- #1

- 11

- 0

## 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.