A60-w, 120-v light bulb and a 200-w, 120-v light bulb are connected in series across a 240-v line. assume that the resistance of each bulb does not vary with current. (note: this description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-w, 120-v light bulb dissipates 25 w when connected to a 120-v line.) a. find the current through the bulbs. b. find the power dissipated in the 60 w bulb. c. find the power dissipated in the 200 w bulb. d. one bulb burns out very quickly. which one? 60-w bulb, 200-w bulb.

A. 0.77 A

Using the relationship:

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

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

The two light bulbs are connected in series, so their equivalent resistance is

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:

B. 142.3 W

The power dissipated in the first bulb is given by:

where

I = 0.77 A is the current

is the resistance of the bulb

Substituting numbers, we get

C. 42.7 W

The power dissipated in the second bulb is given by:

where

I = 0.77 A is the current

is the resistance of the bulb

Substituting numbers, we get

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

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.

battery with hi vtge high val resist in series