What Is the Internal Resistance of the Used Battery in This Circuit Problem?

AI Thread Summary
The discussion revolves around solving a physics problem related to a circuit with a new and a used battery. The new battery has an emf of 6.00 V and an internal resistance of 1.00 ohms, while the used battery has the same emf but a different internal resistance. The resistor R dissipates 1.75 W when the used battery is connected, prompting the need to calculate the internal resistance of the used battery. To find the current through the circuit, Ohm's Law and the power formula P=IV are suggested as methods. The final answer for the internal resistance of the used battery is determined to be 2.57 ohms.
liajohnson
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Hello! I'm just preparing to write my physics 12 final and can't seem to solve a problem form a past exam that i was doing

Here is the question:

a circuit using a new battery which has an emf of 6.00 V and an internal resistnace of 1.00 ohms is shown on the left( In the diagram there also is a current of 0.375 A and a resistor place in series labeled R)
THe battery is then replaced with a used one that has the same emf of 6.00 V bu ta different internal resistance.
If resistor R now dissipates 1.75 W, waht is the internal resistance of the used battery?

(diagram the same but with 1.75 W at the R (resistor in series placed in circuit)

THe answer is 2.57 ohms

i tryed using Vterminal =6.00- (.375)(1) in regards to the orginal battery
and v terninal is then equal to 5.625
You then need to figure out I in the next situiation I'm assuming using Power. P=IV... i don't know i got lost and tryed to work backwards but no dice

any help woudl be awesome
email me at liajohnson55@hotmail.com
 
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liajohnson said:
a circuit using a new battery which has an emf of 6.00 V and an internal resistnace of 1.00 ohms is shown on the left( In the diagram there also is a current of 0.375 A and a resistor place in series labeled R)
Use this information to find the value of R.
THe battery is then replaced with a used one that has the same emf of 6.00 V bu ta different internal resistance.
If resistor R now dissipates 1.75 W, waht is the internal resistance of the used battery?
Start by finding the current.
 
1)Since the resistor R is in series with internal resistance of the battery , add them up in series , apply Ohm's Law to get the value of internal resistance.

2)Power generated by a resistor is given by i^2R

.BJ.
 
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