1) Current in each bulb: 0.1 A
The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:
[tex]R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega[/tex]
And so, the current through the circuit is (using Ohm's law):
[tex]I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A[/tex]
And since the two bulbs are connected in series, the current through each bulb is the same.
2) 4 W and 8 W
The power dissipated by each bulb is given by the formula:
[tex]P=I^2 R[/tex]
where I is the current and R is the resistance.
For the first bulb:
[tex]P_1 = (0.1 A)^2 (400 \Omega)=4 W[/tex]
For the second bulb:
[tex]P_1 = (0.1 A)^2 (800 \Omega)=8 W[/tex]
3) 12 W
The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:
[tex]P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W[/tex]
