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Two light bulbs each having a resistance of 3ohms are connected in series to a battery marked 12 volts. how much current will flow in the circuit

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jcherry99

Givens

Total Resistance (the bulbs are connected in series)

  • R = r1 + r2
  • r1 = 3 ohms
  • r2 = 3 ohms
  • R = 3 + 3 = 6 ohms
  • E = 12 volts
  • I = ??

Formula

E = I * R

Solution

12 = I * 6         Divide both sides by 6

12/6 = I*6/6

2 = I  

The current = 2 amperes. Answer

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