View the thread, titled "What will V1, V2, V3 and V4 in the example be?" which is posted in UK Electrical Forum on Electricians Forums.
4V=0V and many amps / warm battery.
That part of circuit is a little short of components!
That part of circuit is a little short of components!
It's a scientific question that has blurred my mind for years. I need to know what what happens to the voltage and the lamp. (specially if we omit the burning of wire because of overheating possiblities) what will the voltages in such cases be?⚘
The lamp will light, assuming it is suitable for 12VDC. The second (40V) battery will overheat/discharge.
If you are taking R as 3 Ohms then you can use Ohm's law to calculate current flow in the left hand part of the circuit.
No current will flow from the second battery to the left hand side of the circuit because all current will flow down the short circuit (centre).
If you are taking R as 3 Ohms then you can use Ohm's law to calculate current flow in the left hand part of the circuit.
No current will flow from the second battery to the left hand side of the circuit because all current will flow down the short circuit (centre).
Wow really? It's hard for me to accept that, cause according the kirchhoff 'slaw, at the upper part of the short circuit cable (cable in the middle) the electrons should flow in both left and right directions, and so i think the battery on the right should effect the lamp. ?Yes, battery on RHS contributes nothing to LHS due to the perfect short circuit
???
'The resistance in wires are not considerable' - so even though they are small they are not zero. Is that what you assume? If this is indeed what you assume then draw the circuit out showing the resistances of these wires and then do the mesh current analysis.
There is another assumption you have made without knowing it - or maybe you know it but have not stated it and are ignoring it - something about each of the batteries ;-) Again, include this in your circuit diagram and then do the mesh current analysis.
I think you confusion arises from assumptions you are making or not aware of.
There is another assumption you have made without knowing it - or maybe you know it but have not stated it and are ignoring it - something about each of the batteries ;-) Again, include this in your circuit diagram and then do the mesh current analysis.
I think you confusion arises from assumptions you are making or not aware of.
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My problem is not the resistance of the wires and they are ignorable for me (in real world it is less than 2 ohms for such a circuit if we do not assume the wires to be too long).
The batteries are as stated in the question. In such a question i have many more question marks about the direction and amount of flowing electrons, which is much more important for me.
So please get to the main point of the question and see if you can solve it.
With no switches I stand by my original post #3
..the 4V battery is now flat.
( The initial circuit has no steady state as the battery depletes ,So defies analysis.)-- Nor fuses if we are dealing with lead-acid etc.
..the 4V battery is now flat.
( The initial circuit has no steady state as the battery depletes ,So defies analysis.)-- Nor fuses if we are dealing with lead-acid etc.
Ah that might explain things. Yes, not very clear if that is the intention.
Prince. I have drawn out and attached some of my explanation for you to study. You will note Kirchoff's current law is obeyed.
For the battery in Mesh M2 to influence the current Mesh 1 (and vice versa) the two meshes must share a branch - in this case resistor R where R is not zero.
When R is zero Ohms then the I bar junction in the middle collapses to a point as I have shown. A shared point between M1 and M2 obeys Kirchoffs current law but a shared point connection will not impress any of the electromotive force in Mesh 1 in M2 nor any of the electromotive force of M2 in M1.
You know that Kirchoff's Voltage states that the sum of the emfs in a mesh loop minus the voltage drops in that mesh loop equals zero. This equality for Mesh 1 includes (I1-I2) x R term as it also does for Mesh 2. If R is zero then this term becomes zero.
electromotive forces cause electrons to move to create currents. Thus there needs to be a shared branch of non-zero resistance for the 40V battery's emf to affect the current through the lamp.
It would be instructive for you to work out the mesh currents I1 and I2 for the circuit I have provided in the second attachment using general values R1, R2 and R3, V1 and V2 and assuming the wires are perfect conductors and the batteries have no internal resistance. Then do it for (R1, R2, R3) = (10,20,30) Ohms and (V1, V2) = (10,20)V; then do it again with R3=0 Ohms.
There is of course the case when R is infinite resistance ie: there is no middle branch - this collapses the circuit to one mess and the analysis becomes very easy.
For the battery in Mesh M2 to influence the current Mesh 1 (and vice versa) the two meshes must share a branch - in this case resistor R where R is not zero.
When R is zero Ohms then the I bar junction in the middle collapses to a point as I have shown. A shared point between M1 and M2 obeys Kirchoffs current law but a shared point connection will not impress any of the electromotive force in Mesh 1 in M2 nor any of the electromotive force of M2 in M1.
You know that Kirchoff's Voltage states that the sum of the emfs in a mesh loop minus the voltage drops in that mesh loop equals zero. This equality for Mesh 1 includes (I1-I2) x R term as it also does for Mesh 2. If R is zero then this term becomes zero.
electromotive forces cause electrons to move to create currents. Thus there needs to be a shared branch of non-zero resistance for the 40V battery's emf to affect the current through the lamp.
It would be instructive for you to work out the mesh currents I1 and I2 for the circuit I have provided in the second attachment using general values R1, R2 and R3, V1 and V2 and assuming the wires are perfect conductors and the batteries have no internal resistance. Then do it for (R1, R2, R3) = (10,20,30) Ohms and (V1, V2) = (10,20)V; then do it again with R3=0 Ohms.
There is of course the case when R is infinite resistance ie: there is no middle branch - this collapses the circuit to one mess and the analysis becomes very easy.
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