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Physics Power Question

  • #1
A 60W,220V bulb and a 40W,220V bulb are connected in series to a supply of 220V. Find the resistance of each bulb and the current flowing through each bulb.

My Solution:
I took the current to be same as the bulbs are connected in series.
As the current will be the same in each bulb, I took the current as 'i'
P=V*i
For bulb with power 60W
60=v*i
v=60/i
For bulb with power 40W
40=V*i
V=40/i

Adding V and v
V+v=220
40/i + 60/i = 220
100/i = 220
i = 100/220 = 10/22 = 5/11

Then I calculated the resistance by using the formula P=i^2 * r

Have I done the question correctly?
If no, then please point out the mistake
 

Answers and Replies

  • #2
1,948
200
No, not correct. The bulbs only provide their nominal power when connected directly to 220V. When connected in series, they will provide less power.
 
  • #3
140
17
Not provide but consume. Only a power source will provide.
 
  • #4
is the current correct?
i have read on an online source that the power in a series circuit is the sum of all the conductors connected in series
first reply to this
then i will reply furthur
 
  • #5
Doc Al
Mentor
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is the current correct?
No.

i have read on an online source that the power in a series circuit is the sum of all the conductors connected in series
The total power will be the sum of the power dissipated in each bulb.

Hint: First find the resistance of each bulb.
 
  • #6
oh sorry!
I forgot to mention the word "power" i.e. I meant to say that "the power in a series circuit is the sum of the powers of all the conductors connected in series"

No.


The total power will be the sum of the power dissipated in each bulb.

Hint: First find the resistance of each bulb.
consider this
the power in this case will be the powers of both the bulbs
P = 60W + 40W = 100W
Now, P= VI
100W = 220*I
I = 100/220

This should be correct
Shouldn't it be?
 
  • #7
Doc Al
Mentor
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consider this
the power in this case will be the powers of both the bulbs
P = 60W + 40W = 100W
Nope.

As dauto said in his post above, the bulbs will only have their rated power when they are connected directly to the full voltage source (not when they are in series with other bulbs). So the "60 W" bulb does not consume 60 W when in series with another bulb.

So, figure out the resistance of each bulb.
 
  • #8
  • #9
Doc Al
Mentor
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It seems to me that u are right but still I would like to mention this website. It shows the resultant power as sum of all the other powers
I don't need a website to tell me that. I even said it above.

Does the total power equal the sum of the individual powers? Yes!

Your mistake: Thinking that the power dissipated in the "60 W" bulb is still 60 Watts, even when it's in series with another bulb.

Once again I suggest that you first figure out the resistance of each bulb. Unlike the power, which will change depending on how you hook the bulbs together, the resistance of each bulb will remain the same whether they are in series or not.
 
  • #10
Agree with whatever you said above
In the website I mentioned above they have used bulbs as an example to illustrate the concept. So, in that case if the bulbs are connected in series they won't get the rated voltage and hence they won't dissipate the maximum power
So knowing this fact why do we still add up individual powers to get the resultant power in series?
 
  • #11
Doc Al
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Agree with whatever you said above
In the website I mentioned above they have used bulbs as an example to illustrate the concept. So, in that case if the bulbs are connected in series they won't get the rated voltage and hence they won't dissipate the maximum power
Right.

So knowing this fact why do we still add up individual powers to get the resultant power in series?
Because it's true regardless of whether the power dissipated is the rated power or not.

You agree that Resultant power = Total Voltage * current, right?

Well, the total voltage is just the sum of the voltage drops across each bulb, which must add up to the total voltage.

Total voltage = V1 + V2
Thus:
Resultant Power = (V1 + V2)*current = V1*current + V2*current = Power1 + Power2

Make sense?
 
  • #12
In the same way, in the question I mentioned above, why can't we directly find the current in the circuit and as the bulbs are connected in series, each will have the same current
 
  • #13
Doc Al
Mentor
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In the same way, in the question I mentioned above, why can't we directly find the current in the circuit and as the bulbs are connected in series, each will have the same current
You can find the current, but first you'll need the resistance.
 
  • #14
Let me get this straight
To find the current FOR THE ENTIRE CIRCUIT
P = 40W +60W
P =100W
do you agree with that?

Then, do you agree that P= VI
100W = 220V*I FOR THE ENTIRE CIRCUIT
 
Last edited:
  • #15
Doc Al
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Let me get this straight
To find the current FOR THE ENTIRE CIRCUIT
P = 40W +60W
P =100W
do you agree with that?
No.

For the nth time, when connected in series, the individual powers are not equal to 40W and 60W and thus the total power does not equal 100W.


Then, do you agree that P= VI
100W = 220V*I FOR THE ENTIRE CIRCUIT
No.
 
  • #16
Does the total power equal the sum of the individual powers? Yes!
.

.
when a formula has been derived that power can be added in series then why can't we use it?
when will we use that formula or how do we use it in this question?
please throw some light on the above questions
 
  • #17
Doc Al
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when a formula has been derived that power can be added in series then why can't we use it?
Because you don't know the power used by each bulb. So how can you add them?

when will we use that formula or how do we use it in this question?
You won't need it, but it's nice to know.
 
  • #18
2,463
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If this question is about tungsten filament lamps it should be remembered that the change of filament resistance in going from room temperature to normal operating temperature is huge ( about 500% if memory serves). It's a bit of a stretch to assume that any resistance changes can be ignored.
 
  • #19
Doc Al
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If this question is about tungsten filament lamps it should be remembered that the change of filament resistance in going from room temperature to normal operating temperature is huge ( about 500% if memory serves). It's a bit of a stretch to assume that any resistance changes can be ignored.
Yes, that's true. But that's the (very crude) assumption often made. (That for the range of currents in the problem, the resistance can be treated as fixed.)

I think the resistance change from a cold to a hot filament is even greater than you suggest.
 
Last edited:

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