Batteries in series, parallel, and internal resistance

In summary: Yes.So what is the zero-load voltage of each...I'm unsure about the zero-load voltage. So if no current is flowing through the circuit, resistance would be infinite.If no current is flowing through the circuit, resistance would be infinite.
  • #1
ddobre
33
2

Homework Statement


Three identical batteries are first connected in parallel to a resistor. The power dissipated by the resistor is measured to be P. After that, the batteries are connected to the same resistor in series and the dissipated power is measured to be 4P (four times larger than for the parallel connection). Find the internal resistance of each of the batteries if the resistor has resistance R.
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Homework Equations


P = IV = I^2R = V^2/R
V = IR
(Parallel): 1/Req = 1/R1 + 1/R2
(Series): Req = R1 +R2

The Attempt at a Solution


With this problem, it is the fact that instead of resistors, I have batteries with resistances that confuses me. Nevertheless, I tried using the standard parallel resistor equation to solve for each of the individual internal resistances of the batteries. This is what I tried to do, knowing current would vary in parallel: P = I^2R, P = (I1)2 * r1. Then I would solve for r1. Similarly, I tried P = V^2/R to find the resistances of the series circuit: P = V2/r1. But I can't help thinking something is missing is wrong. I have a feeling that since I'm analyzing battery internal resistances, these equations won't work. Any advice?
 

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  • #2
ddobre said:
identical batteries
 
  • #3

If the batteries are identical then they should all have the same internal resistance, as well as current flowing through them.
 
  • #4
ddobre said:
If the batteries are identical then they should all have the same internal resistance, as well as current flowing through them.
Right. Does that get you there?
 
  • #5
haruspex said:
Right. Does that get you there?

So what I've found is that if they are all identical, then r1 = r2 = r3 = r, and the total current in parallel is the sum of the currents through each battery, so i1 + i2 + i3 = I total. In conjunction with the equation P = I^2R, I should find, for the parallel case, that r123 = P/(3I2)
 
  • #6
ddobre said:
for the parallel case, that r123 = P/(3I2)
In what resistor is P dissipated?
 
  • #7
haruspex said:
In what resistor is P dissipated?

The power dissipated must have come mainly from the larger resistor, R
 
  • #8
ddobre said:
The power dissipated must have come mainly from the larger resistor, R
ddobre said:
The power dissipated by the resistor is measured to be P.
 
  • #9

So if the power is dissipated in R, then there is some relationship between R and the internal resistances of the batteries.
 
  • #10
ddobre said:
there is some relationship between R and the internal resistances of the batteries.
Not directly.
What is the current through R in the parallel case?
What equations can you write regarding voltage drops?
 
  • #11
haruspex said:
Not directly.
What is the current through R in the parallel case?
What equations can you write regarding voltage drops?
In parallel, the current is the sum of the currents through each component. So some current goes through r1, r2, r3, and the sum of these currents reaches R. So the current through R would be equal to, using V = IR, V/(I1+I2+I3). And Voltage remains constant in the parallel case. But, since there is internal resistance, there will be a slight voltage drop determined by Vab = (emf) - Ir
 
  • #12
ddobre said:
So the current through R would be equal to, using V = IR,
What about using the given variables?
 
  • #13
haruspex said:
What about using the given variables?

It would be equal to: R = P/(I1+I2+I3)
 
  • #14
ddobre said:
It would be equal to: R = P/(I1+I2+I3)

But each I would be I2
 
  • #15
ddobre said:
It would be equal to: R = P/(I1+I2+I3)
You forgot the 2
ddobre said:
But each I would be I2
I don't think that is what you meant. What is the relationship between I1 and I?
 
  • #16
haruspex said:
You forgot the 2

I don't think that is what you meant. What is the relationship between I1 and I?

It should be: R = P/(I1+I2+I3)2
 
  • #17
ddobre said:
It should be: R = P/(I1+I2+I3)2
Right, where those three currents each equal I/3.
That gives you an expression for the currents.
What is the total resistance of the circuit? What zero-load voltage does that imply for the each battery?
 
  • #18
If I use the equation for parallel resistors for the batteries, and they each have the same internal resistance, I end up with r/3, which would leave only two resistors in the circuit. But from the diagram, I think that they would be parallel to each other. With R equal to the above equation, and the sum of the currents equal to I: Req = 3I2/(3P +rI2).

I'm unsure about the zero-load voltage. So if no current is flowing through the circuit, resistance would be infinite
 
  • #19
ddobre said:
from the diagram, I think that they would be parallel to each other.
Three r resistances are parallel, but they are in series with R.
 
  • #20
haruspex said:
Three r resistances are parallel, but they are in series with R.

In that case, Req = (3P + rI2)/(3I2)
 
  • #21
ddobre said:
In that case, Req = (3P + rI2)/(3I2)
Yes.
So what is the zero-load voltage of each battery?
 
  • #22
If the system is open, then Vab = emf
 
  • #23
ddobre said:
If the system is open, then Vab = emf

Each battery should have the same voltage because of the parallel configuration
 
  • #24
ddobre said:
Each battery should have the same voltage because of the parallel configuration

Could I solve for r using the equation I derived for the total resistance by relating it to P = V2/R, and then setting Req = V2/P
 
  • #25
ddobre said:
Could I solve for r using the equation I derived for the total resistance by relating it to P = V2/R, and then setting Req = V2/P
Let's say the emf of each battery is E.
i)For parallel connection:
Suppose the voltage across R is V.
Find the expression for V in terms of E, R and r (using nodal analysis).

ii)For Series connection:
Suppose the voltage across R is V1.
Find the expression for V1 in terms of E, R and r (simple series circuit).

iii)What is the relation between V and V1?
 

Related to Batteries in series, parallel, and internal resistance

1. What is the difference between batteries in series and parallel?

Batteries in series are connected end to end, with the positive terminal of one battery connected to the negative terminal of the next. This increases the voltage of the overall circuit, but the current remains the same. Batteries in parallel are connected side by side, with all positive terminals connected to each other and all negative terminals connected to each other. This increases the current of the overall circuit, but the voltage remains the same.

2. How do batteries in series affect the overall voltage?

When batteries are connected in series, their voltages add together. For example, if two 1.5V batteries are connected in series, the overall voltage will be 3V. This is because the positive terminal of one battery is connected to the negative terminal of the other, effectively creating a larger battery with double the voltage.

3. What is the purpose of connecting batteries in parallel?

Connecting batteries in parallel increases the overall current of a circuit. This is useful for applications that require a higher current, such as powering electronic devices or starting a car.

4. How does internal resistance affect the performance of batteries?

All batteries have some amount of internal resistance, which is the resistance to the flow of electrons within the battery itself. This resistance can cause the battery to lose some of its voltage, resulting in a decrease in overall performance. As batteries age, their internal resistance increases, causing them to hold less charge and have a shorter lifespan.

5. Is it better to connect batteries in series or parallel?

The answer depends on the specific application. Connecting batteries in series is ideal for increasing voltage, while connecting them in parallel is better for increasing current. It is important to consider the needs of the circuit and choose the appropriate battery configuration to meet those needs.

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