Asked by Spencer

An ammeter is connected in series to a battery of voltage V_b and a resistor of unknown resistance R_u The ammeter reads a current I_0. Next, a resistor of unknown resistance R_r is connected in series to the ammeter, and the ammeter's reading drops to I_1. Finally, a second resistor, also of resistance R_r, is connected in series as well. Now the ammeter reads I_2.

If I_1/I_0= 4/5, find I_2/I_0.

express numerically. I hate when they ask us this stuff without numbers. I could do it if I had the resistances.
Answered by bobpursley
Its not so difficult...

Io=Vb/Ru I1=Vb/(Ru+Rr)

so IoRu=I1(Ru+Rr)

then 4/5=Ru/(Ru+Rr) or solving...
Ru=4Rr. or Rr=1/4 Ru. If you add a second Ru, then Rtotal=Ru+ .5Ru or 3/2 Ru

I2/Io= Ru/(1.5Ru)= 2/3

check my thinking
Answered by Spencer
wow your right i feel stupid. i thought the other resistor added was different. now it makes sense thanks.
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