Since the 3 lamps are connected in series, their equivalent resistance is equal to the sum of the three resistors:
[tex]R_{eq} =R_1+R_2+R_3=1.2 \Omega + 1.2 \Omega +1.2 \Omega = 3.6 \Omega[/tex]
And the total current flowing in the circuit can be calculated by using Ohm's law:
[tex]I= \frac{V}{R_{eq}} [/tex]
where V is the voltage of the battery. By using V=3 V, we find
[tex]I= \frac{3 V}{3.6 \Omega}= 0.83 A[/tex]
