Electronic – voltage divider circuit for dc motor

The motor will require a startup current which is quite a bit higher than running current, so it's very possible 600mA is not enough (I can't determine the size from the photo - some measurements would be useful so a rough comparison with similar sized motors can be done)
Is there a voltage/current rating plate on the motor? Does it have a model number printed on it?

What makes you think it is this specific motor? There are many that look exactly the same.

Anyway, if you really can't find out any more about it, then you will have to do a few tests:

If you have a scope (or multimeter but will be a bit trickier) then measure the voltage across the power supply, then connect the motor. If the voltage sags significantly (e.g. more than a volt or two) then it's quite likely the motor is trying to sink more current than the supply can provide.

Also, test the winding resistance. Divide the operating voltage by this and you will get an idea of the startup current. It's possible a large capacitor or two across the supply (e.g. let cap charge, then use a series switch to turn motor on) might help it get started if it's not too far off the mark power wise.

Here is a table of data for a 6VDC motor picked at random:

You can see that the starting current is 3.51A, which is about 4 times the max continuous current of 0.84A. If we look at the winding resistance of 1.71Ω, we can see that 6V / 1.71 = 3.51A, so this if your peak startup current (before any back EMF develops)

The measurements for this motor are 22mm diameter and 32mm length. If your motor is a similar size or bigger then it is quite possible it is of a similar or higher power rating and too much for the power supply (although this is obviously far from certain given the widely varying type/quality of motors)

Basically, this won't really work - or work well. An alternative is presented at the end of this answer.

Before that, the calculations, and an explanation of why it won't work.

From Wikipedia:

The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating.

• Assuming the resistance of the filament was measured when the bulb was not lit, this gives a rough estimate of 77 x 15 = 1155 Ohms.
• This implies a ~45 Watt bulb, let's go with 50 Watts for convenience. This back-calculates to a hot filament current of around I = P / V = 217 mA = ~220 mA.
• To obtain ~5 Volts at 220 mA, R = V / I = 22.727 Ohms = ~ 22 Ohms, the nearest standard resistor value.
• Power dissipated by this resistor: P = V x I = 1.1 Watts, so let's go with a 5 Watt resistor to be safe.
• LED choice: Let's say red LED, 1.8 Volt forward voltage, 20 mA nominal current. Such an LED will light up quite well at 6 mA (230 Volt situation) and will also stay within current limits at 12 mA (460 Volts situation).
• Therefore a suitable LED current limiting resistor is 560 Ohms

Using two LEDs wired in anti-parallel would be recommended, because LEDs typically have low reverse voltage ratings (often as low as 5 Volts). With two LEDs facing opposite ways, the maximum reverse voltage appearing across the non-conducting LED will be the forward voltage of the conducting one, at any point in the AC cycle. It'll work fine in DC as well, so all is good.

Now, the problems:

What happens when the bulb is just switched on, and the filament is cold?

• Resistance = 77 Ohms, therefore current = ~ 3 Amperes! That's before adding the bleed resistor of 22 Ohms
• Current with 22 Ohms added in series: 2.323 Amperes. Power across resistor: 119 Watts!
• Result: Fireworks.

What happens when the bulb is supplied with 460 Volts?

• Assuming it is a 230 Volts rated bulb: The filament burns out.
• If it is a 460 Volts rated bulb: Not much goes wrong other than the sense resistor blowing up anyway.

Solutions:

1. Use the light bulb to illuminate the LDR directly.
2. Use a high wattage bulb in series with the 50 Watt bulb, for the voltage divider arrangement.
3. Use an alternative, such as a current sensing transformer, then use that CT's output in place of the LDR for sensing. A CT can be made by winding several turns of thin enamel wire around the wire supplying current to the light-bulb.
4. For AC and DC current sensing, use a prebuilt module such as this one ($4.35 from eBay including international shipping) that incorporates a nice "lil black IC", the Allegro ACS712, and will provide safe, nicely isolated current sensing: