Power Dissipated by 60W Bulb in 20V Source[/B]

[estike]

Homework Statement

What is the power dissipated by a 60 Watt (rated at 120 V) light bulb if it is connected to a 20 V power source (assume the resistance of the bulb is constant)?

Homework Equations

The Attempt at a Solution

If it [/B]

 

[BvU]

Hello E, welcome to PF !

You were saying ?

 

[estike]

I don't know where to start i know that the power dissipated is in Joules but i don't know what equation to use

 

[CWatters]

Power is actually measured in Watts.

You initially have a 60W bulb running on 120V. What equation relates Power, Voltage and Current?

 

[estike]

Ohm´s law V=RI

 

[estike]

W = VC right ?

 

[CWatters]

Yes Ok but better to write it as P = I * V.

Now using that and Ohms law write a new equation relating Power, Voltage and Resistance.

 

[estike]

if 60 = 120 * I = 0.5 A

I use this current and know that is 120V*0.5A = W right ? and W is 60 right

 

[CWatters]

Yes Ok the current is 0.5A.

You actually need to calculate the resistance as well.

 

[estike]

hum ok i understand know so if V= RI since 220 = R *0,5A = so the resistance is 110 ohms and since is constant

20 = 110*i = so I =0.182 A

So P = I*V = 0.182 A*20 = 3.6 Watts ? Am i right ?

 

[CWatters]

hum ok i understand know so if V= RI since 220= R *0,5A

should be 120V.

 

[estike]

so is 120 = R * 0.5A = 60 ohms

And then 20 = 60*I = 1/3 A

so W = 1/3A*20 V = 6,66 watts right

 

[CWatters]

Back later.

 

[estike]

but the answers are 60W , 1.67 W , 20W or 120W i really don't understand then :x

 

[estike]

i know what i did wrong it was 240 ohms i made a stupid error

W = 1/12A*20V and then it is 1,67 W

 

[CWatters]

Sorry I got called away..

Ok you got to the right answer but there is an easier route that I was trying to lead you down in post #7. I think you're substituting values too soon. There is no need to work out the current in either circuit. Here is how I would do it...

P = I * V
V = I * R

Substitute to eliminate I..

P = V²/R (Aside: I memorised that one and also P = I² * R)

Rearrange to give R...

R = V²/P

Substitute values..

= 14400/60
= 240 Ohms

For second circuit R is the same. V changes to 20V...

P = V²/R
= 20²/240
= 1.7W

 

[CWatters]

You can also do it without working out R...

P₁ = V₁²/R
P₂ = V₂²/R

P₂ / P₁ = V₂² / V₁²

P₂ = P₁ * V₂² / V₁²

= 60 * 20²/120²
= 1.7W

 

[theodoros.mihos]

@estike you are right, the value is 5/3 = 1.666...