A 150-watt lightbulb is brighter than a 60.-watt lightbulb when both are operating at a potential difference of 110 volts. Compared to the resistance of and the current drawn by the 150-watt lightbulb, the 60.-watt lightbulb has

Current, I = 0.54 A

Resistance, R = 201.6 ohms

Explanation:

Electric power is given in terms of current, I, and voltage, V, as:

P = I*V

I = P/V

The current flowing in the 60 W bulb will be:

I = 60/110 = 0.54 Amps

The resistance can be gotten by using an alternate formula for Power in terms of resistance:

P = V²/R

R = V²/P

R = 110²/60

R = 201.6 ohms

The 60W bulb has a current 0.54A flowing through it and a resistance of 201.6 ohms.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j

it is b. absorbs thermal energy