# Internal Resistance of a battery

## Homework Statement

When a 10 ohm load is placed across the terminals of a battery, the terminal voltage is 11.0V. When a 100 ohm load is used instead, the terminal voltage is 11.9V. What is the internal resistance of the battery?

V=IR

## The Attempt at a Solution

I do not know how to approach this question but I do know that I have to use V = IR. I also know that the terminal voltage occurs after the voltage drop from the internal resistance. Is this right? A hint on how to approach this question would be nice! Thank you

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NascentOxygen
Staff Emeritus
Draw the circuit you are dealing with, representing the battery itself as an ideal voltage source in series with a resistor, the resistor value at this stage being unknown, so call it Ri.

Draw the circuit you are dealing with, representing the battery itself as an ideal voltage source in series with a resistor, the resistor value at this stage being unknown, so call it Ri.
I did do that. I feel like this question is impossible?
So, terminal voltage = IR

when using 10ohm resistor
terminal voltage = 11 = I(10)

when using 100ohm resistor
terminal voltage = 11.9 = I(100)

current should be the same in both cases, but when solving for current it isnt the same?

phinds
Gold Member
2019 Award
I did do that. I feel like this question is impossible?
So, terminal voltage = IR

when using 10ohm resistor
terminal voltage = 11 = I(10)

when using 100ohm resistor
terminal voltage = 11.9 = I(100)

current should be the same in both cases, but when solving for current it isnt the same?
Why would you think the current would be the same in both cases?

When Nascent said draw the circuit, what he meant was DRAW THE CIRCUIT. And put it here where we can see it. And show your work relative to the drawing.

also, the answer to this question is 0.917

Why would you think the current would be the same in both cases?

When Nascent said draw the circuit, what he meant was DRAW THE CIRCUIT. And put it here where we can see it. And show your work relative to the drawing.
If the same battery is used, shouldn't the current in both cases be the same?
Also here is a drawing of the circuit after finding the currents for both cases, use the relationship you listed, V=IR. The internal resistance should be equal to $\frac{ΔV}{ΔI}$ in this case. which will get you the answer you had.

• 1 person
phinds
Gold Member
2019 Award
If the same battery is used, shouldn't the current in both cases be the same?
Also here is a drawing of the circuit

View attachment 67052
I continue to be dumbfounded at how you can imagine that would be the case. Do you believe that a battery feeding a million ohm resistor draws as much current as the same battery feeding a 1 ohm resistor?

• 1 person
NascentOxygen
Staff Emeritus
If the same battery is used, shouldn't the current in both cases be the same?
If it is the same battery, then its model will be the same: same ideal voltage, same internal resistance.

Also here is a drawing of the circuit

View attachment 67052
Sorry, my mistake. I should have said, draw the circuits, because as you'll appreciate there are two circuits to deal with here. The first "when a 10 Ω load ..." and another "when a 100 Ω load ...". Two different circuits.

So draw those two circuits, marking on each all the information you know.

NascentOxygen
Staff Emeritus
also, the answer to this question is 0.917
It must have units.

• 1 person
after finding the currents for both cases, use the relationship you listed, V=IR. The internal resistance should be equal to $\frac{ΔV}{ΔI}$ in this case. which will get you the answer you had.
Okay i got the answer, thank you. the internal resistance is 0.917 ohms,
but can you please derive how you got $\frac{ΔV}{ΔI}$ ?

thank you

If anyone still wants to know how here:
V = terminal voltage, E = EMF of battery, I = current from tests (will be different depending of resistances), r = internal resistance.
V = E - Ir....... This eqn states that the voltage exiting and entering the terminals is equal to the total emf the battery would have without the internal resistance, minus the little bit of voltage drop in the battery due to internal resistance.
Now subtract these equations derived from both scenarios (E is a constant so it cancels out)
(V1 = E - I1r) - (V2 = E - I21r)
Now:
V1 - V2 = -I1r + I2r
Factor out the r:
V1 - V2 = (-I1 + I2)r
divide:
(V1 - V2)/(-I1 + I2) = r
or:
∆V/∆I = r
To Find I by the way, use V = IR, make it I = V/R, use the values given in the question, and plug em in.