Batteries in series, parallel, and internal resistance

[ddobre]

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.

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?

 

[haruspex]

identical batteries

 

[ddobre]

If the batteries are identical then they should all have the same internal resistance, as well as current flowing through them.

 

[haruspex]

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?

 

[ddobre]

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)

 

[haruspex]

for the parallel case, that r123 = P/(3I²)

In what resistor is P dissipated?

 

[ddobre]

In what resistor is P dissipated?

The power dissipated must have come mainly from the larger resistor, R

 

[haruspex]

The power dissipated must have come mainly from the larger resistor, R

The power dissipated by the resistor is measured to be P.

 

[ddobre]

So if the power is dissipated in R, then there is some relationship between R and the internal resistances of the batteries.

 

[haruspex]

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?

 

[ddobre]

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

 

[haruspex]

So the current through R would be equal to, using V = IR,

What about using the given variables?

 

[ddobre]

What about using the given variables?

It would be equal to: R = P/(I₁+I₂+I₃)

 

[ddobre]

It would be equal to: R = P/(I₁+I₂+I₃)

But each I would be I²

 

[haruspex]

It would be equal to: R = P/(I₁+I₂+I₃)

You forgot the ²

But each I would be I²

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

 

[ddobre]

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

 

[haruspex]

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?

 

[ddobre]

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

 

[haruspex]

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.

 

[ddobre]

Three r resistances are parallel, but they are in series with R.

In that case, Req = (3P + rI²)/(3I²)

 

[haruspex]

In that case, Req = (3P + rI²)/(3I²)

Yes.
So what is the zero-load voltage of each battery?

 

[ddobre]

If the system is open, then Vab = emf

 

[ddobre]

If the system is open, then Vab = emf

Each battery should have the same voltage because of the parallel configuration

 

[ddobre]

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

 

[cnh1995]

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?