How to solve for EMF and internal resistance in a circuit?

In summary, you solved for the current in the closed circuit by multiplying the current through the battery by the resistance of the closed circuit (1.565). This gave you a current of 5.51. Next, you solved for the current in the open circuit by multiplying the current through the battery by the resistance of the open circuit (1.636). This gave you an current of 8.18. Finally, you used these two values to set up the equations using V=E-ir.
  • #1
np115
5
3
Homework Statement
What is the emf of the battery in the figure(Figure 1)?
What is the internal resistance of the battery in the figure?
Relevant Equations
V=iR
V=E-ir
I know the answers are 9 V and .5 ohms but I have no idea how to get there. Originally I solved for Delta V in the open circuit through (1.636)(5) which gave me 8.18V. Then, I solved for Delta V in the closed circuit through (1.565)(1/((1/5)+(1/10)) which gave me 5.51 V. I used these voltages to set up the equations using V=E-ir. *this is something that really confuses me! is this equation referring to the total resistance in the circuit or just the internal resistance in the battery? If it is total, how would I put this into the equation?* Using that equation I got 8.18 V = E -(1.636 (r+5)) and 5.21 = E - (1.565(r+3.33)). Then, I used elimination to solve but got really confused because things do not add up. I have a quiz tomorrow over this and I have NO IDEA WHAT I"M DOING! Really appreciate some help!
 

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  • #2
np115 said:
I know the answers are 9 V and .5 ohms but I have no idea how to get there. Originally I solved for Delta V in the open circuit through (1.636)(5) which gave me 8.18V.
How did you do this? You don't know R of the open circuit because you don't know r of the battery. So it's V=IR=1.565*(5+r), which can't yet be solved.
Similar situation for the closed circuit.

Try setting up your two voltage equations and making them equal to each other. That gets rid of V, leaving you with two equations and 1 unknown variable, which is solvable.
 
  • #3
wait why would it be 1.565(r+5)? that's the current value for the open circuit. So shouldn't it be 1.636(r+5)? and when I solved that by setting both equations equal to each other, i got r=-41.8 which doesn't make any sense. And when I tried solving it with your equations, I got r=33.45.
 
  • #4
Could you solve for the two currents if instead you were told the battery V and r and not the two currents?
 
  • #5
I'm sure you could since the change in voltage for both circuits would still be the same so if you put both equations equal to each other, you could solve for i.
 
  • #6
np115 said:
ait why would it be 1.565(r+5)? that's the current value for the open circuit. So shouldn't it be 1.636(r+5)?
My apologies, I mixed up the values and for some reason thought the ammeter was located on the top wire before it branched off to the 5 ohm resistor. Getting only 1 current value for the 2-branch circuit makes things much more difficult. It's been a while since I did basic circuits. Let me give it some thought and get back to you.
 
  • #7
Ok, here we go.
First, find all the info you can for your first circuit (when switch is open). You know the current, so you know the voltage across the resistor.

Now, do the same thing for the circuit when the switch is closed. Notice that when the switch is closed, you have ##R_5## and ##R_{10}## in parallel, and both are in series with ##R_1##, the battery's resistor. Remember that parallel resistors have something in common (hint: it starts with a V). We want the current through both branches (we have one already).

Once you have the current through both branches, you know the current through the battery for both circuits (they gave us the current for the first one). Use these, try some KVL and Ohm's law, and if you need more help let me know.
 
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  • #8
I got it, thank you guys so much for your help!
 
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  • #9
np115 said:
I got it, thank you guys so much for your help!
Np. Sorry for the confusion at first. My EE skills are very rusty.
 

1. What is EMF in a circuit?

EMF stands for electromotive force, which is the energy per unit charge that is supplied by a source, such as a battery, to move electric charge through a circuit.

2. How do you calculate EMF in a circuit?

EMF can be calculated by multiplying the current in the circuit by the resistance of the circuit. This is known as Ohm's law: E = IR, where E is the EMF, I is the current, and R is the resistance.

3. What is internal resistance in a circuit?

Internal resistance is the resistance within a source, such as a battery, that opposes the flow of current. It is caused by the physical properties of the source and can affect the overall performance of the circuit.

4. How do you determine the internal resistance in a circuit?

The internal resistance can be determined by measuring the voltage across the source when there is no load connected, and then when a load is connected. The difference in voltage can be used to calculate the internal resistance using Ohm's law: r = (E1-E2)/I, where r is the internal resistance, E1 is the voltage without the load, E2 is the voltage with the load, and I is the current.

5. How does the internal resistance affect the EMF in a circuit?

The internal resistance can affect the EMF in a circuit by causing a voltage drop and reducing the overall output of the source. This means that the EMF measured across the source will be less than the theoretical value calculated using Ohm's law. The internal resistance also affects the maximum current that can be drawn from the source.

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