To increase the brightness of a desk lamp, a student replaces a 60-watt light bulb with a 100-watt bulb. Compared to the 60-watt bulb, the 100-watt bulb has

1.less resistance and draws more current

2.less resistance and draws less current

3.more resistance and draws more current

4.more resistance and draws less current

Option 1. Less resistance and draws more current

Explanation:

To better understand this concept, let us determine the resistance and current of each bulb assuming a 240V supply.

Recall:

Power = current (I) x voltage (V)

P = IV

Power = square voltage/ resistance

P = V^2 /R

For the 60watt bulb:

Power (P) = 60watts

Voltage = 240V

Current (I) =?

Resistance =?

A. P= IV

60 = I x 240

Divide both side by 240

I = 60/240

I = 0.25A

B. P = V^2 /R

60 = (240)^2 / R

Cross multiply to express in linear form

60 x R = (240)^2

Divide both side by 60

R = (240)^2 / 60

R = 960 ohms

For 100watts bulb:

Power (P) = 100watts

Voltage = 240V

Current (I) =?

Resistance =?

A. P= IV

100 = I x 240

Divide both side by 240

I = 100/240

I = 0.42A

B. P = V^2 /R

100 = (240)^2 / R

Cross multiply to express in linear form

100 x R = (240)^2

Divide both side by 100

R = (240)^2 / 100

R = 576 ohms

Summary:

60watts bulb has a current of 0.25A and a resistance of 960 ohms

100watts bulb has a current of 0.42A and a resistance of 576 ohms.

Comparing the resistance and current of the 60watts and 100watts bulb under the same supply of 240V,

the 100watts bulb has lesser resistance than the 60watts bulb and draws more current than the 60watts bulb.

electric field is a function of 1/r^2

yes, the force per unit charge for a point charge is inverse square law.