# Solve Battery Parallel Problem: Get Help Now

• pantera1441
In summary, the problem is trying to find the current through a series circuit with unknowns like the battery voltages. You solve for the current by dropping voltage drops across the batteries and resistors.
pantera1441
anyone help me out with this probelem?

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What you want to do is to attack this problem by using nodal equations, as you can see from the drawing this circuit has only one node placed in the junction between the three resistors, you can start by assuming the following:

1. By KCL the current at the node is equal to 0

Since we don't know anything about the direction of the currents, we can safely let the node voltage determine the direction.

2. Let the $$V_{x}$$ denote the node voltage
Let $$I_{1}$$ denote the current from 28.3 V source
Let $$I_{2}$$ denote the current from 14.15 V source
Let $$I_{3}$$ denote the current in the 16.5Ohm resistance

3. Applying the nodal equations we get the following:

$$I_{1}+I_{2}+I_{3}=0$$

Note that the current can take any direction. Now we replace currents with voltage drops divided by resistor values:

$$\frac{V_{x}-28.3}{30.6}+\frac{V_{x}-14.15}{4.32}+\frac{V_{x}}{16.5}=0$$

Solving for $$V_{x}$$ we get 12.9

Now we can solve for the current in the leftmost resistor:

$$I_{3}=\frac{V_{x}}{16.5}=0.784A$$

DONE!

thanks a lot for the help...
much obliged!

pantera1441 said:
thanks a lot for the help...
much obliged!

No problem, I hope you understood the steps towards the solution.

Yeah, you use KCL bc all the current is conserved, and you find the actual current that enters the bottom resistor, after that you can find the current across any of the resistors, using the way you set it up

In the future, it is better to not just do the problem for the OP but to give little hints to push him/her in the right direction...

pantera1441 said:
Yeah, you use KCL bc all the current is conserved, and you find the actual current that enters the bottom resistor, after that you can find the current across any of the resistors, using the way you set it up

Correct and remember if the batteries were switched the other way around, then you would use $$V_{x}+V_{battery}$$ instead

russ_watters said:
In the future, it is better to not just do the problem for the OP but to give little hints to push him/her in the right direction...

I won't argue about that, but i thought that making the complete description of the steps including the description of what rules i applied would help him better, so he/she will know what to do next time he/she is faced with the problem.

do you think you can "push me in the right direction" on this one, I know they're looking for the difference between A and B, but i forget how to do the voltage drop and I'm clueless when there's more than one battery

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pantera1441 said:
do you think you can "push me in the right direction" on this one, I know they're looking for the difference between A and B, but i forget how to do the voltage drop and I'm clueless when there's more than one battery
Hehe, shure. First you have to determine the current flowing through the loop, note that the current will be the same at all places. When You've determined the current in the loop, start by writing the equation expressed as voltage drops , you should start at point a and move towards point b, note that the first voltage drop across the battery will be negative, since you're moving from -to +, and it will be positive across the resistors and positive across the battery that is near the b point, since the battery is going from + to -.

do you mean a positive drop...I know the current is 3.614mA, and that it drops almost 16V in between the two batteries, but what do I do about the negative voltage on the other side of the bottom right battery, add it to get a negative number?

I tried 40.86V and that was wrong, is it 25V - 15.86V = 9.14V as the difference

Last edited:
As all the components are in series you can rearrange the circuit to group all the cells and all the resistors.
This should make it easier to work out the circuit voltage and hence the current.

got it...take the current of -3.614mA and multiply it by the 2 resistors added together on the bottom to get -15.86V

Thanks,

You'll be hearing from me again this week about RC Circuits...totally new to me

LOL, ok ;)

## 1. What is a battery parallel problem?

A battery parallel problem occurs when two or more batteries are connected together in parallel to increase the total energy output. However, this can lead to issues such as unequal charging, which can cause damage to the batteries and decrease their overall lifespan.

## 2. How can I solve a battery parallel problem?

The best way to solve a battery parallel problem is to seek help from a qualified professional. They will be able to assess the situation and provide a solution that is tailored to your specific needs. This may involve adjusting the wiring, replacing damaged batteries, or implementing a different charging method.

## 3. Can I solve a battery parallel problem on my own?

While it is possible to try and solve a battery parallel problem on your own, it is not recommended. Without proper knowledge and experience, you may end up causing further damage to the batteries or putting yourself at risk. It is always best to seek help from a professional.

## 4. How long does it take to solve a battery parallel problem?

The time it takes to solve a battery parallel problem will vary depending on the severity of the issue and the complexity of the solution. In some cases, it may only take a few hours to resolve the problem, while in others it could take days. It is important to be patient and allow the professional to properly address the issue.

## 5. How can I prevent battery parallel problems in the future?

The best way to prevent battery parallel problems is to properly maintain and care for your batteries. This includes following the manufacturer's instructions for charging and storing them, regularly checking for any damage or wear, and replacing them when necessary. It is also important to seek professional help if you notice any issues with your batteries.

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