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.

good for her? ? where is the question?

temp rises, and expansion