I want to build voltage source of 5V capable enough to supply more than rated current of 7805 ic.
Is that all you want to do? You want a 5V supply that can provide more than 1A current? That circuit won't do that; all it does is feed current into the base of the transistor, which makes the transistor conduct, but nothing more.
You need to look at an earlier post on this thread by CDRIVE. He wrote:
In the event that you're trying to obtain 5V regulated at a current higher than the 7805's 1 Amp limit. This might be what you're looking for. If not, nothing lost.
That post includes a schematic that will do what you want. It uses a power transistor as an emitter follower (aka common collector), to boost the amount of current available. This circuit does not provide current limiting, so if you short out the output, the transistor will dissipate a LOT of heat. It will also dissipate heat during normal operation. You can calculate its power dissipation from the formula P = V * I where P is power in watts, V is the voltage ACROSS the transistor (i.e. the input supply voltage minus the output voltage; the voltage that is being DROPPED by the transistor) in volts, and I is the output current in amps.
For example, if your input voltage is 12V and your output current is 3A, power dissipation in the transistor will be 21 watts. This is calculated from 7 volts multiplied by 3 amps (7 volts is the voltage dropped across the transistor). The transistor will definitely need a heatsink, otherwise it will overheat and be damaged. Transistors don't have any kind of protection feature; they're "dumb".
This circuit also has poor output accuracy and regulation, but this shouldn't be a problem unless you're powering something that needs a very exact voltage.
The circuit you show with a second transistor, a load resistor, and a second supply doesn't do anything useful. You just need the 7805, the diode, and the power transistor.
In the circuit I have connected 7805 ic to ensure 5V supply. This means if I connect 4.3 ohms across the o/p of 7805 ic and base of Q2 then I would be able to get 1A current.One more thing I want to clarify is that in which mode Q1 is operating?
That's right, if you connect a 4.3 ohm resistor from the output of a 7805 to the base of a transistor, with its emitter grounded, you will get about 1A of current flowing into the base-emitter junction. This will make the transistor conduct. So current will flow in the collector circuit. It doesn't really do anything useful; it doesn't give you a regulated voltage source. The transistor would be operating in common emitter mode.
As Steve suggests, you should download the data sheet for the 7805 regulator. It will include circuits for boosting the regulator output voltage (try to get data sheets from several manufacturers; the 7805 is made by several companies). This is probably where the emitter follower design came from originally.