Automatic Street Light using LDR Project ( Need someone's assistance)

From left to right:
To the left is your AC input, a voltage somehere between 9 V and 12 V, typically.
The 4 diodes form a bridge rectifier to convert AC input into DC.
The capacitor is a smoothing capacitor to create a smooth Dc from the pulsating Dc after the bridge rectifier.
The LED is useless. It is the wrong way round and lacks a current limiting resistor.
The component labeled 7806 is a linear voltage regulator. It creates a regulated 6 V DC output from the unregulated input. Note that the typical application note for this type of regulator asks for a e.g. 100 nF ceramic capacitor at input and output. These capacitors are missing here. See the datasheet of an 7806 fpr more information.
The LDR (top resistor) and the fixed resistor (bottom) form a resistive voltage divider to drive the base of the NPN transistor.
When it is bright, the resistance of the LDR is low which will create a high base voltage and turn on the transistor. Th current through the transistor will turn on the relay and the relay output in turn will activate or decativate the light depending on the connection of the light to the relay contacts: as the relay is a switchover relay the light can be turned on or off when the LDR is lit.
The diode across the relay contacts is a freewheeling diode or flyback diode to protect the transistor from the inverse back-emf of the relay coil when turned off.
Again the LED as shown in parellel to the diode is useless as it is connected in the wrong polarity and lacks a current limiting resistor, see above.

Google the terms in italic for more infomration on the marked terms.

 

Any add-on to a conventional light will consume power which is in addition to the power used by the lamp itself.
Make your own calculationn:

• What if the lights are on for exactly the same length of time?
• What if the automatic optimizes (minimizes) the time the lamp is on compared to the time a conventional light is on?
• What if the automatic is badly adjusted and increases the time the lamp is on compared to the time a conventional light is on?

To express this in percent you'd have to know the power dissipated by the sensor circuit and that dissipated by the lamp.
Note that you will have to take into account also the time where the light is off. A converntional lamp would not require any power in this state, the sensor will require power to keep alive.