current division in circuit

[duke37]

Try using ohms law.
With ideal components, in the first circuit, you have a 1V source and a zero ohm short. What is the current?
Is this possible?

 

[scientifico]

There is 1/0 that isn't possible, does that mean that with real components nearly all the current would pass in the wire and nothing in the 2 resistances?

 

[duke37]

In a practical circuit, there will be resistance in the voltage supply and resistance in the wire. The voltage across the resistors will depend on the values of these.

 

[scientifico]

if the wire resistance were 2 ohms and the voltage supply 1 ohm will the current be:

1/((1/100)+(1/100)+(1/2)+(1/1)) = 0.66 ?

Thanks

 

[duke37]

Its not what I get. The three resistors in parallel have a resistance of 1.92R.
Add 1R for the source gives total resistance 2.92R.
Current =1/2.92

 

[scientifico]

Yes it's just because i putted the battery internal resistance as it was a fourth resistor in parallel with the other three... for example if a connect 2 same batteries in parallel both with an internal resistance of 1 ohm, will the total resistance be 2 ohm?

 

[(*steve*)]

Battery internal resistance is modelled as a resistance in series with a voltage source, not one in parallel with it. You can't have other resistances directly in parallel with it.