Finding Voltage Using Junction & Loop Rules

In summary,The voltage between a and d and the terminal of each battery can be found using three equations.
  • #1
Pao44445
47
0

Homework Statement


find the voltage between a and d and terminal of each batteries
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Homework Equations


Junction Rule and Loop Rule

The Attempt at a Solution


I got three equation
JR : I3 = I1+I2
LR :
45-I3-47I3-34I1 = 0
45-I3-47I3+85-I2-18I2 = 0
I can't solve these variables :(
 
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  • #2
Pao44445 said:
I can't solve these variables :(
There are three equations with three unknowns. You can use the substitution method.
 
  • #3
You have 3 linear equations in 3 unknowns (I1, I2, I3). So the equations can be solved unless they are self-contradictory.
I assume that you have not dealt with matrices yet, but you can still solve these.
1) Use the first equation to replace every I3 in the other 2 equations.
2) Solve the modified second equation for I2 in terms of I1.
3) Use that result to replace all the I2s in the modified last equation.
4) Solve the modified last equation for I1. That should give you a number for I1.
5) Plug the I1 number into the result of step 2 to get a number for I2.
6) Plug the I1 and I2 numbers into the first equation to get a number for I3.
So now you have the values of all I1, I2, I3.
 
  • #4
ok, I give up, is there other way to solve this?
some says I can actually use ohm's law directly

Vad=(ΣI) (Σ1/R)
 
  • #5
Pao44445 said:
ok, I give up, is there other way to solve this?
some says I can actually use ohm's law directly

Vad=(ΣI) (Σ1/R)
There are other methods of analysis, some of which will yield fewer equations to solve. But you haven't learned of these methods yet, and they won't be taught until you've learned the fundamentals of Kirchhoff's laws. All other methods are based upon this foundation.

But most importantly, in general all methods involve solving simultaneous equations when the circuit is more complex than a pair of nodes. You won't be able to avoid this; You will have to become familiar with solving simultaneous equations.

Regarding what "some says", you can't use Ohm's law directly to solve this circuit, but there is a method that incorporates Ohm's law along with Kirchhoff's laws. An analysis technique that you will eventually learn of is called Nodal Analysis. It is based upon a clever use of Kirchhoff's laws to find the potentials at the nodes of a circuit, and involves writing equations for each essential node in a circuit (you'll learn what an essential node is, too), one equation for each node, and solving the simultaneous equations to find the potentials at each node with respect to a reference node. Again, simultaneous equations are involved in general, but in the case of this particular circuit there is only one essential node!

So for this circuit in particular it is possible to write a single equation with one unknown to find the potential of the one essential node with respect to a reference node. From there you could find the individual currents in the branches connecting the essential node to the reference node. Rather than solving simultaneous equations you would solve a single equation for one node potential and then do several individual calculations to find currents and other potentials around the circuit. It would be more work overall to find the branch currents, but would involve no simultaneous equations (for this particular circuit).
 
  • #6
You can do it by hand. If you study matrices and linear algebra, there are ways to automate the process.
FactChecker said:
You have 3 linear equations in 3 unknowns (I1, I2, I3). So the equations can be solved unless they are self-contradictory.
I assume that you have not dealt with matrices yet, but you can still solve these.
1) Use the first equation to replace every I3 in the other 2 equations.
Modify Equ 2: 45-I3-47*I3-34*I1 = 0;
simplify: -48*I3 - 34*I1 = -45;
use equation 1: -48*(I1 + I2) -34*I1 =45;
simplify: -82*I1 - 48*I2 = -45;
Modify Equ 3: 45-I3-47*I3+85-I2-18*I2 = 0;
simplify: -47*I3 - 19*I2 = -130;
use equation 1: -47*(I1 + I2) -19*I2 = -130;
simplify: -47*I1 - 66*I2 = -130;
2) Solve the modified second equation for I2 in terms of I1.
I2 = (-45 + 82*I1) / -82 = 45/82 - I1 = 0.55 - I1
3) Use that result to replace all the I2s in the modified last equation.
Modified last equ: -47*I1 - 66*I2 = -130;
replace I2s: -47*I1 - 66*(0.55 - I1) = -130;
simplify: 19*I1 + 36.3 = -130
4) Solve the modified last equation for I1. That should give you a number for I1.
19*I1 + 36.3 = -130;
19*I1 = -166.3;
I1 = -166.3/19 = -8.75
5) Plug the I1 number into the result of step 2 to get a number for I2.
I2 = 0.55 - I1 = 0.55 - (-8.75) = 9.30
6) Plug the I1 and I2 numbers into the first equation to get a number for I3.
I3 = I1 + I2 = -8.75 + 9.30 = 0.55
So now you have the values of all I1, I2, I3.
I1 = -8.75;
I2 = 9.30;
I3 = 0.55

