Electronics - Parallel Circuits - Discussion

You explained well, thanks @Pankaj.

You explained well, Thanks @Pankaj.

When R4 will short circuited then total circuit will be series circuit and current will be the same in the whole circuit. R1 R2, R3 will not conduct current.

very well said @Pankaj.

As per the logic the galvanometer is saying 2.4 A is going next, so all three galvanometers are denoting 2.4A,
So, all three branches don't take current so, 2.4 A chooses path 4 I mean R4. And the low resistance path is available in R4 only right?

Correct me If I am wrong.

If one leg of the parallel circuit shorts all of the current available will flow out of the power supply for a brief time and then sense an overload and shut down (blow the fuse).

Let me simplify this answer.... Look at the schematic and realize that "G" stands for conductance... So if you look at all of the branches, if either of the branches were to short other than R4, since current normally takes the easiest path, then there is no way that the current would be constant throughout every branch leading up to the final one in this circuit. So since R4 still has a the equivalent amount of the total conductance ( or current) then R4 would have to be the problematic component in this circuit.. This is a clear indication that current has taken the easiest pass at the very last branch of this circuit.

But applying kvl gives different answer.

1st convert G into R and 0.02A pass through R1 and current increases as resistance, so high current pass through R4.

What is G here ?

I hope G is a Galvanometer here.

A galvanometer is a type of ammeter: an instrument for detecting and measuring electric current.