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I don't know how to resolve this problem.

A complete circuit consists of a 18,0 V battery (internal resistance r) and a resistor R. The terminal voltage of the circuit is 15,8 V and the current is 4,00 A. What is:

a) the internal resistor r of the battery.

b) the resistance R of the circuit resistor.

I am not sure how to resolve it. I know the Ohm's Law, but not sure how to apply it here.

Sorry for the basic question (just starting with Physics again after very long time).

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closed as too localized by David Z Dec 29 '11 at 8:04

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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Hi John, and welcome to Physics Stack Exchange! This is not a site for general homework help, and in particular not a site where you can just post your homework questions and ask for solutions. You need to narrow down your problem to the specific physical concept that is giving you trouble and ask about that. I see that you've already gotten a complete answer to this one, but in the future, please take the guidelines in this meta question into account before posting homework questions. –  David Z Dec 29 '11 at 8:05

2 Answers 2

up vote 1 down vote accepted

The total resistance is $R+r = \frac{18}{4} = 4.5 \Omega$. $R$ itself is $\frac{15.8}{4} = 3.95 \Omega$. The difference $r$ is $0.55 \Omega$.

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Total resistance: 4.5 Ohm For R -> 15,8/4 = 3.95 Ohm For r -> 4.5 - 3.95 Ohm = 0.55 Ohm So... a) -> 0.55 Ohm b) -> 3.95 Ohm Am I right? –  John Dec 29 '11 at 1:10
    
@John: Right. Ohms = Volts / Amps for the whole circuit and for each part of it. Resistance in series just adds, as does voltage. The current is the same in every part of the circuit because it's a series circuit - there's no other place for it to go. –  Mike Dunlavey Dec 29 '11 at 1:21
    
Ahhh okay, now I understand all the results... thank you very much, you were very helpful! –  John Dec 29 '11 at 1:24
    
@John: You're welcome. If you get into a circuit with parallel resistance, then it's the current that adds, and 1/resistance that adds. –  Mike Dunlavey Dec 29 '11 at 1:31
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Mike: bear in mind that it's against our site policy to provide complete answers to homework questions, and it is strongly discouraged to provide any answer at all when the poster hasn't shown sufficient effort to make it a conceptual question. Ordinarily I would (temporarily) delete this, but since John has already seen and accepted it, I don't think there's much point, so I'll leave it alone this time. But please keep this in mind in the future. –  David Z Dec 29 '11 at 8:08

So, if 18V battery gives out only 15.8V voltage, than there is a voltage drop (18-15.8) on battery's internal parasitic resistance. Gived the current, which is the same across all items you can calculate internal battery resistance.

For R - you have remaining voltage drop 15.8V and same current.

After you calculate both resistances, verify that 18V will give you 4A with their combined resistance.

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