if I have DC circuit (example standalon bulb) rated at 55 watts and 110 volts , 0.5 Ampere?

(C) None of the above. The power drawn will be the square of the voltage reduction, or 1/4 the original power.

The bulb draws 55 W when the voltage is 110. The bulb has constant resistance. So lowering the voltage will reduce both current and power.

We can use this relation to find the resistance in the bulb.

R = V^2 / P

R = (110V)^2 / 55W

R = 220 ohm

And at full voltage the current is:

V = I R

I = V/R

I = 110V / 220ohm

I = 0.5 A

If you reduce the voltage to 55V, the resistance is nearly constant and new current should be half the old current:

I = V/R

I = 55W/220ohm

I = 0.25A

And that will reduce the power drawn by the square:

P = I^2 R

P = (0.25A)^2 * 220ohm

P = 13.75W (one-quarter of 55W)

First of all the resistance of the bulb changes a lot with the applied voltage. However, if you ignore that, then you can assume the resistance is R=110/.5=220 ohms.

Neither answer is correct. For a constant resistance P=V^2/R or P=I^2*R. Reducing the voltage to 1/2 reduces the power to 1/4.

Watts equals Volts times Amps. If you lower the power available and you cannot increase the current to compensate.

Ohm's Law