Dissipation, resistance, and power.

[exohemmx3]

Homework Statement

A string of 19 identical Christmas tree lights
are connected in series to a 120 V source. The
string dissipates 60 W.
What is the equivalent resistance of the
light string? Answer in units of ohms.
What is the resistance of a single light? An-
swer in units of .

What power is dissipated in a single light?
Answer in units of W.

One of the bulbs burns out. The lamp has a
wire that shorts out the bulb filament when it
burns out, dropping the resistance of the bulb
to zero.
What is the resistance Rnew of the light
string now? Answer in units of ohms.

Find the power Pnew dissipated by the string
now. Answer in units of W.

Homework Equations

P=IV
I=V/R
P=(V/R)V
R=V2/R

The Attempt at a Solution

I'm not sure how to start, since the dissipation of power is in watts. thanks for any hints or help you can give !

 

[LowlyPion]

I'm not sure how to start, since the dissipation of power is in watts. thanks for any hints or help you can give !

Welcome to PF.

Well you have Power dissipated, and you have the Voltage don't you?

Can't you figure the resistance of the 19 bulbs together?

And if then the 19 become 18 what happens if figure power again from the same formula?

 

[nlightner]

I would approach this problem by first understanding the concepts of dissipation, resistance, and power. Dissipation refers to the loss of energy in a system, usually in the form of heat. Resistance is a measure of how much a material or component impedes the flow of electrical current. Power is the rate at which energy is transferred or used.

In this problem, we are given a string of 19 Christmas tree lights connected in series to a 120 V source. We are also given that the string dissipates 60 W of power. From this information, we can use the equation P=IV to calculate the current (I) flowing through the string. Rearranging this equation, we get I=P/V. Plugging in the values, we get I=60/120=0.5 A.

To find the equivalent resistance of the light string, we can use the formula R=V/I. Plugging in the values, we get R=120/0.5=240 ohms.

The resistance of a single light can be found by dividing the equivalent resistance of the string by the number of lights, which in this case is 19. So the resistance of a single light is 240/19=12.63 ohms.

To calculate the power dissipated by a single light, we can again use the formula P=IV. Plugging in the values, we get P=0.5*120=60 W.

Now, if one of the bulbs burns out and the resistance drops to zero, the equivalent resistance of the string will also change. Without the burnt out bulb, the resistance of the string would be 240 ohms. But with the shorted out bulb, the resistance will be much lower. To find the new resistance, we can use the formula Rnew=R1+R2+...+Rn, where R1, R2, etc. are the resistances of each individual light. Since one of the resistances is now zero, the new resistance will be 240-12.63=227.37 ohms.

Finally, to find the new power dissipated by the string, we can use the formula P=IV. Plugging in the new resistance and the voltage, we get P=120*0.5=60 W, which is the same as the original power dissipated. This makes sense since the total voltage and current have not changed, only the individual