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If V1 is ideal voltage source, the current value in a loop will be 1.2V/resistance of a loop. If you assume that this is ideal resistance of 0Ω (not existing in practice) the theoretical current will be infinite ..
and if you assume it is practical, you now that the battery has its internal resistance and a maximum current that it can give, and also you know that the battery consists of a chemecal compounds that will finish the interaction and the battery will be dead. so you will see a drop in voltage and current and the current will have an exponentiual decaying from maximum of battery to zero.
if you want to test this circuit you can but be aware of using liquid battery because it will explode since it has large capacity of current, and it will explode due to heat and may harm you.
Each voltage source has an internal resistance so it must be included in ur circuit diagram ,according to ur draw it's in series with the battery assume its symbol is (Ri) so the value of the current is
I=V1/Ri
hi
Iguess if the wires are very long (transmission line)so a capacitor effect may appear then current will rise exponentially , and reflections may exist olso.
First of all ,The solution to this question is simple,but has two theoritcal and practical assumptions:
1- theoritcal :
the current will be infinity ,but this impossible to achieve.Unless we use
Superconductor Material instead of the shorted wire that u put.and be asure
that if u use Sper.Cond Material u can not reach the infinity.
2-Practical:
the infinity current is Desirable ,but impossible to achieve until now .
actual result of ur circuit dependant on many things,for instance,the battery
types and the wire length.
*each battery has internal Resistance.
*each wire has also internal resistance.
the summary,the current will be infinity in the theory field .Practically, Very High Current .
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