A circuit contains a 12-volt battery and two 3-ohm bulbs in series.
a. Calculate the total resistance of the circuit.
b. Calculate the current in the circuit.
c. Calculate the power of each bulb.
d. Calculate the power supplied by the battery.

a). Since the bulbs are in series, their total resistance is the sum of their individual resistances. The total resistance in the series circuit is 6 ohms.b). The current in the circuit is (voltage between the ends) / (total resistance) .Current = 12 V / 6 ohms = 2 Amperes.c). The power dissipated by any component can be expressed in different ways,and if we're smart, we pick the formula that makes the problem easy for us.-- Power = (voltage) x (current)-- Power = (current)² x (resistance)-- Power = (voltage)² / (resistance)For this one, I think the middle formula is easiest.Power = (current)² x (resistance) = (2 A)² x (3 ohms) = 12 watts for each bulbd). Each bulb dissipates 12 watts,so for both bulbs, the battery has to supply 24 watts.Check this solution with the first formula for power, above:-- Power = (voltage) x (current)Battery power = (battery voltage) x (current) = (12 V) x (2 A) = 24 watts    Check. yay!