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

[liajohnson]

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, what 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 would be awesome
email me at liajohnson55@hotmail.com

 

[Doc Al]

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, what is the internal resistance of the used battery?

Start by finding the current.

 

[Dr.Brain]

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.