A battery of 40V and internal resistance of 2ohm is connected in parallel with a second
battery of 44V and internal resistance of 4ohm A load resistance of 6ohm is connected across
the ends of the parallel circuit. Caleulate the current in each battery and in the load.
Answers
Explanation:
The homework writers are at it again in the electronics world.
This time it is two baatteries , that are connected in parrallel.
But the problem is the unbaalanced voltages being aapplied to the load.
As the second battery gets connected the current in the higher voltage battery will flow to the lower voltage battery.
The internal resistance of the two baatteries will limit the current, with the maaximum being the combination of the series resistaance and the load.
Most lead-acid batteries have a cell voltage of 2V per cell. A 24 V Battery would consist of 12 cells each with a internal resistance of 0.033 ohms.
This gives the 24 volt battery an effective series resistance of 0.40 Ohms.
The problem is the 20 Volt battery placed in parralel.
This will cause a current flow into the 20V battery across the 0.25 series resistance.
A Balancing resistaance of 7 diodes in series gives the required forward voltge drop from 24 to 20 volts. The current in the diodes is to drop the voltage down from 24 volts down to 20, and keep the current blocked from flowing into the lover voltage battery,
The 20 Volt battery then is what is presented to the load resistnce of 4 ohms.
This then with Ohms Law mens the current in the load is 5 Amperes.
The 20 Amper bttery will supply half of the 5 ampers or 2.5 Amps.
The 24 volt battery will have 2.5 amps through the diodes that drop 0.67 Volts per Silicon diode.
The current of 2.5 amps will be dropped over the 0.4 ohm resistance , this yeilds a voltge of 1.0 volt.
The curent in the 20 volt battery will be dropped on the 0.25 ohm resistnce, and has 2.5 Amps through it, this give a voltge of 0.625 volts.