current produced by two batteries connected in parallel

[PG1995]

Hi

Please have a look on the attachment. You can find my questions there. Please help me with them. Thank you.

Regards
PG

PS: The batteries in Q2 have the same emf, say, 5V.

 

[Raza]

Yes PG1995, you are right in both the assumptions.
And a nice thing " You answered your both questions yourself" :wink:

 

[danishdeshmuk]

Yes PG1995, you are right in both the assumptions.
And a nice thing " You answered your both questions yourself" :wink:

which configuration is better in terms of longer time period ,

2 or more batteries connected in series
2 or more batteries connected in parallel

in series voltage increase & current remains same
in parallel voltage remains same but the current increases

but P = v * I

= 12 * 50

= 600 W

for 2 batteries in series (same type , same voltage & same current rating batteries)

P = v * I

= 24 * 50

= 1200 W

For 2 batteries in parallel (same type , same voltage & same current rating batteries)

P = v * I

= 12 * 100

= 1200 W

so whats the difference then ?

thanks

 

[Raza]

n series voltage increase & current remains same
in parallel voltage remains same but the current increases

Yes it is.

but P = v * I

= 12 * 50

= 600 W

But here you calculating wrong way.

P = v * I

= 24 * 50

= 1200 W

Here you know the P and you will be calculating the I where as the V is also known to you, so
P=600W
V=24
I=P/V=600/24= 25 A
So the current is gone half as you doubled the voltage for same power circuit.
Hope it clarifies.

 

[ark5230]

Calculating power is a different exercise.
What is responsible for the flow of current is:
1. Potential difference.
2. Conducting path.
In the circuit shown there is a net PD of 5V.
It is like charging a 5V battery using a 10V charging source.
Current will flow through 5 V battery in the CCW direction.
The arrangement is like that of charging a battery.
If one cell is 5V and the other one is 10V as above then
Series lamp may glow if the current is sufficient to make it glow.
The lamp will not glow if the current is not sufficient for it to glow.
Lamp will blow off it the current is more that what it can withstand.
Calculations can be carried using equivalent circuit including the internal resistance of the cells.

 

[danishdeshmuk]

Yes it is.

But here you calculating wrong way.

Here you know the P and you will be calculating the I where as the V is also known to you, so
P=600W
V=24
I=P/V=600/24= 25 A
So the current is gone half as you doubled the voltage for same power circuit.
Hope it clarifies.

so, which would you prefer RAZA then ?

thanks

 

[Raza]

Connecting two batteries in series will raise the voltage to double as of the one battery. Hence you can not use two batteries in parallel for the same load as two batteries in parallel. But if you are designing a circuit then (my suggestion) go for two batteries in Series. Means it is better to use high voltage compared to low, strictly depends on the design and circuit demand.

 

[danishdeshmuk]

means if you have a choice then you would use two batteries in series rather than in parallel ..... ?

but can i ask for what reason(s) ?

thanks

 

[Raza]

When in series the voltage is added to each other and the current goes half for the same power rating. As the current reduces the batteries remain at easy state. Increasing the battery life low charging current, etc.

 

[danishdeshmuk]

When in series the voltage is added to each other and the current goes half for the same power rating. As the current reduces the batteries remain at easy state. Increasing the battery life low charging current, etc.

but people almost all of them prefer to use batteries in parallel to get the greater back-up ....... so, reducing the life of batteries & increasing the load on the transformer & charging circuit ............ ??? Am i correct

thanks