Wattage Question

[dbabn7]

Good Evening:
I hope some one can help me understand the following question:

The Lamps are rated 60 and 100 watts respectively. When Connected in series,
a. Each lamp takes the same power
b. The 60 watt lamp takes more then the 100 watt lamp
c The 100 watt takes more than the 60 watt lamp
d. Both lamps take more than 60 but less than 100 watts.

I picked letter C but the answer key says its B. Thank you in advanced for your help.

 

[Eddie702]

You don't have enough information.

you have to know at least two things out of 4 Amperage, wattage voltage or ohms. You only gave us watts, but at what voltage?? What voltage are the lamps rated at and what is the supplied voltage to the circuit?

 

[Hv&Lv]

Good Evening:
I hope some one can help me understand the following question:

The Lamps are rated 60 and 100 watts respectively. When Connected in series,
a. Each lamp takes the same power
b. The 60 watt lamp takes more then the 100 watt lamp
c The 100 watt takes more than the 60 watt lamp
d. Both lamps take more than 60 but less than 100 watts.

I picked letter C but the answer key says its B. Thank you in advanced for your help.

Kind of a silly question actually.
It assumes some things.
one is don’t think of it like a table lamp but as a series resistor.

A source is at one end.
There is a 60W bulb and a 100W bulb AFTER the 60W in series.
The power to run the 100W bulb has to go through the resistance of the 60W bulb.
The 60W bulb won’t allow it due to resistance, so the 100W bulb will get less power than the 60.
Bulbs aren’t constant power devices.

 

[synchro]

Good Evening:
I hope some one can help me understand the following question:

The Lamps are rated 60 and 100 watts respectively. When Connected in series,
a. Each lamp takes the same power
b. The 60 watt lamp takes more then the 100 watt lamp
c The 100 watt takes more than the 60 watt lamp
d. Both lamps take more than 60 but less than 100 watts.

I picked letter C but the answer key says its B. Thank you in advanced for your help.

The 100 watt lamp has a lower resistance than the 60 watt lamp because it draws more power when they are both fed with the same voltage. But when the two lamps are in series, the same amount of current must flow through both of them. And so because the power dissipated in a resistor is I² x R, the one with the higher resistance (i.e., the 60 watt one) will dissipate more power.

 

[dbabn7]

Thank so based on this info I broke it down as follows:
1. 60w and 100w are the ratings without any connections;
2. I found the resistance of each bulb separately assuming I’m putting 120v, for the 60w I got 240ohms, and 144ohms for the 100w;
3. Then I added the total resistance of 240 to 144 I got 384 ohms;
4. Now that I have resistance I can calculate them in a series circuit;
5. With 120v total and 384ohm total I get .3125amps in the circuit.
6. Now I went to each bulb with the amperage to see the power so for the 240 ohms(60w) I got 75 watts (IxI x R = w) and for the 144ohm (100watts) I got 14 watts.
Thank you for your help and please correct me if my understanding is flawed in any part.

 

[synchro]

Thank so based on this info I broke it down as follows:
1. 60w and 100w are the ratings without any connections;
2. I found the resistance of each bulb separately assuming I’m putting 120v, for the 60w I got 240ohms, and 144ohms for the 100w;
3. Then I added the total resistance of 240 to 144 I got 384 ohms;
4. Now that I have resistance I can calculate them in a series circuit;
5. With 120v total and 384ohm total I get .3125amps in the circuit.
6. Now I went to each bulb with the amperage to see the power so for the 240 ohms(60w) I got 75 watts (IxI x R = w) and for the 144ohm (100watts) I got 14 watts.
Thank you for your help and please correct me if my understanding is flawed in any part.

Looks good except it should be 23.4 watts for 240 ohms (the 60W lamp).
In reality the resistance of an incandescent bulb is not a constant, and it increases as the filament temperature rises at higher applied voltages. But that's getting beyond the scope of questions like this, and it would not change the correct answer from the choice b).

If you're taking a multiple choice test where the time is limited, sometimes like in this one you don't need to calculate an exact numerical answer to select the correct answer. But it's always helpful to practice using all the appropriate calculations.

 

[dbabn7]

Thank you very much

 

[Carultch]

You don't have enough information.

you have to know at least two things out of 4 Amperage, wattage voltage or ohms. You only gave us watts, but at what voltage?? What voltage are the lamps rated at and what is the supplied voltage to the circuit?

You can solve this problem without knowing the voltage at which the bulbs are rated at the given Wattage, as long as you assume they are both rated at the same voltage. Obviously if you were to try this in reality, you would have to know what voltage it is, so you don't damage the bulbs.

Construct two copies of P=V^2/R, and solve it for R in both cases (R = V^2/P). Call it V0, the voltage at which they both get nominal power.
R1 = V0^2/P1nom
R2 = V0^2/P2nom

Put them in series, the resistance adds up conventionally:
Rnet = R1 + R2
Rnet = V0^2 * (1/P1nom + 1/P2nom)

Apply voltage V across this combination, to get the current:
I = V/Rnet
I = V/(V0^2 * (1/P1nom + 1/P2nom))

Use P=I^2*R to find the power delivered to each bulb:
P1 = I^2*R1
P2 = I^2*R2

Substitute for R1 & R2:
P1 = I^2*V0^2/P1nom
P2 = I^2*V0^2/P2nom

I^2*V0^2 is a constant, we'll call K.
P1 = K/P1nom
P2 = K/P2nom

Now plug in 60W and 100W for P1nom and P2nom respectively.
P1 = K/60
P2 = K/100

You can see that no matter what we make K, power P1 value will be larger than power P2, when these two bulbs are wired in series. To me, it is counter-intuitive that the bulb with the lower power rating, will have the highest power delivered to it, but that is the way this would work, assuming Ohm's law does a good job at representing the behavior of the bulbs.

 

[drcampbell]

Sometimes the best you can do is choose the most-correct (or least-incorrect) answer.

If they're both incandescent lamps and both rated for the same voltage, (b) is probably the most-correct answer. If they're in series, the current is the same everywhere, and the lamp with the higher resistance will have the greater voltage drop across it and dissipate the most power.

But none of the calculations above are warranted.
An incandescent lamp's resistance is HIGHLY variable with the filament's temperature. When they're in series, the voltage on each lamp isn't the same and it isn't the same fraction of the design voltage. The filament temperatures will be radically different and their resistances will be radically different fractions of their nominal hot resistance.

The only case in which the equations above are valid is when the lamp resistances are constant.
- - -
That said, when was the last time anybody ever saw -- let alone installed -- a 60- or 100-watt incandescent lamp?
The question is as obsolete as the technology.