# Parallel Circuit

Discussion in 'Homework Help' started by Marksripp, Sep 27, 2011.

1. ### Marksripp Thread Starter New Member

Sep 27, 2011
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With a 8.5VDC battery hooked to 2 parallel groups of 5 lamps drawing .2 AMPS each what is the total load being applied?

2. ### Georacer Moderator

Nov 25, 2009
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In the Homework Help forum we don't give you the answers to your homework. We ask that you first post your efforts in tackling your problem.

Please post your work and thoughts so far.

3. ### Marksripp Thread Starter New Member

Sep 27, 2011
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0
If I add the current dropped across each string @ .2Amps per bulb that would be 1 Amp per string for a total output load of 2 Amps.

4. ### Georacer Moderator

Nov 25, 2009
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Not quite. You see, each lamp needs 0.2A, but since the string has 5 lamps in series, each string needs exactly 0.2A.

The question asks you to find the load attached on the voltage source, so you must utilize the formula:
$R_{load}=\frac{V_{source}}{I_{total}}$

5. ### Marksripp Thread Starter New Member

Sep 27, 2011
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If I use E=IxR with E=8.5, I=.2, R should be 42.5 then placeing those in the total formula would make it a .4Amp total load. But that does not sound sufficient enough of a load.

6. ### Georacer Moderator

Nov 25, 2009
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The load is expressed in Ohms, not Amperes, so it's impedance you 're looking for, not current.

You know that your source is stable at 8.5V and it provides 0.2Amps per string for two strings. Thus you can calculate the load as:
$R=\frac{V}{I}=\frac{8.5}{2 \cdot 0.2}=21.25Ohms$

7. ### JoeJester AAC Fanatic!

Apr 26, 2005
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I read the question as the lamps were in parallel making it a 2 ampere load or 4.25 ohms.

8. ### Georacer Moderator

Nov 25, 2009
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Each lamp in each string needs 0.2A. That means each string needs 0.2A too, am I wrong here?
Two strings in parallel sum to 0.4A.

9. ### Marksripp Thread Starter New Member

Sep 27, 2011
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I believe that I am asking this question the wrong way and am increasing the confusion factor at least to me. The battery is a 6 cell ni-cad, What I am trying to determine is the load that will be placed on the battery, it must supply a lamp load of two parallel groups of five number 328 lamps (each lamp draws .2 Amps) for at least 15 minutes.

10. ### Georacer Moderator

Nov 25, 2009
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So you are trying to determine the necessary AmpHours? If that is so, you have a 0.4Amp consumption that you want to last for 0.25hours. Thus you need a battery cell pack of a total 100mAh. Make sure your batteries can provide that instantaneous current, but I thing ni-cad batteries have a high such rating.

11. ### JoeJester AAC Fanatic!

Apr 26, 2005
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328 lamp
Voltage: 6V, 6 Volts
Amperage: .2A, 200mA

6 cell ni-cad = 7.2 V

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If you wire five lamps in series, each string would require 30 volts to operate properly at the 0.2 Ampere standard. Your nicad is a six cell battery.

The problem calls for each string of five to be wired parallel for proper operation. Kirchoff's current law will give you the total amperes in the load.

Minimum battery rating ... ampere hour / 4 or ampere hour * 0.25

12. ### Georacer Moderator

Nov 25, 2009
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Should we take the problem too seriously and try to match the problem to the reality? I suggest we stick to the initial statement in the first post and solve it theoretically.

Of course, the OP has the last word.

13. ### Marksripp Thread Starter New Member

Sep 27, 2011
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0
Thank you for all your assistance. Have a great day

14. ### JoeJester AAC Fanatic!

Apr 26, 2005
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No we shouldn't. It did however require clarification which was provided by the OP. At that point, the proper guidance should have been provided illustrating where the OP understands that incomplete questions will get varying responses as the members will try to interpet what was asked.

What I failed to do was require the OP to provide the exact question from their instructor or their attempts at the solution in which case the misunderstanding of the question might have be clarified.