Kirchoff Rules: Solving Current for Two Batteries in Parallel

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In summary, the problem involves finding the current provided to a wheelchair motor connected to two automobile batteries in parallel. The electromotive force of each battery is 12V and the internal resistance is 0.025Ohms, while the motor's resistance is 1Ohms. The correct answer is 11.5A, but the attempt at solving the equations resulted in 11.9A. After further discussion and trying different equations, it was determined that the equations were set up correctly and the discrepancy may be due to rounding. The same problem was then considered with the batteries and motor connected in series, resulting in a single equation for the current.
  • #1
Alouette
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Problem:
Two automobile batteries are connected in parallel to power a wheelchair. If each of the batteries has electromotive force 12V and internal resistance r=0.025Ohms, and the the wheelchair motor has resistance R=1Ohms, find the current provided to the motor.

Attempt at solution:
So I have emf=12 V, r = 0.025 ohms, and R = 1 ohms...

I set up the current routes such that
I1 + I2 = I3
-I1r1 + emf - I3R = 0
-I2r2 + emf - I3R = 0

Then when solving for I3, I get r(I2 - I1) = I3 *made one "r" since both are same*

Where have I gone wrong in setting up the equations?
 
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  • #2
Can you show how you tried to solve the equations? Just by looking at them I can see that adding your two loop equations together would be promising.
 
  • #3
Actually I'm not even certain now, I think I got tangled between equations yesterday...

I tried again, just adding both like you said, and ended up with
-I1r1 - I2r2 + 2emf - 2I3R = 0

Factored out the r, both I's equal I3,

Then after moving I3 to one side and solving for it, I eventually got I3 = 11.9, but the answer is supposed to be 11.5. Thoughts?
 
  • #4
Alouette said:
Actually I'm not even certain now, I think I got tangled between equations yesterday...

I tried again, just adding both like you said, and ended up with
-I1r1 - I2r2 + 2emf - 2I3R = 0

Factored out the r, both I's equal I3,

Then after moving I3 to one side and solving for it, I eventually got I3 = 11.9, but the answer is supposed to be 11.5. Thoughts?

Your answer looks better. I find I3 = 11.85 A.
 
  • #5
Yup, I just rounded. So was there a mistake in my equations? I'm not that sure they're right since the algebra seems to be fine..
 
  • #6
Alouette said:
Yup, I just rounded. So was there a mistake in my equations? I'm not that sure they're right since the algebra seems to be fine..

Your equations look fine. Could be whoever set the problem rounded down to the nearest .5A (for unknown reasons).
 
  • #7
I'll have to figure what went wrong there.

I'm confused as to how to set up equations for series (same question but instead of parallel) though, would it be the same equations but signs flipped?
 
Last edited:
  • #8
Alouette said:
I'll have to figure what went wrong there.

I'm confused as to how to set up equations for series (same question but instead of parallel) though, would it be the same equations but signs flipped?

No, you'd have a single equation for a single loop.
 
  • #9
So would the equation be
I1r1 + I2r2 + 2emf = I3R?

I'm just trying to think that the current would just accumulate across the batteries, right?
 
  • #10
Alouette said:
So would the equation be
I1r1 + I2r2 + 2emf = I3R?

I'm just trying to think that the current would just accumulate across the batteries, right?
If the batteries and motor are connected in series then there is a single loop and only one current. So change all the various current variables to one variable name.

The same amount of current flows through each component in a series connection.
 
  • #11
Right, current would be one variable, so what about this equation:
-Ir + emf - Ir +emf - IR = 0

then I would just have to find I?
 
  • #12
Alouette said:
Right, current would be one variable, so what about this equation:
-Ir + emf - Ir +emf - IR = 0

then I would just have to find I?

Yes, that should do it.
 
  • #13
Thanks for your help!
 
  • #14
Alouette said:
Thanks for your help!

No problem!
 

FAQ: Kirchoff Rules: Solving Current for Two Batteries in Parallel

What are Kirchoff's rules?

Kirchoff's rules, also known as Kirchoff's laws, are two principles used in circuit analysis: Kirchoff's current law (KCL) and Kirchoff's voltage law (KVL). KCL states that the sum of currents entering a node in a circuit must equal the sum of currents leaving that node. KVL states that the sum of voltage drops around a closed loop in a circuit must equal the sum of voltage sources in that loop.

What is a parallel circuit?

A parallel circuit is a type of electrical circuit in which the components are connected in such a way that there are multiple paths for the current to flow. This means that the current is divided between the different branches of the circuit, and the voltage across each branch is equal.

How do you solve for current in a parallel circuit with two batteries?

To solve for current in a parallel circuit with two batteries, you can use Kirchoff's current law. This law states that the total current entering a node must equal the total current leaving that node. In this case, you can set up a system of equations using the battery voltages and the resistances in the circuit. By solving for the unknown currents, you can determine the current through each branch of the circuit.

What is the difference between series and parallel circuits?

In a series circuit, the components are connected end to end, so that the current has only one path to flow through. This means that the current is the same at all points in the circuit. In a parallel circuit, the components are connected side by side, so that the current can flow through multiple paths. This means that the current is divided between the different branches of the circuit.

How do Kirchoff's rules apply to real-world circuits?

Kirchoff's rules are fundamental principles that apply to all electrical circuits, including real-world circuits. They are used to analyze and solve complex circuits by simplifying them into smaller, more manageable parts. By applying these rules, scientists and engineers can design and troubleshoot various electrical systems and devices, such as power grids, electronic circuits, and more.

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