A. 0.77 A
Using the relationship:
[tex]P=\frac{V^2}{R}[/tex]
where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.
For the first light bulb, P = 60 W and V = 120 V, so the resistance is
[tex]R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega[/tex]
For the second light bulb, P = 200 W and V = 120 V, so the resistance is
[tex]R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega[/tex]
The two light bulbs are connected in series, so their equivalent resistance is
[tex]R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega[/tex]
The two light bulbs are connected to a voltage of
V = 240 V
So we can find the current through the two bulbs by using Ohm's law:
[tex]I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A[/tex]
B. 142.3 W
The power dissipated in the first bulb is given by:
[tex]P_1=I^2 R_1[/tex]
where
I = 0.77 A is the current
[tex]R_1 = 240 \Omega[/tex] is the resistance of the bulb
Substituting numbers, we get
[tex]P_1 = (0.77 A)^2 (240 \Omega)=142.3 W[/tex]
C. 42.7 W
The power dissipated in the second bulb is given by:
[tex]P_2=I^2 R_2[/tex]
where
I = 0.77 A is the current
[tex]R_2 = 72 \Omega[/tex] is the resistance of the bulb
Substituting numbers, we get
[tex]P_2 = (0.77 A)^2 (72 \Omega)=42.7 W[/tex]
D. The 60-W bulb burns out very quickly
The power dissipated by the resistance of each light bulb is equal to:
[tex]P=\frac{E}{t}[/tex]
where
E is the amount of energy dissipated
t is the time interval
From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.
