Am reading pull-down resister herehttp://www.doctronics.co.uk/voltage.htm. PART-1 The circuit schematic of a light sensor using a voltage divider circuit is as shown: http://www.doctronics.co.uk/images/vdiv4.gif [Broken] LDR has a resistance of 500Ω, 0.5kΩ , in bright light, and 200kΩ in the shade. in the shade, V_out will be: (10/210)×9=0.42V in the bright light, V_out will be: (10/10.5)×9=8.57V by voltage divider rule so by applying voltage divider rule we came to know that the circuit will give a high output voltage in the bright light and low output in the dark. So if we use this circuit with a bulb connected at the output then in the afternoon the bulb should glow. But there is a problem the bulb which is to be connected has its own resistance which might be lower than 10kΩ. Let's say the bulb to be connected is of resistance 100Ω. Since the bulb is in parallel with the 10kΩ resister the equivalent resistance of this parallel combination is approximately 100Ω. Applying the voltage divider rule we will find that the bulb will not glow in both dark and bright light. So the circuit is impractical and is of no use. My question is: Does the voltage divider circuit has no practical importance? PART-2 Here(http://www.doctronics.co.uk/voltage.htm#switches [Broken]) it is explained that we usually use a pull down resister of very high resistance of nearly 10kΩ. Rather using a high resistance resister we can left open the terminals at which pull-down resister is connected. By doing so we will obtain ∞ resistance for pull-down resister and whole of the Vin will appear at Vout. Why we use a resister of 10Ω not an open circuit for pulling whole of the Vin at Vout.