I am building a mega project with Arduino Mega. I have 3-5 components including sensors and motor drivers. All that components requires 5-volt power supply. I have connected a power supply row of the breadboard with Arduino 5volt power supply and i have connected the vcc pin of these components with that breadboard row where I have connected the Arduino 5V pin. I am using 1.2V*10 = 12Volt battery(with 10 1.2 volt AA battery). But while I am powering the Arduino in the vin pin by the 12-volt battery then the 5-volt row of the breadboard doesn't supply power more than 3.5-4.0 Volt. But if I power the Arduino with USB, then the 5volt pin of the Arduino supplies 4.45 volt to the breadboard power supply row. How can I supply the same 4.45 volt or more than 4.30 volt to the breadboard power supply row using my battery without using usb connection so that my sensors and servo motor can work properly?
You don't need 12V, much of the power is just wasted as heat by the 5V regulator.
AA battery is usually 1.5V, not 1.2V. Are you using some kind of rechargeable that only puts out 1.2V?
5 x 1.5V = 7.5V, use that into the regulator, it will run much cooler.
Or 6 x 1.2V.
The regulator will only supply 800mA maximum before it overheats and shuts down. With 12V in, it will overheat sooner. With 7.2V to 7.5V, you have a chance for getting closer to 800mA.
How much current do your sensors and motor drivers need?
When they are not connected, do you see 5V output?
"The regulator will only supply 800mA maximum before it overheats and shuts down. "
It would be better to use maximum 'power/watts' instead.
If you're using 1.2V nominal AA batteries they are probably NiMH rechargeables. Note that these are likely to be 1.4V or more when fully charged. Pushing 14V+ into the Vin pin is a bad idea.
Running motor drivers from the 5V pin is also bad unless they are VERY small VERY low current motors.
And powering any relatively high current devices via breadboard can also cause all sorts of problem.
I think a knowledgeable person can do the math:
(Vin - 5V) x current drawn = excess heat to be dissipated.
(12V - 5V) x 200mA for example = 1.4W of excess heat to dissipate with very little heatsinking to do it.