I want to ask something about voltage.

Referring to the figure, there's no drop of P.D. across E ,D through EFD.

But, through EDCB, there's a P.D difference between them.

The voltage across E and D in fact= e.m.f.

But why we neglect the first approach? Can anyone reconcile the two approachs.

By the looks of the diagram there is 0 Resistance through EFD

In theory you would create an infinite current.

 

In reality the source is an ideal source with a resistor.

and the wire wil also be a resistor.

If you add these resitors your circuit will make sense again.

What will the voltmeter show if it is put at A?

What will the voltmeter show if it is put at A?

In theory you don't have a solution.

your voltage is 0*inf.

 

In reality it wil be lower than your open circuit e.m.f. the value will depend on those resistors. (which wire is used, which battery is used)

You can't really put a voltmeter at A, you need to measure the emf or pd across something.

 

In many circuits the resistance of a wire is considered 0 for simplicity. In a circuit such as this should we assume that the resistance of the wire does count?

 

Calculating voltages would probably require some details on the battery/cell itself.

 

Voltage across EFD will equal voltage across EAD assuming that the resistance through the 2 wires between E and D is the same.

It's supposed that p.d. of D and E across DCBE is not equal to that across EFD, right? so how can I determine the p.d.?

This is a shorted capacitor. If there is no PD across ED at one path, then there can be none on a different path, if the system is at equilibrium.

Why is it a shorted cap? It's drawn as a cell.

 

Otherwise like swansont said, if there's no pd ED for one path there can be no pd across ED in another path.

 

The only way there can be any pd in that circuit is if the cell is supplying a pd, in which case it will be the same pd across ED no matter which path you take.

 

Do you have no other information in the question? How can you calculate the pd if you don't know the pd of the cell supplying the pd across the circuit.

 

This is seemingly a simple parallel circuit. The voltage across all branches; EBCD, EAD and EFD is the same.

Why is it a shorted cap? It's drawn as a cell.

 

Otherwise like swansont said' date=' if there's no pd ED for one path there can be no pd across ED in another path.

 

The only way there can be any pd in that circuit is if the cell is supplying a pd, in which case it will be the same pd across ED no matter which path you take.

 

Do you have no other information in the question? How can you calculate the pd if you don't know the pd of the cell supplying the pd across the circuit.

 

This is seemingly a simple parallel circuit. The voltage across all branches; EBCD, EAD and EFD is the same.[/quote']

 

 

Sorry, I took that to be a capacitor. Ok, it's an emf with no load. All of the parallel paths have the same PD.

I'm probably missing something, but I view your schematic as short across a battery.

Yeah, a shorted battery, pretty much.

 

What he has is a setup & question which presumably tests the whether you know your parallel & series circuits and how the voltage and current pattern applies to each type of circuit.