The current I ( in Amps) in a circuit with a resistance R (in Ohms) and a battery whose voltage is E and whose internal resistance is r (in Ohms) is given by:
I = E / (R + r)
If R changes at the rate of 0.250 Ohms/min, how fast is the current changing when R = 6.25 Ohms, if E = 3.10 V and r = 0.230 Ohms?
2007-02-25 01:56:06 · 3 answers · asked by babblefish186 3 in Education & Reference ➔ Homework Help
Why is it people always have a smart-*** comment when someone is trying to get homework help in here? Did you happen to notice that the catagory here is HOMEWORK HELP!
2007-02-25 02:36:06 · update #1
assuming E and r are fixed
dI / dt = -{E / (R+r)^2 } * {dR/dt}
with the given values
dI / dt = - {3.10 / (6.25+.230)^2 } * {0.250}
an extra + or - for dR/dt = 0.250 will be good since it is not specified if R is increasing or decreasing
2007-02-25 03:20:36 · answer #1 · answered by mth2006to 3 · 0⤊ 0⤋
Just plug all the numbers in and solve for I, the units should all cancel out and give you and answer in Ohms.
2007-02-25 01:58:28 · answer #2 · answered by crzywriter 5 · 0⤊ 0⤋
Have you tried it yourself first or are you looking for a swift ride with your homework?
2007-02-25 01:59:16 · answer #3 · answered by Sami V 7 · 0⤊ 0⤋