How to prevent Buzzer circuit to drain battery
You have connected a 3 V buzzer to a 9 V battery. Voltage is three times too high. This is likely to cause three times the correct current to flow. Since power is V x I (volts x amps) this will cause 9 x rated power to be dissipated in the buzzer.

You need to add series resistance in circuit with the buzzer to bring the voltage down to 3 V. This means the resistance needs to have 6 V across it so from Ohm's law:

$$ R = \frac {V}{I} = \frac {6}{0.025} = 240~ \Omega $$.

220 Ω is the nearest standard value and will be fine.

Your LED resistors are marked as 10 kΩ. Using Ohm's law we can calculate that the maximum current they can provide is 

$$ I = \frac {V}{R} = \frac {9}{10k} = 0.9~mA $$

Most LEDs need about 5 - 20 mA to give a good daylight brightness level. You have enough information to calculate the correct resistance values.



You can further improve the efficiency of the LEDs by connecting them in series as the "one current" will drive the two LEDs. They will drop about 2 V each leaving 5 V across the resistor.

You can improve the efficiency AND have bright LEDs! 







simulate this circuit &ndash; Schematic created using CircuitLab

The LEDs can handle 25 mA (check the datasheet) so connect the lot in series and you want 7 V or so at the top of L1 which means R1 needs to drop 2 V at 25 mA. You can work that out?