Understanding Power Dissipation in Series Circuits

[steven barnett]

Hey guys I made an account here because I cannot understand this for the life of me.

Two 95 W (120V ) lightbulbs are wired in series, then the combination is connected to a 120 V supply.
How much power is dissipated by each bulb?
Answer 24W

The part I don't understand is why 95=(120^2)/R Shouldn't there be a voltage drop of 60V at each light bulb so you use (60^2) Or you could look at the equivalent resistance and use 190=(120^2)/2R ? I cannot understand why you use the first equation for the life of me. That's saying that the power of one lightbulb is the voltage of the whole system^2 / the resistance of one bulb.
Thank you sorry for ranting this is upsetting me .

 

[BvU]

Hello Steven,

Take it easy ! Keep it cool ! etc etc.

The 120 V / 95 W on the bulb means that 120 Volt times the current I is 95 W.
With uncle Ohm's formula $I = V/R$ that makes 120 Volt times $V/R$ = 95 W, in which $V$ = 120 Volt.
So $\displaystyle {{V^2\over R} = 95 }$ Watt when $V = 120$ Volt.

With only half the volts you get one quarter the Watts, so 24 W.

 

[Baluncore]

Two 95 W (120V ) lightbulbs are wired in series, then the combination is connected to a 120 V supply.
How much power is dissipated by each bulb?

Light bulbs are difficult because they change resistance with temperature. Let us ignore that and assume they are fixed resistors.

A 95 watt bulb connected to 120 volt will have a current of ( 95 watt / 120 volt ) = 0.79167 amp.
The resistance of the bulb will be ( 120 volt / 0.79167 amp ) = 151.57 ohm.

The resistance of two globes in series will be 303.156 ohm.
If 120V is applied to the pair of globes in series, the current will be 0.3958 amp.
Each will dissipate 0.3958 amp * 60 volt = 23.75 watt.

That's saying that the power of one lightbulb is the voltage of the whole system^2 / the resistance of one bulb.

That is correct. For a single globe; W = V^2 / R. or; R = V^2 / W
R ohms = 120 volt * 120 volt / 95 watt = 151.579 ohms

 

[steven barnett]

A 95 watt bulb connected to 120 volt will have a current of ( 95 watt / 120 volt ) = 0.79167 amp.
The resistance of the bulb will be ( 120 volt / 0.79167 amp ) = 151.57 ohm.

I guess what I don't understand is why you use 120V here and not 60V. Since you are doing this for 1 bulb and each bulb only uses 60V why would you use 120V?

 

[Baluncore]

One bulb alone, not two in series.
Two bulbs in series have half the voltage each. Total resistance is twice, so current is then half.
Power = (voltage / 2) * (current / 2) = watt / 4 = 23.75 watt

 

[steven barnett]

I just don't understand why in P=V^2/R you use 120V... It makes absolutely no sense to me that for one of these bulbs in the series you use 120V and not the voltage drop of 60V... I really really really appreciate you trying to help me with this

 

[Baluncore]

A single 95 W globe on 120 V is specified. From that we must work out the resistance of one globe.
Using; R = V² / W;
One globe on 120 V.
R = 120² / 95 = 151.57 ohms.

Then two globes on 120 V.
W = V² / R .
Two globes on 120 V = one globe on 60 V .
W = V² / R = 60² / 151.57 = 23.75 watt .

 

[Asymptotic]

The part I don't understand is why 95=(120^2)/R

Rated bulb voltage and rated bulb wattage are used to calculate hot filament resistance. R=E²/P = 120²/95 = 151.58Ω

Two bulbs in series is a total resistance of 303.16Ω. A current of 0.396 amps flows when 120V is connected across them.

• Bulb resistances are equal, and two bulbs are in series, thus one-half of the source voltage (60V) is dropped across each. For each bulb, P=V²/R, 60²/151.58Ω = 3600/151.58 = 23.75 watts
• Current through each bulb is 0.396 amps. P=VI. 60V * 0.396A = 23.75 watts. Looking at it a third way, P=I²R, 0.396²*151.58 = 0.1568*151.58Ω = 23.75 watts.

The trick of it is to think of each light bulb as a resistor, calculate it's resistance at rated voltage and wattage, calculate what I current will flow through this resistance when V voltage is placed across it, then figure out (V*I) watts.

 

[NascentOxygen]

I just don't understand why in P=V^2/R you use 120V... It makes absolutely no sense to me that for one of these bulbs in the series you use 120V and not the voltage drop of 60V... I really really really appreciate you trying to help me with this

The bulbs are designed to deliver white light without the filament burning out when powered by 120V. So when operated at ½ their rated voltage they won't glow much, perhaps in the dark you'll see a dull red filament, so their emission will be limited to the red and infra red spectrum.

One advantage of operation off a reduced voltage is that you can expect them to last a very long time before the filament breaks!

 

[jim hardy]

@steven barnett

I have difficulty figuring out just what is the question.

I guess what I don't understand is why you use 120V here and not 60V. Since you are doing this for 1 bulb and each bulb only uses 60V why would you use 120V?

I think several folks answered that one.

I just don't understand why in P=V^2/R you use 120V...

In one case you're solving P=V^2/R for R , to figure out what is the resistance R of the bulb when you're given its ratings of P= 95 watts and V= 120 volts.
In the other case you're solving P=V^2/R for P, to figure out how much power P such a lamp with resistance R will produce when subjected to 60 volts. You got R from previous calc, and you already said you know when two are in series each gets half of the 120.

So that's why you use different numbers in the two places. You're solving for different properties under different conditions.
It's a two step solution not one.
Computers have accustomed us to instant answers. Slow down and take one step at a time.

see if this old thread helps you with thinking about KVL and KCL.
https://www.physicsforums.com/threa...t-live-and-neutral-wires.892309/#post-5613568

old jim

 

[russ_watters]

I just don't understand why in P=V^2/R you use 120V...

The bulb is rated for a certain wattage at a certain voltage. You can use that information to find its resistance before getting into the specifics of the problem at hand.