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

 

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.

 

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.

 

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:
\(R_{load}=\frac{V_{source}}{I_{total}}\)

 

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.

 

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:
\(R=\frac{V}{I}=\frac{8.5}{2 \cdot 0.2}=21.25Ohms\)

 

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

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

 

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

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.

 

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.

 

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.

 

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

 

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

6 cell ni-cad = 7.2 V

------------------------------------------------------------------
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

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.

 

Thank you for all your assistance. Have a great day

 

Should we take the problem too seriously and try to match the problem to the reality?

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.