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Electric Circuit Problem - Help Asap Final In 3 Days

  • Thread starter liajohnson
  • Start date
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 dunno i got lost and tryed to work backwards but no dice

any help woudl be awesome
email me at liajohnson55@hotmail.com
 

Doc Al

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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 [itex]i^2R[/itex]

.BJ.
 
Last edited:

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