# help!

Discussion in 'Homework Help' started by GMChandio, Feb 26, 2015.

1. ### GMChandio Thread Starter New Member

Feb 26, 2015
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A 12V, 25W tungsten filament bulb is supplied with current from n cells connected in series. Each cell has an emf of 1.5V and internal resistance of 0.25. What is the value of n in order that the bulb runs at its rated power?

2. ### MrChips Moderator

Oct 2, 2009
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Take a look at the titles of all the other threads.

You can see that a title of help! is of no help to anybody.

3. ### GMChandio Thread Starter New Member

Feb 26, 2015
28
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What is that supposed to mean?
Deffinatly not the answer to my question.

Apr 5, 2008
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5. ### GMChandio Thread Starter New Member

Feb 26, 2015
28
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Sorry I just saw that.
Thanks BTW.

6. ### GMChandio Thread Starter New Member

Feb 26, 2015
28
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How do I solve it further....

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Apr 5, 2008
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Hello,

When you must make a battery of 12 Volts, out of 1.5 Volt cells.
How many cells are needed?
The internal resistance of the cells can be added.

Bertus

8. ### GMChandio Thread Starter New Member

Feb 26, 2015
28
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Can you solve it?
And provide me the solution?
Because iam not getting it.

Apr 5, 2008
15,806
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Hello,

Did you read this before posing in the homework section?

Bertus

10. ### GMChandio Thread Starter New Member

Feb 26, 2015
28
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I've tried it again and again but iam not getting anywhere near answer.
You said add the internal resistance, but then again there is another question how many times do I add the internal resistance?
Or can I just simply divide 12V/1.5V?

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Apr 5, 2008
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Hello,

For the number of cells you can indeed divide the 12 Volts by the 1.5 Volts.
Then you can also calculate the internal resistance.

Bertus

GMChandio likes this.
12. ### GMChandio Thread Starter New Member

Feb 26, 2015
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Ok, thankyou.

13. ### kubeek AAC Fanatic!

Sep 20, 2005
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806
Well that is not entirely true, if the resistance of each battery was for example 0,33 ohm, then the output power would be closest to 25w with 15 cells instead of 12, so you need to take that into account.

14. ### GMChandio Thread Starter New Member

Feb 26, 2015
28
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Didn't got that can you elaborate?

15. ### kubeek AAC Fanatic!

Sep 20, 2005
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Lets say you have 8 batteries in series, that gives you 12V. But since each battery has 0.25 ohms, the total resistance inside the battery will be 0.25*8=2 ohms. If you draw 0.5A from that battery you will have only 12-2*0.5=11V volts available at the battery terminals.
25W at 12V is a bit over 2A, so you can imagine it will take more than 8 batteries to get the output voltage near 12V when providing 2A of current.

The big question is how many cells do you actually need to get that output voltage under load.

16. ### GMChandio Thread Starter New Member

Feb 26, 2015
28
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What you are saying is that I can't divide 12 by 1.5?
So how do I find the number of cells?

17. ### kubeek AAC Fanatic!

Sep 20, 2005
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First you need to find what the current is when you have n cells. Then use P=R*I^2 to find out the power going into the bulb.

18. ### GMChandio Thread Starter New Member

Feb 26, 2015
28
0
I get what you are saying but in order to find the current, shouldn't I know the value of n?
Or the current would remain same throughout.
Can I do it like this.
P=V^2/R
25=12^2/R
R=5.76ohm
Now
V=IR
12/5.76=I
I=2.08amp
Is this the value of current?

19. ### kubeek AAC Fanatic!

Sep 20, 2005
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You need to start from the other side. The voltage will be n*1.5V and the internal resistance will be n*0.25ohms.
To this voltage you have connected the internal resistance and load resistance in series, what will the current be? This current is then used to calculate the power across the load resistor, you should get an equation with n as the only variable, then you solve for n that gives 25W. Then find the two adhacent whole numbers and pick the one that gives closer power.

20. ### WBahn Moderator

Mar 31, 2012
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Q1) How much current does the bulb require?

Q2) If a single cell is delivering the current in Q1, what will the voltage of that cell be?

Q3) What is the smallest number of cells, each at the voltage in Q2, that can supply at least 12V to the load.