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Three resistors with resistances of 5 ohm, 9 ohm, and 12 ohm, respectively, are connected in a series circuit with a 6.0 V battery. Find the potential difference across the 5 ohm resistor
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if R1 and R2 are the resistances connected in series then effective resiatcne Rnet = R1+R2

if R1 and R2 are connected in parallel, then 1/Rnet =1/R1 +1/R2 or Rnet =   R1R2/(R1+R2)

same current passes through each of the ressitor in series combination and
same votage passes through eacf of the resistor in parallel combination

apply ohms law V = iR
where V = volatge and i = current

so here Rnet = 5+9+12 = 26 ohms

i = V/Rnet = 6/26 = 0.2307 Amps

so Vacross 5ohms = iR = 0.2307 * 5 = 1.153 Volts ------answer

other volatages are

V acrosss 9 ohms = 9*0.2307 = 2.0763 Volts

V across 12 ohms = 12* 0.2307 = 2.7684 Volts
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