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
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