[julie_m]

The bulb being used as a fuse for the motor probably is long-since NLA, but you could use a fusible resistor in its place. You will need to use Ohm's Law V = I * R and the Power Law P = I * R to select a value which will be dissipating as nearly as possible its rated power in operation.

If you use a 10 ohm resistor as a temporary substitute for the bulb, you can measure the voltage V across this and then get the current I using I = V / R = V / 10. Then assume the motor resistance is large compared to the safety device, so the current won't change much with a different resistor. Just choose a fusible resistor with a suitable value to give a power dissipation at that current which is just within its rating. If the motor current increases, the power dissipated in the fusible resistor will increase twice as fast (since P = V * I and V = I * R, so P = I ** 2 * R. If the current increases to 1.1 times as much as it was before, the power will increase to 1.1 times 1.1 times as much as it was before = 1.21 times). Careful manufacture ensures the fusible resistor will fail (or, from the point of view of whatever it's protecting, succeed .....) open-circuit when its rated power dissipation is exceeded, and without sending shrapnel flying nor setting fire to anything in its vicinity.