = 1 V; R_1 = 5 \Omega; R_2 = 2 \Omega; R_3 = 4 \Omega[/tex]

Both batteries are ideal. a.) What is the rate at which energy is dissipated in [tex]R_1[/tex]?

Since [tex]P = i^2R[/tex], I figured that I should find the current. I used the loop rule (I'm not sure if I should have done this or not). I started at a point between [tex]emf_1[/tex] and [tex]R_3[/tex]. I pretended the current was pointed downward at [tex]emf_1[/tex] and upward through [tex]R_1[/tex] and [tex]R_2[/tex]. I called those last two currents [tex]i_1[/tex] and [tex]i_2[/tex], respectively. [tex]i_{total} = i_1 + i_2[/tex].

For the inside (left) loop:

[tex]R_3i_{total}+R_1i_1+emf_1 = 0[/tex]

For the outside loop:

[tex]R_3i_{total} + R_2i_2-emf_2+emf_1 = 0[/tex]

[tex]+R_3(i_1+i_2)+R_1i_1 = -emf_1[/tex]

[tex]-R_3(i_1+i_2)-R_2i_2+emf_2=-emf_1[/tex]

[tex](5\Omega)i_1-(2\Omega)i_2 = -7 V[/tex]

[tex]-(4\Omega)(i_1+i_2)-(2\Omega)i_2 = -3 V[/tex]

[tex]-(4\Omega)i_1-(6\Omega)i_2 = -3 V[/tex]

I solved the system of equations and ended up with [tex]i_1 = 0.95 A[/tex]. This isn't the correct answer, because [tex]i^2R[/tex] ends not equalling 0.346 W, which is the right answer.

I have the feeling I did something really wrong. Did I make it way too complicated? I have a tendency to overdo things :smile:

~Silimay~
 

Attachments

  • 28-31circut.jpg
    28-31circut.jpg
    11.4 KB · Views: 435
Physics news on Phys.org
Hi,

Two things I can see:

- With the way you defined things and the way you are going around your loops, the signs on your voltages should be switched (e.g. first loop eq should be R3itotal1+R1ii = 0).

- I think you made some algebra mistakes after that.

Try again and I think you'll get it.
 


Hi Silimay,

Thank you for sharing your circuit problem and your approach to solving it. It seems like you have the right idea in using the loop rule to find the current, but there are a few things that may have caused your incorrect answer.

Firstly, when applying the loop rule, it is important to make sure that the direction of current and the signs of the emf and resistances are consistent. In your equations, it seems like you have used the opposite sign for the emf_1 term in the first equation, which would result in a different answer for i_1.

Secondly, when solving a system of equations, it is important to double check your calculations and make sure all the variables are accounted for. In your final equation, you have only solved for i_1, but the question asks for the rate at which energy is dissipated in R_1, which involves both the current and the resistance. You will need to use the value of i_1 you have found and the resistance of R_1 to calculate the power dissipated in R_1.

Lastly, it is always a good idea to check your answer using basic physics principles. In this case, you can use the formula P = i^2R to calculate the power dissipated in R_1 and compare it to the answer given. If they do not match, it is likely that there is an error in your calculations.

I hope this helps you to understand where you may have gone wrong in your approach. Remember to always double check your work and use the correct signs and directions when applying equations in physics problems. Keep practicing and you will become more confident in solving circuit problems!
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
2K