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

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