You will have to double check all the steps and calculations. It is always good to plug the final answers into the original equations to check that they work and there were no calculation mistakes.
 
  • #7
FactChecker said:
I1 = -8.75;
I2 = 9.30;
I3 = 0.55

You will have to double check all the steps and calculations. It is always good to plug the final answers into the original equations to check that they work and there were no calculation mistakes.
I can affirm that these values are not correct :smile:

And FactChecker, you should be aware that doing the work for the member requesting help is against the rules. We can only guide the member to a solution via hints, suggestions for research, pointing out errors, and so forth. They have to do the heavy lifting themselves.
 
  • #8
gneill said:
I can affirm that these values are not correct :smile:

And FactChecker, you should be aware that doing the work for the member requesting help is against the rules. We can only guide the member to a solution via hints, suggestions for research, pointing out errors, and so forth. They have to do the heavy lifting themselves.
Sorry. I will limit my answers more in the future. I wanted to give a detailed outline of the substitution method. I knew that the answer was wrong and said that he should check the work so that he would go through it in detail and correct it. I know the correct values.
 
  • #9
FactChecker said:
Sorry. I will limit my answers more in the future. I wanted to give a detailed outline of the substitution method. I knew that the answer was wrong and said that he should check the work so that he would go through it in detail and correct it. I know the correct values.
A better approach would be to present a correct but unrelated example, or steer the member to a tutorial on the web.
 
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1. What are junction and loop rules?

Junction and loop rules are fundamental principles used in circuit analysis to determine the voltage and current at different points in a circuit. The junction rule, also known as Kirchhoff's current law, states that the sum of currents entering a junction must equal the sum of currents leaving the junction. The loop rule, also known as Kirchhoff's voltage law, states that the sum of voltage drops in a closed loop must equal the sum of voltage rises in that loop.

2. How do I use junction and loop rules to find voltage?

To find voltage using junction and loop rules, you must first draw a circuit diagram and label all the known and unknown values. Then, using the junction rule, you can write an equation for each junction in the circuit. Similarly, using the loop rule, you can write an equation for each closed loop in the circuit. Solving these equations simultaneously will give you the values of the unknown voltages.

3. Can junction and loop rules be applied to any type of circuit?

Yes, junction and loop rules can be applied to any type of circuit, whether it is a simple circuit with just a few components or a complex circuit with multiple branches and loops. These rules are based on fundamental principles of electricity and are universally applicable.

4. Are there any limitations to using junction and loop rules?

While junction and loop rules are powerful tools for circuit analysis, they do have some limitations. These rules assume that the circuit is in a steady state and that the components are ideal, without any internal resistance. In reality, circuits may have non-ideal components and may not be in a steady state, which can affect the accuracy of the results obtained using junction and loop rules.

5. Is there a simpler method for finding voltage in a circuit?

Yes, there are other methods for finding voltage in a circuit, such as using Ohm's law or using a voltmeter to directly measure the voltage at a specific point. However, junction and loop rules are often the preferred method as they allow for the analysis of complex circuits and can be applied to circuits with non-ideal components and varying current and voltage values.

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