pulldownrelaytransistors
This is a follow up Question to What is use of R41 (base to emitter) resistor in Transistor Relay Circuit?
My doubt is can we connect the resistor R41 between emitter of optocouler and GND?
The pulldown resistor is usually connected to one end of a switch. 
and it would be more consistent to connect R41 (in original schematic) to the emitter of optocoupler (SWITCH).
Without a circuit diagram its impossible to answer the question specifically.
If you use the emitter follower circuit with the load between emitter and ground the maximum voltage across the load will be base voltage - 0.6V due to the base-emitter junction. For a 5V input you get 4.4V output.
If you use the common emitter circuit with the load between the collector and +12V you will get a voltage at the collector between 12V (transistor OFF) and 0.1V (approx) which is the saturation voltage (Vsat) when the transistor in turned ON.
Edit: To answer comment.
The value of R1 sets the size of base current. R1 is simply a label and may have different values in different circuits. In this case there are two different (separate) circuits so R1 will have a different value in each.
The transistors are actually acting in the same way.
The base-emitter junction acts like any silicon diode ( its the 'PN' part of the NPN transistor). For current to flow through the junction the base must be about 0.6V higher than the emitter. This is true in both circuits.
The differences.
In the case of the emitter follower circuit the voltage across the load can RISE until this condition (Vbe = 0.6V) is satisfied. The transistor is not fully turned ON and the voltage across the transistor (Vce) will be 12 - 4.4 = 7.6V. This means the transistor will have to dissipate POWER (7.6 x Ic) and could get quite warm or possibly overheat.
In the case of the common emitter circuit the emitter voltage is FIXED at 0V and will limit the base voltage to 0.6V. When fully turned ON the voltage across the transistor is the saturation voltage (Vsat) - typically about 0.1 - 0.3V depending on transistor type and collector current. The rest of the voltage is dropped across the load. The power dissipated by the transistor is minimal (Vsat * Ic).
To calculate the value of R1 we use Ohm's law (V=IR) and the relationship between base and collector current for the transistor, Ic = current gain x Ib.
Example calculations:
Suppose the load current is 500mA and the typical current gain of the transistor is 250. That would give a value of 500/250 = 2mA for the base current. (you get this value from the datasheet)
In the case of the emitter follower the voltage drop needs to be a small as possible. It has 5V at the input but we want as much voltage as possible at the base. Lets say 4.9V at the base so R1 has to drop 0.1V @ 2mA. This gives a value for R1 of 50 ohms.
(In theory you may not even need a resistor but in the event of a short circuit load the maximum current into the base will be 88mA.)
In the case of the common emitter with a 5V input R1 needs to drop 5 - 0.6 volts. (= 4.4V) because the base voltage is fixed at 0.6V. At a base current of 2mA this gives a value for R1 = 2200 ohms.
Yes, that's essentially correct.
However, you should actually over-drive the base of the transistor in order to make sure that it stays in saturation even if conditions change a bit (particularly transistor current transfer ratio, which is difficult to control precisely). This is called a "design margin".
So, assume that in the worst case, the hFE of your transistor is 1/2 or even 1/3 the "nominal" value, and redo your resistor calculation accordingly. There's no harm in introducing more current than needed (within limits, of course) to the base of a transistor.
Best Answer
While it looks superficially like a voltage divider, the voltage on the base of the transitor is constrained to the range 0 to 1V by the transistor itself. The value for this part is usually such that only a very small current flows through it.
The purpose of R41 is to prevent any current that leaks through the opto-coupler or the transistor C-B junction from biasing the transisor partly on and causing it to overheat. 20uA leakage mutiplied by worst-case beta or 300 is 6mA, at about 24V that's 144mW of heat generated in the transistor.