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assume the following configuration: battery connected to 2 parallel resistors with an ammeter in series with the battery... observe the current measurement ... remove one of the resistors .... observe the current again, it will have decreased: if the resistors were of equal value, the current will decrease to half of its original value when one of the resistors is removed. Mathematics:

V=IR (V- voltage, I - current, R - resistance

in a parallel circuit, R=(R1*R2)/(R1+R2) where R1 and R2 are the values of resistance of the resistors. Before removal- Ib=V*(R1+R2)/(R1*R2)

After removal (assume R2 is removed)- Ia=V/R1

so Ia/Ib=(R1*R2)/(R1*(R1+R2))

or Ia=Ib*(R2/(R1+R2)

if R1=R2 then Ia=Ib*R2/(2*R2) or Ia=Ib/2 so the current after is 1/2 of that before.

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