Physics circuit question..?

Question

4) A battery with and EMF of 1.900V is connected to an external resistance of 120.0W and a current of 15.80 mA flows. What is the internal resistance of the battery?

oh i pasted the wrong problem the problem is actually

5) What is the internal resistance of a 12 V car battery whose terminal voltage drops to 8.4 V when the starter uses 75 A?

Answer 1

answer for the question # 4)

according to KVL we have

Vo=Vb-R.I ;
[Vo = outer voltage =8.5(v)
Vb = battery voltage =12(v)
R= battery's internal resistantce = ?
I = current of flow = 75 (A) ]

so we have :

8.5 = 12 - R*75 ==> R = (12-8.5)/75 (ohm) ==> R = 0.047 (ohm)
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Answer 2

Assuming you meant the total battery EMF to be 1.9 V and the external resistance to be 120 ohms. The sum of the voltages drops around a loop is equal to zero. Then 1.9V -[internal resistance x .0158A] - [120 x .0158A] = 0. Solving this equation results in an internal resistance of .253 ohms. I hope this is correct. It has been a while since I did any home work.

Answer 3

Assuming you mean the external resistance to be 120 OHMs, the problem is:
(1.90)V = (120.0 + R)(15.80) 10**(-3)

{1.90/15.80} 1000 = ( R + 120.0 )= 120.25 or R = 0.25 OHM

Now, of course, you have quoted the voltage to only three significant figures and, although you have quoted the resistance and amperage to four, the result is not going to be better than three ( or four if you assume 1.900 ) so you really can't determine the internal resistance to better than 0.25 +/- 0.5 .

Answer 4

Figure out the total resistance from ohms law.

Rtotal = V / I

Now Rtotal = Rext + Rint

You can do the math.