Find emf and internal resistance of a battery

In summary, the conversation discusses the problem of finding the equivalent resistance combination and internal resistance of a battery when a 60 Ω and 120 Ω resistor are connected in parallel. The current through the battery is given for two cases, and using simultaneous equations, the internal resistance is found to be 10 Ω and the emf to be 35 V.
  • #1
Blu3eyes
29
0

Homework Statement


When a 60 [tex]\Omega[/tex] resistor and a 120 [tex]\Omega[/tex] are connected in parallel across a battery there is a current of 700 mA flowing through the battery. When the 120 Ohms resistor is disconnected from the circuit the current through the battery drops to 500 mA. What is the emf and the internal resistance of the battery?

Homework Equations


emf= I(R load + ri)
V terminal = emf - I*ri

The Attempt at a Solution

 
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  • #2
What is the equivalent resistance combination of the 60R||120R load?

Lets call it x and note that it must be less than 60R. Let's call the internal resistance in series with this parallel combo y (and I use R for Ohms).

So we know that yR + xR with a constant battery voltage generates a current of 70mA.

When we remove the 120R we are left with only 60R as the circuit load resistance. So now, the internal resistance yR + 60R generates a current of 500mA? But we know xR must be less than 60R, so the total circuit resistance must have increased, which should cause a decrease in current for a constant voltage.

Are the two current values (70mA and 500mA) correct or is my logic wrong?

Anyhow, to help solve this problem, use a drawing and simultaneous equations!
 

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  • #3
Something's fishy with the currents. If you increase the load resistance (by removing a parallel path -- the 120 Ohm resistor), the current through the circuit should decrease, not jump from 70 mA to 500 mA.
 
  • #4
Zryn said:
What is the equivalent resistance combination of the 60R||120R load?

Lets call it x and note that it must be less than 60R. Let's call the internal resistance in series with this parallel combo y (and I use R for Ohms).

So we know that yR + xR with a constant battery voltage generates a current of 70mA.

When we remove the 120R we are left with only 60R as the circuit load resistance. So now, the internal resistance yR + 60R generates a current of 500mA? But we know xR must be less than 60R, so the total circuit resistance must have increased, which should cause a decrease in current for a constant voltage.

Are the two current values (70mA and 500mA) correct or is my logic wrong?

Anyhow, to help solve this problem, use a drawing and simultaneous equations!

gneill said:
Something's fishy with the currents. If you increase the load resistance (by removing a parallel path -- the 120 Ohm resistor), the current through the circuit should decrease, not jump from 70 mA to 500 mA.

Sorry folks. My bad. It actually is 700 mA and then drops to 500 mA.
Thanks for helping me.
 
  • #5
Blu3eyes said:
Sorry folks. My bad. It actually is 700 mA and then drops to 500 mA.
Thanks for helping me.

Ah. That change makes sense. So, are you good? Have you worked through the problem?
 
  • #6
gneill said:
Ah. That change makes sense. So, are you good? Have you worked through the problem?

I understand the problem now, but still something is not clear to me.
I figured out x=40 R then,
I(y+40)=I(y+60)
Iy+40I =Iy +60I
0 =20Iy
this is where I got stuck, I am not sure I'm heading the right way
 
  • #7
It's also true that Vterminal=I·Rload .

So, EMF = I·Rload + I·rinternal
 
  • #8
Okay. If y is your internal resistance, and if 40 Ω is the equivalent of the 60 Ω and 120 Ω resistors in parallel, and if we let E represent the internal battery voltage, and further, if I1 and I2 are the two currents noted for the two cases, then you've got the two equations:

E = I1(y + 40 Ω)
E = I2(y + 60 Ω)

Can you run from there?
 
  • #9
gneill said:
Okay. If y is your internal resistance, and if 40 Ω is the equivalent of the 60 Ω and 120 Ω resistors in parallel, and if we let E represent the internal battery voltage, and further, if I1 and I2 are the two currents noted for the two cases, then you've got the two equations:

E = I1(y + 40 Ω)
E = I2(y + 60 Ω)

Can you run from there?

Thanks, I've got it. I totally forgot that I was given the two currents, so I was stuck.
After solving everything, ri=10 [tex]\Omega[/tex].
emf=35 V
Thank you folks!
 

1. How do I find the EMF of a battery?

The EMF (electromotive force) of a battery can be found by measuring the potential difference between the positive and negative terminals of the battery using a voltmeter. Make sure the battery is not connected to any external circuit when taking this measurement.

2. What is the formula for calculating the internal resistance of a battery?

The formula for calculating the internal resistance of a battery is:
Rint = (V0 - V1) / I
Where Rint is the internal resistance, V0 is the EMF of the battery, V1 is the potential difference when the battery is connected to a load, and I is the current flowing through the circuit.

3. Can I use any type of voltmeter to measure the EMF of a battery?

No, it is important to use a high-impedance voltmeter (typically with an internal resistance of at least 10 MΩ) to accurately measure the EMF of a battery. Using a low-impedance voltmeter can affect the measurement and give an incorrect value.

4. How does the internal resistance of a battery affect its performance?

The internal resistance of a battery affects its performance by causing a voltage drop when the battery is connected to a load. This voltage drop reduces the amount of usable voltage available to the circuit and can also cause the battery to heat up. A higher internal resistance can result in a decrease in the battery's efficiency and overall performance.

5. Can the internal resistance of a battery change over time?

Yes, the internal resistance of a battery can change over time due to factors such as age, usage, and temperature. As a battery ages, its internal resistance may increase, leading to a decrease in performance. Additionally, extreme temperatures can also affect the internal resistance of a battery.

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