The 120-W bulb has a greater resistance and will have a greater voltage drop.
Electrical power is the rate at which work is done in a circuit. It is obtained by; P = V^2/R
For the 60-W bulb;
R = P/V^2 = 60-W/(120)^2
R = 0.0042 ohm
For the 120-W bulb;
R = P/V^2 = 120-W /(120)^2
R = 0.0083 ohm
Hence, we can see that when we compare the two bulbs, the 120-W bulb has a greater resistance and will have a greater voltage drop.
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