# parallel power supply

#### recca02

Joined Apr 2, 2007
1,214
this trivial question just popped up in my mind ,its not a hw question
if we connect 2 ideal power sources ( 2 batteries for example of same voltage V)
to a common load (a bulb of some power rating) how much current will
flow through the bulb . if we consider that I amp flows if only one battery was
there. if R is resistance of bulb assumed. (I=V/R).also assume conductors to have 0 resistance.

i did not try it on paper but have given some thought to it
-- parallel batteries should share the load and hence equal current should flow
case1- since parallel batteries are connected voltage is equal to one battery
acting alone hence current= V/R=I
so shud I/2 current flow in each loop?
case 2-
since two batteries are connected each provides V/R=I
and current thru bulb equals --2I
case 3-
apply superposition theorem
since voltage sources are ideal while considering each source separately
the removed voltage crkt is shorted hence current through bulb is zero in
both consideration. so is the current thru bulb 0+0=0 amps

#### beenthere

Joined Apr 20, 2004
15,819
If the batteries are in parallel, then the current through the bulb won't change. In series, the voltage will double. The current will not, as the filament has a non-linear response to current. It will most likely burn out, but, if it doesn't, the current will not be twice that for the original voltage.
If the batteries are in series opposing, then not surrent flows through the bulb.

#### thingmaker3

Joined May 16, 2005
5,084
With two identical batteries in parallel, half as much current is sourced by each battery. Ohm's Law still governs the current through the bulb.

#### recca02

Joined Apr 2, 2007
1,214
thanx everybody. so for a series battery connection with a load having a linear response
the current must double ,right?

#### thingmaker3

Joined May 16, 2005
5,084
Yes. Current will always and forever be the product of voltage divided by impedence. That's why we call Ohm's Law a "Law" instead of a "suggestion." #### John Luciani

Joined Apr 3, 2007
477
Your case three (superpostion) is correct.
The "ideal" battery has a resistance of 0 and will shunt
the current away from the bulb.

(* jcl *)

---

http://www.luciani.org

#### recca02

Joined Apr 2, 2007
1,214
sir, if the circuit gets shorted and currrent is shunted will the bulb glow? (in view of my case 3 assumption)
ne ways thanks again.

#### John Luciani

Joined Apr 3, 2007
477
The current takes the path of least resistance.

An "ideal" short and an "ideal" voltage source each have a resistance of 0
and will shunt current away from the bulb (which I am assuming has
a resistance > 0).

NB: If you try this with a "real" battery you will probably get a hot or exploding
battery since your battery has a resistance > 0.

(* jcl *)

-------
www.luciani.org

#### recca02

Joined Apr 2, 2007
1,214
then what abt the case of two infinite buses (i mean main supplies )
synchronized properly to work in parallel connected to same load
i believe they can act as ideal sources (probably huge wire resistance
cant be neglected so let us consider the load is having a very high resistance)
will the load not get any supply, i might be wrong but for huge loads connecting
two generators or two transformers (on distribution side) in parallel is a common practice.

#### John Luciani

Joined Apr 3, 2007
477
Paralleling transformer outputs is done to increase the output
current.

For superposition you replace an "ideal" voltage source with a short
circuit. You do not replace a transformer output with a short circuit.

(* jcl *)

----
http://www.luciani.org

#### thingmaker3

Joined May 16, 2005
5,084
It must be noted, though, that the Superposition Theorem works only for circuits that are reducible to series/parallel combinations for each of the power sources at a time
So, no, reducing a circuit to an ideal shorted battery won't tell us anything. If we use an estimate of the resistance of the wire, superposition will work.