Internal resistance in a battery

[lDanny]

1. I am confused on how do you know if the battery is supplying a current to the circuit or if the circuit is supplying a current to the battery?

2. In a circuit with 200V (battery) from positive terminal travels thru a parallel resistor (40 ohm and 20 ohm) then to a 70ohm resistor then to the negative terminal. There is an ammeter on the 120 ohm resistor. What is the reading of that ammeter?

The answer in TPR was the parallel resistors equal 30ohm + 70 ohm= 100 ohm. 200V/100ohm = 2A. therefore the voltage drop is 60V. So 60V/120ohm = .5A.

My question is, from the diagram it looks like the battery + terminal is directly connected to the parallel resistor. Then shouldn't the voltage just be 200V? Is it because there is a series resistor in there that you would have to consider? But the ammeter is on the 120 ohms resistor and the + terminal is connected straight to the parallel resistor.

thanks

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[GoodmanBrown]

Hey, Danny. Let me see if I can shed some light on this.

1) Batteries always supply current to the circuit. A circuit cannot provide current to anything. Think of the battery as a water pump and the current as a flow of water. In a hose without any pressure, water won't go anywhere, it will just stay where it is. Electricity's the same.

2) The TPR book does this problem totally wrong. Parallel resistors aren't added together. That's series resistors. To sum parallel resistors, use R_total = (1/R1 + 1/R2)^-1. So, here we have R_total = (1/30 + 1/70)^-1 = 21 ohms. So that's your total resistance through the parallel part. Now, because the next resistor is just in a series with the parallel ones, we can add those. So, the total resistance of the circuit is 141 ohms.

So, ohm's law states that V=IR. So, we know V = 200V from the battery, so that's our change in volts over the entire circuit. (Like a water pump, the voltage must be equal around the entire circuit. So, the battery "pumps" it 200V up, and the restistors must bring it down by 200V before it reaches the battery again.) And we know R = 141 ohms. So, I = 200/141 = 1.418 amps. That's what the total current through the circuit is, and because there's only one branch at the 120 ohm resistor, it must be that current there. So, the answer is 1.418 amps.

 

[inaccensa]

1. I am confused on how do you know if the battery is supplying a current to the circuit or if the circuit is supplying a current to the battery?

2. In a circuit with 200V (battery) from positive terminal travels thru a parallel resistor (40 ohm and 20 ohm) then to a 70ohm resistor then to the negative terminal. There is an ammeter on the 120 ohm resistor. What is the reading of that ammeter?

The answer in TPR was the parallel resistors equal 30ohm + 70 ohm= 100 ohm. 200V/100ohm = 2A. therefore the voltage drop is 60V. So 60V/120ohm = .5A.

My question is, from the diagram it looks like the battery + terminal is directly connected to the parallel resistor. Then shouldn't the voltage just be 200V? Is it because there is a series resistor in there that you would have to consider? But the ammeter is on the 120 ohms resistor and the + terminal is connected straight to the parallel resistor.

thanks

Wait didn't he say that the resistors 40 ohm and 30 ohm are parallel and the 70ohm is in series

 

[GoodmanBrown]

Wait didn't he say that the resistors 40 ohm and 30 ohm are parallel and the 70ohm is in series

Wait. Now I'm confused as well. He did say a 20 ohm and 40 ohm parallel. But that would be 13.3 ohms total resistance, not 30 ohms. Then he did say 70 ohms in series. So, yeah, I was wrong, it's 83.3 ohms total in the circuit for a current of 2.4 amps. But then he says:

There is an ammeter on the 120 ohm resistor. What is the reading of that ammeter?

Does he mean the 20 ohm resistor? I'm confused here. He mentions 120 ohms again, though so it seems like it's not a mistake...

 

[lDanny]

sorry it should have been "parallel resistor (40 ohm and 120 ohm)." In TPR solutions reads, "The two parallel lines are equivalent to asingle 30ohm resistor. The 70ohm R is now in series with the 30ohm R, so the current is I=V/R= (200V)/100ohm = 2A. The voltage drop across the 30ohm R is therefore (2A)(60ohm)= 60V. The ammeter is on the branch with 120ohm, so the current is V/R = (60V)/(120 ohm) = .5A.

But since the voltage is going straight to the 120ohm R, would it be 200V? Because it is going there first?

Referring back to my first question, in TPR is says that if the battery is supplying the current to the circuit then its V= emf -Ir. But if the circuit is supplying the current to the battery (as in charging) then its V= emf +Ir. So how do you know when which is supplying the current?

Thanks again for the help.

I just snapped some pictures of the problem. And I added another example they had will with the battery going straight to the resistor.

I think its something simple that I am not getting.. sigh..
grr.. dont know why its coming out so small.. but here is the direct links.
http://picasaweb.google.com/lh/photo/Uqp2RQw2y86E5kcXKexl4A?feat=directlink
http://picasaweb.google.com/lh/photo/CET9AS1L8qbtzxPBDN6BGA?feat=directlink
http://picasaweb.google.com/lh/photo/7WwwICMgZd6bzfw--c1KCA?feat=directlink

 

[GoodmanBrown]

Hey, Danny. It's all clear now for me, so let me try to explain. Here are a couple of points that helped make things clear for me.

1) Think of voltage as "potential energy" of a marble rolling. The battery is a lift that brings the marble "up" 200V. The entire circuit has to start and stop at the same spot. So, the parallel resistors cannot bring the marble down 200V worth because then, the 70 ohm resistor would bring it down even lower and the battery would have a break. So, the battery brings the marble up 200V and all the resistors bring it down 200V. So, the two parallel resistors and the 70 ohm resistor work together to go down 200V. This brings the path back to the starting point before the battery.

2) Now, for circuit vs. battery supplying the current. In a circuit that only has one battery, the battery is supplying current. It's doing the work that is required to push electrons around. The instance of a battery charging is still one really strong battery pushing current the wrong way through another battery. In both cases, the circuit itself isn't doing work, but some battery or other provider of force is pushing the electrons. The circuit is just a path. I.e., when you see a mom pushing her baby stroller down a sidewalk, it's not the sidewalk doing any work, it's just providing a place to push. Same with a circuit.

 

[lDanny]

thanks for the reply..

so there is a voltage drop in 200V question because there is a parallel and series R? And not in the 3 parallel R because they are all parallel?


thanks again... man, i am really hating physics. lol

 

[GoodmanBrown]

No, anytime you have a resistor, there will be a voltage drop. Voltage does the work of pushing electrons through the resistor. So, anytime there is a current through a resistor, there will be a voltage drop through it.