Answer was 27w.
Let me explain why using Kirchoff's first and second law.
First, (conservation of charge):
6 Amps flowed around the circuit. This implies, because it is parallel, each lamp would get 3A. Using I^2 x R for Power, it would be 3^2 x 3 = 27
Now with 2nd law (conservation of energy)
Voltage produced by battery : 12 V. V = E - Ir E = 12 I = 6 and r = 0.5. 12-6x0.5= 9
Using the equation V^2/R, it is 9^2 / 3 = 27w.
There you go ^^
Let me explain why using Kirchoff's first and second law.
First, (conservation of charge):
6 Amps flowed around the circuit. This implies, because it is parallel, each lamp would get 3A. Using I^2 x R for Power, it would be 3^2 x 3 = 27
Now with 2nd law (conservation of energy)
Voltage produced by battery : 12 V. V = E - Ir E = 12 I = 6 and r = 0.5. 12-6x0.5= 9
Using the equation V^2/R, it is 9^2 / 3 = 27w.
There you go ^^