Resistor circuit

  • Engineering
  • Thread starter casanova2528
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  • #1
v1 of battery1 is higher than v2 of battery2

The circuit is arranged in a parallel way with 3 layers where the only
Resistor is in the middle layer.
battery1 is on the top layer and the only resistor is on middle layer
Battery2 is on bottom layer.
The terminals of the battery is not facing each other.


It appears like the batteries are arranged in series, but
It is not! How does one calculate the current on resistor?
Does all the current from battery1 skip the resistor on the middle layer
And go to the negative terminal of the battery2 on third level?
 

Answers and Replies

  • #2
Redbelly98
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Can you post a diagram? I can't picture this circuit from the verbal description.
 
  • #3
can you post a diagram? I can't picture this circuit from the verbal description.
|----1(+ -)----|
|-----R1-------|
|----2(- +)----|


there it is. battery 1 has a higher emf than battery 2. how does one figure out the current through resistor1?
 
  • #4
The Electrician
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You have to know the internal resistance of the batteries.
 
  • #5
Redbelly98
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Something is wrong here. It could be we need to know the batteries' internal resistance, as The Electrician said.

Or (if we assume ideal batteries with zero resistance) 2 points that must be at different potentials (because of the batteries) must also be at the same potential (because they are shorted together). That makes the problem unsolveable, or suggests the diagram is incorrect.
 
  • #6
Something is wrong here. It could be we need to know the batteries' internal resistance, as The Electrician said.

Or (if we assume ideal batteries with zero resistance) 2 points that must be at different potentials (because of the batteries) must also be at the same potential (because they are shorted together). That makes the problem unsolveable, or suggests the diagram is incorrect.
the diagram is correct. it is a diagram that i made from another problem that was solvable. I was wondering if this diagram is solvable. the batteries are ideal.

so, what else happens beside the circuit being shorted?
 
Last edited:
  • #7
there must be a way to figure out the small amount of current through resistor1.
 
  • #8
The Electrician
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If you know the internal resistance of the batteries, then you can figure out the current through the resistor.

In the conceptual diagram you gave, the batteries are mathematically ideal. If you connect them as shown, then an infinite current will circulate in the batteries and the current in the resistor will be indeterminate.

Imagine if a small resistor were in series with each battery. Then the circuit could be easily solved. Real batteries do in fact behave as if there were a small resistor inside them; its value changes with temperature and state of charge of the batteries. But if you can get an approximate value for the internal resistance, then you can solve the network.
 

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