When two batteries of different magnitudes are connected in parallel, then what will be the potential measured by a multimeter across the two batteries??
well idont know whether it is right but There is allways an internal resistance to a battery that will be connected in series with each one. this resistance will distribute the voltage in such a manner that the resulting voltage is constant in each parallel parts
well thats true that there is always an internal resistance but still the voltage in parallel is always same.....so final output will be always same....so, how can the voltage be divided among parallel combination??think over a bit.......
i dont think that the battery with higher potential will dominate....bcoz this problem when tried on ORCAD PSpice simulation gives a voltage value at the output which corresponds to the smaller potential....we'll have to think over it a bit more....i'll try providing with the answer at the earliest...but till then keep fiddling with your minds and try getting up to the answer
If two potential sources are connected in parallel forming a closed loop, charges will transfer from higher potential to lower potential till the two become equipotential. It implies if V1 & V2 are the two sources then resultant will be average of two i.e. (V1+V2)/2. As louis said that the higher one will dominate, but after some time both will be equal. Now rate of charge transfer i.e. current magnitude and amount of charge will depend on the resistance in the path. Considering the batteries to be ideal rechargeable dry cell with zero internal resistance, current will be infinte for the duration in which both become equipotential.
For eg. if 3V & 9V cells are connected then 9v cell will loose charge to 3v cell till both become 6V. This will happen only if the cells (or that with lower potential) were rechargeable.
If the 9V cell is replaced by 9V power supply, i.e. main source of charge is now the ac mains, then 9V will not loose its magnitude but 3V cell will rise up to 9V and after the transient phase of charge transfer the resultant will be 9V instead of 6V in previous case.
Now, if the cells are not rechargeable, then resultant as per OrCAD simulation would be lower one but we found practically its the higher one. We connected two cells as well as two power supplies also and in both cases the resultant found to be the higher one.
Note: connecting two such sources may result in large currents, resulting in burns or even explosion.
If the power supply of 3 V is connected to a rechargeable cell of 9 V, since the cell is at a higher potential, charge carriers would try and diffuse towards the supply, this would bring down the potential of 9 V cell to some lower potential. However, 3 V power supply already has an abundant source of carriers, theoretically infinite which keeps it at 3 Volts, hence there will be no change in the supply voltage even if a few more carriers are supplied by the cell. Hence the resultant potential in that case should be 3 V, and not 9 or 6 Volts as expected. However, when the 9 Volts cell is removed from the circuit, it would have discharged to a lower value.
This is however my interpretation of the problem. Any other explanations if seeked must be posted here.
9 comments:
i dont know the answer
well idont know whether it is right but
There is allways an internal resistance to a battery that will be connected in series with each one.
this resistance will distribute the voltage in such a manner that the resulting voltage is constant in each parallel parts
well thats true that there is always an internal resistance but still the voltage in parallel is always same.....so final output will be always same....so, how can the voltage be divided among parallel combination??think over a bit.......
i agree with u reena mam . i think battery having higher potential will doninate if the case of being not equal. pls tell if im wrong.
i dont think that the battery with higher potential will dominate....bcoz this problem when tried on ORCAD PSpice simulation gives a voltage value at the output which corresponds to the smaller potential....we'll have to think over it a bit more....i'll try providing with the answer at the earliest...but till then keep fiddling with your minds and try getting up to the answer
Mam when will you tell the answer...??
If two potential sources are connected in parallel forming a closed loop, charges will transfer from higher potential to lower potential till the two become equipotential. It implies if V1 & V2 are the two sources then resultant will be average of two i.e. (V1+V2)/2. As louis said that the higher one will dominate, but after some time both will be equal. Now rate of charge transfer i.e. current magnitude and amount of charge will depend on the resistance in the path. Considering the batteries to be ideal rechargeable dry cell with zero internal resistance, current will be infinte for the duration in which both become equipotential.
For eg. if 3V & 9V cells are connected then 9v cell will loose charge to 3v cell till both become 6V. This will happen only if the cells (or that with lower potential) were rechargeable.
If the 9V cell is replaced by 9V power supply, i.e. main source of charge is now the ac mains, then 9V will not loose its magnitude but 3V cell will rise up to 9V and after the transient phase of charge transfer the resultant will be 9V instead of 6V in previous case.
Now, if the cells are not rechargeable, then resultant as per OrCAD simulation would be lower one but we found practically its the higher one. We connected two cells as well as two power supplies also and in both cases the resultant found to be the higher one.
Note: connecting two such sources may result in large currents, resulting in burns or even explosion.
What if the power supply connected in parallel is of lesser value than rechargeable cell.......
For eg:
3V power supply connected in parallel with 9V rechargeable cell...??????
If the power supply of 3 V is connected to a rechargeable cell of 9 V, since the cell is at a higher potential, charge carriers would try and diffuse towards the supply, this would bring down the potential of 9 V cell to some lower potential. However, 3 V power supply already has an abundant source of carriers, theoretically infinite which keeps it at 3 Volts, hence there will be no change in the supply voltage even if a few more carriers are supplied by the cell. Hence the resultant potential in that case should be 3 V, and not 9 or 6 Volts as expected. However, when the 9 Volts cell is removed from the circuit, it would have discharged to a lower value.
This is however my interpretation of the problem. Any other explanations if seeked must be posted here.
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