# parallel power supply

Discussion in 'Homework Help' started by recca02, Apr 2, 2007.

1. ### recca02 Thread Starter Senior Member

Apr 2, 2007
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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

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
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.

3. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
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.

4. ### recca02 Thread Starter Senior Member

Apr 2, 2007
1,211
0
thanx everybody.
so for a series battery connection with a load having a linear response
the current must double ,right?

5. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
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."

6. ### John Luciani Active Member

Apr 3, 2007
477
0
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

Apr 26, 2005
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8. ### recca02 Thread Starter Senior Member

Apr 2, 2007
1,211
0
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.

9. ### John Luciani Active Member

Apr 3, 2007
477
0
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

10. ### recca02 Thread Starter Senior Member

Apr 2, 2007
1,211
0
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.

11. ### John Luciani Active Member

Apr 3, 2007
477
0
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

12. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
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.