parallel
What happens if i connect 2 or more similar 12 volt and 5 amperage electrical adapters in parallel to a single load to avoid power outage and continuous power supply.
Volt= 24 or 12
Amp = 5 or 10
So, if there is a substitute such as transfer switch then please suggest the component name and it be available economically.
It's good to hear about family projects :)
I'm not sure I understand your description of the circuit topology. I'm assuming that for each 9V battery, he will have four strings of LEDs in parallel to each other, and that each string consists of three LEDs wired in series. Is this correct?
First, you have to add up the Forward Voltages of each string, and make sure the total is lower than your power source (9V). If there are any string that total more than 8.5V (or so), you'll need to arrange them in a different way.
Then, you need to figure out your desired current for each string of three LEDs. The datasheets will show what the acceptable Forward Current levels are. Since all three LEDs are in series, they each flow the same current. So you want to limit this current to lowest-current LED. Since each LED in a string is carrying the same current, it won't matter which order you put them in.
Now, for each string, add up the forward voltages, then divide that sum by your desired current. This will give you an initial resistance value. It's wise to use a larger resistor that you calculated, and then decide if you want to lower it later by measuring the actual current with a multimeter.
You should use one resistor per string of LEDs. Their forward voltages aren't going to match exactly. If you use only one resistor, and you get a large enough mismatch between strings, then you could burn one of them up.
Regarding the switch, it won't effect the performance of the LEDs. It will cause the battery to drain a bit faster. Basically, the switch is adding one more LED and a resistor in parallel to your circuit.
And finally, regarding battery life, that really depends on your LEDs. Lets assume you drive them at 20mA. If you have four strings in parallel, this will take 80mA all together. A typical 9V battery has around 500mA-hours of capacity. Divide 500mA-h by 80mA, and you get over six hours :)
I hope this helps!
Although you've awarded the answer points, I think you've missed a point. This is an excellent example of when to use a Thevenin equivalent, to wit:
simulate this circuit – Schematic created using CircuitLab
ETA: Bugger. Got the V2 equation wrong. Fixed it.
Best Answer
You should never place two voltage sources in parallel. Their voltages are never exactly the same and that combined with a low output impedance gives a very high current from the higher to the lower voltage. The supplies will get very hot and most likely get damaged.
The way to solve this is to put a diode in series with each of them. Then the supply with the higher voltage will supply the power, the diode on the other supply prevents current from the first one from flowing in. Suppose PS1 is 12.5V and PS2 is 12V. Then the connected cathodes of the diodes are at 12.5V - 0.7V = 11.8V. So there's only 200mV across the diode of PS2, too little to let it conduct. If PS1 fails, the voltage drops to 11.3V at which PS2's diode begins conducting and PS2 begins supplying power.