- #1
dhkdeoen
- 11
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Homework Statement
In the figure R1 = 1.29R, the ammeter resistance is zero, and the battery is ideal. What multiple of ε/R gives the current in the ammeter?
Homework Equations
V=IR, 1/R=1/R_1+1/R_2..(when resistances are parallel)
etc
The Attempt at a Solution
since ammeter doesn't have resistance, two top resistances and two bottom resistances are basically parallel. First, I set the currents coming out from the battery is i, and I was getting R_eq
R_top: R_1+R=(119/229)R
R_bottom: R/2
R_eq=R_top+R_bottom=(467/458)R
set the voltage as ε
ε=(467/458)*iR
ε=V_top+V_bottom=V_top+iR
(since two bottom resistances are same, so V_bottom=iR)
(467/458)*iR=V_top+iR
V_top=(467/458)*iR-iR=(9/458)*iR
and
i_topleft=V_top/1.29R=(9/458)iR*1/(1.29R)=(0.015233...)*i
i_A=current through ammeter
i_A+i_topleft=i
i_A=i-(0.015233..)i=(0.9847..)i
so I got 0.9847.. for the answer, which is wrong. where did I go wrong? I spent like 3 hours on this problem, and still can't get it.